name solution department of natural · pdf filenamesolution_ department of natural...

TIME OF COMPLETION_______________

NAME____SOLUTION_________________________

DEPARTMENT OF NATURAL SCIENCES

PHYS 1111 Exam 3 Section 1

Version 1 December 6 2005

Total Weight 100 points

1 Check your examination for completeness prior to starting There are a total of nine (9)

problems on seven (7) pages

2 Authorized references include your calculator with calculator handbook and the Reference

Data Pamphlet (provided by your instructor)

3 You will have 50 minutes to complete the examination

4 The total weight of the examination is 100 points

5 There are five (5) multiple choice and four (4) calculation problems Work all multiple choice

problems and 4 calculation problems Show all work partial credit will be given for correct work

shown

6 If you have any questions during the examination see your instructor who will

be located in the classroom

7 Start 1030 am

Stop 1145 am

PROBLEM

POINTS

CREDIT

1-5

20

6

20

7

20

8

20

9

20

TOTAL

100

PERCENTAGE

CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS IN

MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR

PARTIAL CREDIT

1 Rolling without slipping depends on

a Kinetic friction between the rolling object and the ground

b Static friction between the rolling object and the ground

(4)

c Tension between the rolling object and the ground

d The force of gravity between the rolling object and the Earth

2 A car is negotiating a flat circular curve of radius 50 m with a speed of 20 ms The

maximum centripetal force (provided by static friction) is 12 times 104 N What is the centripetal

acceleration of the car

a 040 ms2

b 080 ms2

(4)

c 40 ms2

d 80 ms2 acp = Vt

2R = (200 ms)

2(500 m)

3 The gravitational force between two objects is proportional to

a The distance between the two objects

b The square of the distance between the two objects

(4)

c The product of the masses of the two objects

d The square of the product of the masses of the two objects

4 Two equal forces are applied to a door at the doorknob The first force is applied

perpendicular to the door the second force is applied at 30deg to the plane of the door Which

force exerts the greater torque

a The first applied perpendicular to the door

b The second applied at an angle

(4)

c Both exert equal non-zero torques

d Both exert zero torques

5 Two uniform solid spheres have the same mass but one has twice the radius of the other The

ratio of the larger spheres moment of inertia to that of the smaller sphere is

a 45

b 85

(4)

c 2

d 4 (25 M (2R)2)(25 M R

2) = 4

6 Three children are trying to balance on a seesaw which consists of a fulcrum rock acting as a

pivot at the center and a very light board 36 m long Two playmates are already on either end

Boy A has a mass of 50 kg and girl B a mass of 35 kg

a Where should girl C whose mass is 25 kg place herself so as to balance the seesaw

= (500kg)(981 ms2)(180 m) = 883 N-m

= - (350kg)(981 ms2)(180 m) = - 618 N-m

C= - (250kg)(981 ms2)(x) = - (245N)(x)

= 0

883 N-m - 618 N-m ndash (245 N)(x) = 0

x = 108 m

b What is the magnitude of the normal force that the rock exerts on the board

Nx = 0

Ny = N

wAx = 0

wAy = - mA g

wBx = 0

wBy = - mB g

wCx = 0

wCy = - mC g

Fx = 0

Fy = 0

N - mA g ndash mB g ndash mC g = 0

N = mA g + mB g + mC g = (mA + mB + mC)g = 1079 N

7 Suppose our Sun eventually collapses into a white dwarf losing about half its mass in the

process and winding up with a radius 10 of its existing radius

a Assuming the lost mass carries away no angular momentum what would the Sunrsquos new

rotation rate be (Take the Sunrsquos current period to be about 30 days)

II I = IF F

II = 25 M0 R02

II = 25 (M02)(R0100)2

(25 M0 R02) I = (25 (M02)(R0100)

2) F

I = (12)(1100)2) F

I = (120000) F

F = (20000) I

I = (2)T = (2)(30 days x (24 hours1 day)(3600 s1hour)) = 242 x 10-6

rads

F = 484 x 10-2

rads

(To practice convert the new angular velocity into the new period)

b What would be its final KE in terms of its initial KE of today

K = frac12 I 2

KI = frac12 (25 M0 R02I

2

KF = frac12 (25 (M02)(R0100)2 F

2

KF KI = [frac12 (25 (M02)(R0100)2 ) F

2] [frac12 (25 (M0)(R0)

2 ) I

2]

KF KI = [(12)(1100)2 ) F

2] [I

2] = (120000)(F

2 I

2) = 20000

8 The rings of Saturn are composed of chunks of ice that orbit the planet The inner radius of the

rings is 73000 km while the outer radius is 170000 km Find the period of an orbiting chunk of

ice at the inner radius and the period of a chunk at the outer radius Compare your numbers with

Saturnrsquos mean rotation period of 10 hours and 39 minutes The mass of Saturn is kg 1075 26

TO2 = ((4

2)(GMS))RO

3

TO2 = ((4

2)(6673 x 10

-11 N-m

2kg

2)(570 x 10

26 kg)))(170000000 m)

3 = 506 x 10

9 s

2

TO = 711 x 104 s = 198 hours

TI2 = ((4

2)(GMS))RI

3

TI2 = ((4

2)(( 6673 x 10

-11 N-m

2kg

2)(570 x 10

26 kg)))(73000000 m)

3 = 400 x 10

8 s

2

TI = 200 x 104 s = 556 hours


NAME_______SOLUTION______________________











5 There are six (6) multiple choice and four (4) calculation problems Work five (5) multiple

choice problems and four (4) calculation problems Show all work partial credit will be given for

correct work shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL 100

PERCENTAGE



PARTIAL CREDIT

1 Tripling the mass of the bob on a simple pendulum will cause a change in the

frequency of the pendulum swing by what factor

a 0330

b 100

(4)

c 300

d 900

2 What condition or conditions are necessary for static equilibrium

a ΣFx = 0

b ΣFx = 0 ΣFy = 0 Σ = 0

(4)

c Σ = 0

d ΣFx = 0 ΣFy = 0






(4)



4 A solid cylinder of mass 10 kg is pivoted about a frictionless axis thought the center O A

rope wrapped around the outer radius R1 = 10 m exerts a force F1 = 50 N to the right A second

rope wrapped around another section of radius R2 = 050 m exerts a force F2 = 60 N downward

What is the angular acceleration of the cylinder

a 10 rads2

b 060 rads2

(4)

c 040 rads2

d 080 rads2

5 A mass on a spring undergoes SHM When the mass is at its maximum displacement from

equilibrium its instantaneous velocity

a Is maximum

b Is less than maximum but not zero

(4)

c Is zero

d Cannot be determined from the information given

2 The moment of inertia of a rigid body

a Depends on the axis of rotation

b Cannot be negative

(4)

c Depends on mass distribution

d All of the above

7 A centrifuge rotor rotating at 10300 rpm is shut off and is eventually brought uniformly to rest

by a frictional torque of Nm 201 If the mass of the rotor is 480 kg and it can be approximated

as a solid cylinder of radius 00710 m

a What is the rotorrsquos moment of inertia

I = frac12 MR2 = frac12 (480 kg)(00710 m)

2 = 00121 kgm

2

b What is the angular acceleration of the rotor as it is brought to rest

I = (120 N-m)(00121 kgm2) = 992 rads

2

c Through how many revolutions will the rotor turn before coming to rest

2 =

2 + 2 )

= 10300 rpm = 1079 rads

= 0

)= - 2 (2 ) = (1079 rads)

2(-2 x 992 rads

2) = 5868 rad = 934 rev

8 A uniform disk turns at srev 42 around a frictionless spindle A nonrotating rod of the same mass

as the disk and length equal to the diskrsquos diameter is dropped onto the freely spinning disk

They then both turn around the spindle with their centers superposed What is the angular

frequency in srev of the combination

Ii = frac12 MR2

If = frac12 MR2 + (112) ML

2 = frac12 MR

2 + (112) M(2R)

2 = frac12 MR

2 + (13) MR

2 = (56) MR

2

Iii = Iff

f = Iii If

f = (240 revs) (frac12 MR2) (

56 MR

2) = 144 revs

9 A 0150-kg toy is undergoing simple harmonic motion on the end of a horizontal spring with

spring constant k = 300 Nm The amplitude of the motion is 0200 m

a Find the total mechanical energy of the toy at any point of its motion

E = frac12 kA2 = frac12 (300 Nm) (0200 m)

2 = 600 J

c Find the maximum speed attained by the toy

frac12 m Vmax2 = frac12 kA

2

m Vmax2 = kA

2

Vmax2 = (km)A

2

Vmax = ((300 Nm)(0150 kg))frac12 (0200 m) = 894 ms

d How many oscillations per second does the toy undergo

T = 2 (mk)frac12

T = 2 ((0150 kg)(300 Nm))frac12

T = 0140 s

F = 1T = 711 Hz

e What is the speed of the toy at 100 cm to the right from the equilibrium position

V = +- (km(A2 ndash x

2))

frac12

V = +- ((300 Nm)(0150 kg)((0200 m)2 ndash (0100 m)

2))

frac12 = +- 775 ms

V = 775 ms

7 Pictured below is a very light wooden plank with two masses 100 kg each on top of it

Find the reaction forces at points A and B

NAx = NA cos (90o) = 0

NAy = NA sin (90o) = NA

NBx = NB cos (90o) = 0

NBy =NB sin (90o) = NB

w1x = w1 cos (-90o) = 0

w1y = w1 sin (-90o) = - m1 g = - (100 kg)(981 ms

2) = - 981 N

w2x = w2 cos (-90o) = 0

w2y = w2 sin (-90o) = - m2 g = - (100 kg)(981 ms

2) = - 981 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt NA + NB ndash 981 N ndash 981 N = 0

NA + NB= 196 N

Assuming that the axis of rotation is at point A

NA = 0

NB = NB (300 m)

w1 = - (981 N) (0750 m) = - 736 N-m

sp = - (981 N) (250 m) = - 245 N-m

= 0

NB (300 m) ndash 736 N-m ndash 245 N-m = 0

NB = 106 N

NA= 196 N ndash 106 N = 900 N


NAME___SOLUTION__________________________



Version 1 November 28 2007


1 Check your examination for completeness prior to starting There are a total of ten (10)






5 There are six (6) multiple choice and four (4) calculation problems Work all calculation

problems and 5 (five) multiple choice Show all work partial credit will be given for correct work

shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT

1 A solid disk rolls down the slope as shown below What is the direction of the discrsquos angular

momentum

2 A very light rod with two identical masses attached to it is rotating in the counterclockwise

direction around the axis which goes through one of the rodrsquos ends as shown below Which one

of the masses has the most kinetic energy

(A) Down the slope

(B) Up the slope

(D) Into the page

(C) Out of the page

(E) Clockwise

3 Given the two masses from Problem 2 as the mass A undergoes the angular displacement of

140 rad the mass B undergoes the angular displacement of

(A) 0 rad

(B) 0700 rad

(C) 140 rad

(D) 280 rad

(E) 560 rad

4 A merry-go-around spins feely when Janice moved quickly to the center along the radial

direction Is it true to say that (A) The moment of inertia of the system decreases and the angular speed increases (B) The moment of inertia of the system decreases and the angular speed decreases (C) The moment of inertia of the system decreases and the angular speed remains the same (D) The moment of inertia of the system increases and the angular speed increases (E) The moment of inertia of the system increases and the angular speed decreases

5 What are the dimensions of the moment of inertia

(A) ML

(B) ML

2

(A) A

(B) B

(C) Both masses have the same amount of

kinetic energy

(D) Neither one of the masses has any

kinetic energy

(C) ML3

(D) ML

2

(E) LT

2

6 What is the correct expression for torque in terms of the magnitude of the force F the radial

distance from the axis of rotation r and the angle between the force and the radial direction

(A) = F r sin

(B) = F r cos

(C) = F r tan

(D) = F r cot

(E) = F r

7 Two disks of identical masses but different radii (r and 2r) are spinning on frictionless

bearings at the same angular speed of 100 rads but in opposite direction The two disks are

brought slowly together The frictional force between the surfaces eventually brings them to a

common angular velocity What is the magnitude of the final angular velocity

LI = LF

LI = (12 m r2 ) + (12 m (2r)

2 ) (-)

LF = (12 m r2 ) F+ (12 m (2r)

2 ) F

(12 m r2 ) + (12 m (2r)

2 ) (-) = (12 m r

2 ) F+ (12 m (2r)

2 ) F

r2 + (2r)

2(-) = r

2 F+ (2r)

2 F

r2 + 4r

2(-) = r

2 F+ 4r

2 F

+ 4(-) = F+ 4F

F= - 35 I

F = - 0600 rads

8 A uniform beam of mass m and length l is hung from two cables one at the end of the beam

and the other 5l8 of the way to the other end as shown below If m = 100 kg and l = 200 m

determine the magnitudes of the forces the cable exerts on the beam

Note we will use standard x-y coordinate system with the positive x-axis pointing to the right and

positive y-axis pointing upward

TLx = 0

TLy = TL

TRx = 0

TRy = TR

wx = 0

wy = - m g

Fx = 0

Fy = 0

TL + TR ndash m g = 0

TL + TR = (100 kg)(981 ms2) = 981 N

TL= 0

TR= TR (125 m)

w= - (100 kg)(981 ms2)(100 m) = - 981 N-m

= 0

TR (125 m) ndash 981 N-m = 0

TR = 785 N

TL = 196 N

9 An automotive engine shaft revs up from 300 x 103 revmin to 450 x 10

3 revmin during a

120 s interval

(a) Calculate the magnitude of the angular acceleration of the shaft assuming it is constant during

the interval

= (300 x 103 revmin)(2 rad1 rev)(1 min60 s) = 0314 x 10

3 rads


3 rads

= + t

= (- ) t = 131 rads2

(b) Determine the number of revolutions of the shaft during the interval

=

+ 2 ()

() = (-

) (2 = (469 rad)(1 rev2 rad) = 746 rev

(c) Assuming that the shaft can be approximated as a solid cylinder with a radius of 500 cm and

mass of 100 kg what is the change in the shaftrsquos kinetic energy during this motion

I = frac12 m R2

I = frac12 (100 kg) (00500 m)2 = 00125 kg-m

2

Ki = frac12 I= frac12 (00125 kg-m

2)(0314 rads)

2 = 616 J

Kf = frac12 I= frac12 (00125 kg-m

2)(0471 rads)

2 = 1387 J

K = 771 J

10 Three forces are applied to a wheel of radius 0350 m as shown below One force is

perpendicular to the rim one is tangent to it and the third makes a 400o angle with the radius

(a) What is the net torque on the wheel due to these three forces for the axis perpendicular

to the wheel and passing through its center

= (119 N) (0350 m) sin(0o) = 0

= (146 N) (0350 m) sin(400o) = -328 N-m

= (850 N) (0350 m) sin(90o) = 298 N-m

= -0310 N-m

(b) Given that the mass of the wheel is 500 kg what is its moment of inertia

I = frac12 m R2

I = frac12 (500 kg) (0350 m)2 = 0306 kg-m

2

(c) Find the angular acceleration that the wheel has as a result of these three forces

= I

I = -101 rads2

(d) Does the wheel rotate clockwise or counterclockwise

Clockwise


NAME____SOLUTION_________________________



Version 1 April 27 2005









choice and four (4) calculation problems Show all work partial credit will be given for correct work

shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

23

8 23 9

17

10

17

TOTAL

100

PERCENTAGE



PARTIAL CREDIT


e Depends on the axis of rotation

f Cannot be negative

(4)

g Depends on mass distribution

h All of the above

2 If the period of a system undergoing a SHM is doubled the frequency of the system is

a Doubled

b Halved

(4)

c For times as large

d One-quarter as large

3 Which one of the following quantities is zero when an object in simple harmonic motion is at its

maximum displacement

a Potential energy

b Acceleration

(4)

c Kinetic energy

d Frequency

4 What condition or conditions are necessary for rotational equilibrium

a ΣFx = 0

b ΣFx = 0 ΣΤ = 0

(4)

c ΣΤ = 0

d ΣFx = 0 ΣFy = 0

5 What is the quantity used to measure an objects resistance to changes in rotation

a Mass

b Moment of inertia

(4)

c Torque

d Angular velocity

6 If a net torque is applied to an object that object will experience

a A constant angular velocity

b An angular acceleration

(4)

c A constant moment of inertia

d An increasing moment of inertia

7 A person of mass 750 kg stands at the center of a rotating merry-go-round of radius 300 m

and moment of inertia of 920 kgm2 The merry-go-round rotates without friction with angular

velocity of 200 rads The person walks radially to the edge of the merry-go-round

a Calculate the angular velocity of the merry-go-round when the person reaches the edge

Ii = 920 kgm2

If = 920 kgm2 + (750 kg)(300 m)

2 = 1595 kgm

2

i rads

Li = Lf

Ii i If f

f = Ii i If

f = (920 kgm2) 200 rads 1595 kgm

2) = 115 rads

b Calculate the rotational kinetic energy of the system before and after the personrsquos walk

KEi = frac12 Ii i 2 frac12 (920 kgm

2) (200 rads)

2 = 1840 J

KEf = frac12 If f 2 frac12 (1595 kgm

2) (115 rads)

2 = 1055 J

d If the merry-go-round rotates in a clockwise direction as viewed from above what is the

direction of its angular momentum

Inward (Right Hand Rule)

8 A shop sign weighing 245 N is supported by a uniform beam as shown below Find the tension

in the wire and the horizontal and vertical forces exerted by the hinge on the beam

Remember to draw a free-body diagram first

(w)x = w cos(- 90o) = 0

(w)y = w sin(- 90o) = - 245 N

(FV)x = FV cos(90o) = 0

(FV)y = FV sin(90o) = FV

(FH)x = FH cos(0o) = FH

(FH)y = FH sin(0o) = 0

(T)x = T cos(145o)

(T)y = T sin(145o)

Fx = 0 FH + T cos(145o) = 0

Fy = 0 -245 N + FV + T sin(145o) = 0

w = - (245 N)(170 m) = - 4165 Nm

T= + (T)(1350 m) sin(35o)

FH = 0

FV = 0

= 0

-4165 Nm + (T)(1350 m) sin(35o) = 0 T = 538 N

FH = 441 N

FV = -636 N

9 Neptune is an average distance of 450 x 109 km from the Sun Estimate the length of the

Neptunian year given that the Earth is 150 x 108 km from the Sun on the average

TN2 = (4

2GMS )RN

3

TE2 = (4

2GMS )RE

3

TN2TE

2 = RN

3RE

3

TNTE = (RN3RE

3)

12

TNTE = ((450 x 109)3(150 x 10

8)

3)12

= 164

TN = 164 years

What is the Neptunersquos orbital speed

V = 2RT

V = 2(450 x 1012

m)(164x365x24x3600 s) = 5467 ms

10 A mass-spring system undergoes SHM in horizontal direction If the mass is 0250 kg the spring

constant is 120 Nm and the amplitude is 150 cm

a What would be the maximum speed of the mass

Vmax = (km) A2

frac12

Vmax = [(120 Nm)(0250 kg)](0150 m)2

frac12 = 103 ms

b Where would it occur

x= 0

c What would be the speed at a half-amplitude position

V = (km)(A2 ndash x

2)

frac12

V = [(120 Nm)(0250 kg)][(0150 m)2 - (00750 m)

2 ]

frac12 = 0900 ms

d How many oscillations per second does the system go through

f = (12 (km)frac12 = (12 (120 Nm)(0250 kg)

frac12 = 110 Hz

e What would be the magnitude of the mass acceleration at the half-amplitude position

a = Fm

F = k x

a = k xm

a = (120 Nm)(00750 m)(0250 kg) = 360 ms2


NAME_____SOLUTION________________________













shown



7 Start 1030 am

Stop 1145 am

PROBLEM

POINTS

CREDIT

1-5

20

6

20

7

20

8

20

9

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT

1 An ice skater performs a pirouette (a fast spin) by pulling in his outstretched arms close to his

body What happens to his moment of inertia about the axis of rotation

a It does not change

b It increases

(4)

c It decreases

d It changes but it is impossible to tell which way

2 A wheel with moment of inertia 300 kg∙m2 has a net torque of 350 N∙m applied to it What

angular acceleration does it experience

a 0857 rads2

b 117 rads2

(4)

c 300 rads2

d 350 rads2

3 A fan has blades 0250 m long rotating at 200 rpm What is the centripetal acceleration of a

point on the outer tip of a blade

a 110 ms2

b 087 ms2

(4)

c 055 ms2

d 023 ms2

4 According to Keplerrsquos first law Halleyrsquos comet circles the sun in an elliptical path with the

Sun at one focus of the ellipse What is at the other focus of the ellipse

a Nothing

b The Earth

(4)

c The comet itself passes through the other focus

d The tail of the comet stays at the other ellipse

5 The quantity ldquomoment of inertiardquo is equivalent in terms of fundamental quantities of mass

length and time to

a ML2T

-2

b ML

(4)

c ML2

d ML-1

T-2

6 A 600-kg painter is on a uniform 25-kg scaffold supported from above by ropes There is a 40-kg pail

of paint to one side as shown What are the magnitudes of the tension forces exerted by the ropes on

the scaffold

(20)

TLx = TL cos (90o) = 0

TLy = TL sin (90o) = TL

TRx = TR cos (90o) = 0

TRy = TR sin (90o) = TR

wpx = wp cos (-90o) = 0

wpy = wp sin (-90o) = - mp g = - (600 kg)(981 ms

2) = - 589 N

wsx = ws cos (-90o) = 0

wsy = ws sin (-90o) = - ms g = - (250 kg)(981 ms

2) = - 245 N

wpail x = wpail cos (-90o) = 0

wpail y = wpail sin (-90o) = - mpail g = - (400 kg)(981 ms

2) = - 392 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt TR + TL ndash 589 N ndash 245 N ndash 392 N = 0

TR + TL= 873 N

Assuming that the axis of rotation is at the point where the left cable is attached to the scaffold

TL = 0

TR = TR (400 m)

wp = - (589 N) (200 m) = - 1178 N-m

sp = - (245 N) (200 m) = - 490 N-m

wp = - (392 N) (100 m) = - 392 N-m

= 0

TR (400 m) - 1178 N-m ndash 490 N-m ndash 392 N-m = 0

TR = 427 N

TL= 873 N ndash 417 N = 446 N

7 A potterrsquos wheel is rotating around a vertical axis through its center at an angular velocity of

150 rads The wheel can be considered a uniform disk of mass 500 kg and diameter 0800 m

The potter then throws a 100-kg chunk of clay onto the rim of the rotating wheel What is the

angular velocity of the wheel after the clay sticks to it

Ii = frac12 mw R2 = frac12 (500 kg)(0400 m)

2 = 0400 kg m

2

If = frac12 mw R2 + mc R

2 = frac12 (500 kg)(0400 m)

2 + (100 kg)(0400 m)

2 = 0560 kg m

2

Ii i = If f

f = (Ii If) i = 107 revs

8 A satellite of mass 5500 kg orbits the Earth kg1060mass 24 and has a period of 6200 s

a Find the magnitude of the Earthrsquos gravitational force on the satellite

Hint find answer to b first and use it to calculate the gravitational force

F = G m1 m2r2

F = G Ms MER2 = (6673 x 10

-11 Nm

2kg

2) (5500 kg) (600 x 10

24 kg )(726 x 10

6 m)

2 = 41779

N

b What is the orbital radius of the satellite

T2 = (4

2GME )R

3

T2 GME = 4

2 R

3

R3 = T

2 GME (4

2)

R = (T2 GME (4

2))

13

R = ((6200 s)2 (6673 x 10

-11 Nm

2kg

2)(600 x 10

24 kg) (4

2))

13 = 726 x 10

6 m

9 A teenager pushes tangentially on a small hand-driven merry-go-round and is able to

accelerate it from rest to an angular velocity of 175 rads in 100 s Assume the merry-go-

round is a uniform disk of radius 250 m and has a mass of 760 kg and two children (each

with a mass of 250 kg) sit opposite each other on the edge

a What angular acceleration is required to reach 175 rads velocity in 100 s

= 0 + t

= ( - 0) t

= (175 rads) (100 s) = 0175 rads2

c What is the moment of inertia of the merry-go-round with two children on it

I = frac12 M R2 + m R

2 + m R

2 = frac12 (760 kg)(250 m)

2 + 2 (250 kg)(250 m)

2 = 2688 kg m

2

b Calculate the torque required to produce the acceleration neglecting frictional torque

= I

= (2688 kgm2)(0175 rads

2) = 470 N-m

Bonus (5 points) What is the magnitude of the tangential force required to produce this torque

= F R

F = R = (470 N-m)(250 m) = 188 N


NAME______SOLUTION_______________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT

1 Two children are riding on a merry-go-round Child A is at greater distance from the axis

of rotation than child B Which child has the larger angular displacement

a Child A

b Child B

(4)

c Both have the same angular displacement

d Cannot be determined based on the given information

2 What is the quantity used to measure an objectrsquos resistance to changes in rotational motion

a Mass

b Moment of inertia

(4)

c Angular velocity

d Angular acceleration

3 Two forces are applied to a doorknob perpendicular to the door The first force is twice as

large as the second force The ratio of the torque of the first force to the torque of the second is

a 14

b 12

(4)

c 2

d 4

4 When you ride a bicycle what is the direction of the angular velocity of the wheels

a To your left

b To your right

(4)

c Forward

d Backward

5 You sit on a freely rotating stool (no friction) When you extend your arms

a Your moment of inertia decreases and your angular velocity increases

b Your moment of inertia increases and your angular velocity increases

(4)

c Your moment of inertia decreases and your angular velocity decreases

d Your moment of inertia increases and your angular velocity decreases

6 Two identical spheres each of mass M and radius R just touch each other What is the

magnitude of the gravitational force that they exert on each other

a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)

7 Shekema and Rodney are carrying a 400 kg block on a 200 m board as shown below

The mass of the board is 500 kg Shekema insists that she is very strong and could have

carried the block all by herself so they place the block 0500 m away from her and 150

m away from Rodney Find the forces (in Newtons) exerted by each to carry the block

NRx = NR cos (90o) = 0

NRy = NR sin (90o) = NR

NSx = NS cos (90o) = 0

NSy = NS sin (90o) = NS

Wboard x = wboard cos (-90o) = 0

Wboard y = wboard sin (-90o) = - mboard g = - (500 kg)(981 ms

2) = - 491 N

Wblock x = wblock cos (-90o) = 0

Wblock y = wblock sin (-90o) = - mblock g = - (400 kg)(981 ms

2) = - 392 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt NR + NS ndash 589 N ndash 491 N ndash 392 N = 0

NR + NS= 442 N

Assuming that the axis of rotation is at the point where Rodney is holding the board

NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0

NS (200 m) ndash 491 N-m ndash 588 N-m = 0

NS = 319 N

NR= 442 N ndash 319 N = 123 N

8 A compact disc rotates from rest to 500 revmin in 550 s

a What is its angular acceleration assuming that it is constant

0 =

= (500 revmin)(1 min600 s)(2 rad1 rev) = 524 rads

= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2

b How many revolutions does it make in 550 s

=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev

c At the moment the disk reaches the angular velocity of 500 revmin what is the tangential

velocity of a point on the rim of the disc 600 cm from the center of rotation

Vt = R = (523 rads)(00600 m) = 314 ms

9 Penny goes bowling with her friends She notices that a bowling ball of radius 110 cm and

mass m = 720 kg is rolling without slipping on a horizontal ball return at 200 ms It then

rolls without slipping up a hill to a height h before momentarily coming to rest Find h

Ei = Ef

m g yi + frac12 m Vi 2 + frac12 I i

2 = m g yf + frac12 m Vf

2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0

frac12 m Vi 2 + frac12 I i

2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m

10 Amanda David and Rotasha play with two children who want a ride on a merry-go-round The

merry-go-round is a solid disc that rotates freely (without friction) around its center and has a

moment of inertia of 500 kg-m2 relative to this axis The children have the same masses of 250 kg

and sit at distances of 100 m and 200 m away from the center of merry-go-round The radius of

marry-go-round is 300 m

a What is the moment of inertia of the merry-go-round with the children on it

relative to the center of rotation

I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2

b Amanda David and Rotasha pull the merry-go-round with tangential forces as

shown in the figure (The magnitudes are A = 300 N D = 150 N R = 200 N)

What is the net torque that they exert on the merry-go-round

= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m

c What is the angular acceleration of the merry-go-round

I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT

1 A particle is traveling in a horizontal circle at a constant speed One can conclude that

__________ is constant

(A) Velocity

(B) Acceleration

(C) Net force

(D) All of the above

(E) None of the above

2 The dimension of torque is the same as

(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8

4 Two children are riding on a merry-go-round Child A is at greater distance from the axis of

rotation than child B Which child has the larger linear displacement

(A) Child A

(B) Child B

(C) Both have the same zero linear displacement

(D) Both have the same non-zero linear displacement

(E) Cannot be determined based on the given information


rotation than child B It is true statement that

(A) Child A has greater angular velocity than child B

(B) Child B has greater angular velocity than child B

(C) Both have the same zero angular velocity

(D) Both have the same non-zero angular velocity


6 You sit on a rotating stool which rotates clockwise (seen from above) What is the direction of

your angular acceleration if the stool is slowing down

(A) Vertically upward

(B) Vertically downward

(C) Horizontally and to your left

(D) Horizontally and to your right

(E) Your angular acceleration has no direction

7 A wheel starts to rotate around its center with a constant angular acceleration of 260 rads2

The wheel starts from rest and rotates for 600 s At the end of that time

a What is its angular velocity

t

260 rads2s) = 156 rads

b Through what angle has the wheel turned

)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad

c How many revolutions has it made

468 rad(2 radrev) = 745 rev

d What is the magnitude of tangential velocity and tangential acceleration of a point 0300 m

from the axis of rotation

Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2

8 To determine the location of his center of mass a physics student lies on a lightweight plank

supported by two scales 250 m apart The left scale reads 290 N and the right scale reads 122 N

(Hint the readings correspond to the normal forces exerted on the plank)

a Draw a free-body diagram

b Find the studentrsquos mass

NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg

c Find the distance from the studentrsquos head to his center of mass

NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m

9 A merry-go-round of radius 200 m is rotating about its center making one revolution every

500 seconds A child of mass 250 kg stands at the rim

a What is the magnitude of the angular velocity of merry-go-round in rads

= (2 rad)(500 s) = 126 rads

b What is the magnitude of centripetal acceleration of the child

acp = r = (126 rads)

2(200 m) = 318 ms

2

c The merry-go-round is very slippery (frictionless) and the child stays on it by holding

onto a rope attached to the pivot at the center of merry-go-round What has to be the

magnitude of tension in the rope for the child to stay on merry-go-round

T = m acp = (250 kg)(319 ms2) = 794 N

d The rope suddenly breaks What is the motion of the child immediately after it

The child continues to move in the direction which was tangential to the circle the moment the

rope broke until some other force changes his velocity

10 A square metal plate 0180 m on each side is pivoted about an axis through point O at its center

and perpendicular to the plate Calculate the net torque about this axis due to the three forces shown

in the figure if the magnitudes of the forces are F1 = 180 N F2 = 260 N and F3 = 140 N The plate

and all forces are in the plane of the page

r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________



Version 1 August 1 2005



problems on eight (8) pages





5 There are six (6) multiple choice and four (4) calculation problems Work all problems

Show all work partial credit will be given for correct work shown



7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT

2 Which one of the following quantities is zero when an object in simple harmonic motion is at

its maximum displacement

e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency

2 Two equal forces are applied to a door The first force is applied at the midpoint of the

door the second force is applied at the doorknob Both forces are applied perpendicular

to the door Which force exerts the greater torque

a The first at the midpoint

(5)

b The second at the doorknob



3 The quantity ldquoangular momentumrdquo (in terms of the fundamental quantities of mass length and

time) is equivalent to

a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T

4 Two moons orbit a planet in nearly circular orbits Moon A has orbital radius r and moon B

PHYS 1111 Exam 3 Version 1

Fall 2005 47

has orbital radius 4r Moon A takes 20 days to complete one orbit How long does it take moon

B to complete an orbit

a 20 days

b 80 days

(5)

c 160 days

d 320 days

5 The moment of inertia of a spinning body about its spin axis depends on its

a Angular speed shape and mass

b Angular acceleration mass and axis position

(5)

c Mass shape axis position and size

d Mass size shape and speed

6 A boy and a girl are riding on a merry-go-round which is turning at a constant rate The boy is

near the outer edge and the girl is closer to the center Who has the greater linear speed

a The boy

b The girl

(5)

c Both have the same non-zero translational velocity

d Both have zero translational velocity

7 A 640-kg person stands on a weightless diving board supported by two pillars one at the enof

the board the other 110-m away The pillar at the end of the board exerts a downward force of

828 N The board has a length of 300


Fall 2005 48

a How far from that pillar is the person standing

N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49

b Find the force exerted by the second pillar

N2 = 1455 N

8 A potterrsquos wheel is rotating around a vertical axis through its center at a frequency of

srev 51 The wheel can be considered a uniform disk of mass 500 kg and diameter 0400

m The potter then throws a 310-kg chunk of clay approximately shaped as a flat disk of

radius 800 cm onto the center of the rotating wheel

a What is the frequency of the wheel after the clay sticks to it


2 = 0100 kg m

2

If = frac12 mw R2 + frac12 mc R

2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f


b (Bonus) If the wheel rotates in clockwise direction as viewed from above what is

the direction of the wheelrsquos angular momentum

9 A 0500-kg mass resting on a horizontal frictionless surface is connected to a fixed spring with


Fall 2005 50

spring constant of 800 Nm The mass is displaced 160 cm from its equilibrium position and

released

a What is the amplitude of oscillations

A = 0160 m

b What is the maximum speed of the mass

Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms

c What is the total mechanical energy of the system

TE = frac12 m Vmax2 = frac12 (0500 kg) (202 ms)

2 = 102 J

d What is the frequency of oscillations

f = (1(2)(km)12

= 201 Hz

e What is the period of oscillations

T = 1f = 0497 s

10 A uniform disk of mass 100 kg and radius 100 m has four symmetrically placed masses each of

mass 0250 kg fastened at positions 500 m from the center The disk is free to rotate around its

center

d What is the moment of inertia of the disk relative to its center

I = frac12 mdisk R2 + 4 m r

2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51

e Two tangential forces are applied to the disk as shown in the figure

What is the net torque acting on the disk

1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm

f What is the angular acceleration of the disk

= I

= I = (300 Nm) (750 kgm2) = 400 rads

2

g Assuming that initially the system was at rest what is the kinetic energy of the

disk 100 s after the start of the motion

KE = frac12 I 2 = frac12 I ( t)

2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54

9 After you pick up a spare your bowling ball rolls without slipping back toward the ball rack with a

linear speed of 285 ms To reach the rack the ball rolls up the ramp that gives the ball a 0530-m

vertical rise What is the speed of the ball when it reaches the top of the ramp

Ei = m g yi + frac12 mVi

2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2

Ef = m g yf + frac12 mVf2

+ frac12 If

2 = m g yf + frac12 mVf

2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55

Bonus (5 points) What is the direction of the ballrsquos angular momentum



PARTIAL CREDIT

1 Rolling without slipping depends on

a Kinetic friction between the rolling object and the ground

b Static friction between the rolling object and the ground

(4)

c Tension between the rolling object and the ground

d The force of gravity between the rolling object and the Earth

2 A car is negotiating a flat circular curve of radius 50 m with a speed of 20 ms The

maximum centripetal force (provided by static friction) is 12 times 104 N What is the centripetal

acceleration of the car

a 040 ms2

b 080 ms2

(4)

c 40 ms2

d 80 ms2 acp = Vt

2R = (200 ms)

2(500 m)

3 The gravitational force between two objects is proportional to

a The distance between the two objects

b The square of the distance between the two objects

(4)

c The product of the masses of the two objects

d The square of the product of the masses of the two objects






(4)





a 45

b 85

(4)

c 2

d 4 (25 M (2R)2)(25 M R

2) = 4





= (500kg)(981 ms2)(180 m) = 883 N-m

= - (350kg)(981 ms2)(180 m) = - 618 N-m

C= - (250kg)(981 ms2)(x) = - (245N)(x)

= 0

883 N-m - 618 N-m ndash (245 N)(x) = 0

x = 108 m


Nx = 0

Ny = N

wAx = 0

wAy = - mA g

wBx = 0

wBy = - mB g

wCx = 0

wCy = - mC g

Fx = 0

Fy = 0







II I = IF F

II = 25 M0 R02

II = 25 (M02)(R0100)2

(25 M0 R02) I = (25 (M02)(R0100)

2) F

I = (12)(1100)2) F

I = (120000) F

F = (20000) I


rads

F = 484 x 10-2

rads



K = frac12 I 2

KI = frac12 (25 M0 R02I

2

KF = frac12 (25 (M02)(R0100)2 F

2

KF KI = [frac12 (25 (M02)(R0100)2 ) F

2] [frac12 (25 (M0)(R0)

2 ) I

2]

KF KI = [(12)(1100)2 ) F

2] [I

2] = (120000)(F

2 I

2) = 20000





TO2 = ((4

2)(GMS))RO

3

TO2 = ((4

2)(6673 x 10

-11 N-m

2kg

2)(570 x 10

26 kg)))(170000000 m)

3 = 506 x 10

9 s

2

TO = 711 x 104 s = 198 hours

TI2 = ((4

2)(GMS))RI

3

TI2 = ((4

2)(( 6673 x 10

-11 N-m

2kg

2)(570 x 10

26 kg)))(73000000 m)

3 = 400 x 10

8 s

2

TI = 200 x 104 s = 556 hours


NAME_______SOLUTION______________________













correct work shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL 100

PERCENTAGE



PARTIAL CREDIT



a 0330

b 100

(4)

c 300

d 900


a ΣFx = 0

b ΣFx = 0 ΣFy = 0 Σ = 0

(4)

c Σ = 0

d ΣFx = 0 ΣFy = 0






(4)







a 10 rads2

b 060 rads2

(4)

c 040 rads2

d 080 rads2



a Is maximum


(4)

c Is zero





(4)


d All of the above





I = frac12 MR2 = frac12 (480 kg)(00710 m)

2 = 00121 kgm

2


I = (120 N-m)(00121 kgm2) = 992 rads

2


2 =

2 + 2 )

= 10300 rpm = 1079 rads

= 0

)= - 2 (2 ) = (1079 rads)

2(-2 x 992 rads

2) = 5868 rad = 934 rev





Ii = frac12 MR2

If = frac12 MR2 + (112) ML

2 = frac12 MR

2 + (112) M(2R)

2 = frac12 MR

2 + (13) MR

2 = (56) MR

2

Iii = Iff

f = Iii If

f = (240 revs) (frac12 MR2) (

56 MR

2) = 144 revs




E = frac12 kA2 = frac12 (300 Nm) (0200 m)

2 = 600 J



2

m Vmax2 = kA

2

Vmax2 = (km)A

2



T = 2 (mk)frac12

T = 2 ((0150 kg)(300 Nm))frac12

T = 0140 s

F = 1T = 711 Hz



2))

frac12

V = +- ((300 Nm)(0150 kg)((0200 m)2 ndash (0100 m)

2))

frac12 = +- 775 ms

V = 775 ms







w1x = w1 cos (-90o) = 0

w1y = w1 sin (-90o) = - m1 g = - (100 kg)(981 ms

2) = - 981 N

w2x = w2 cos (-90o) = 0

w2y = w2 sin (-90o) = - m2 g = - (100 kg)(981 ms

2) = - 981 N

Fx = 0 =gt 0 = 0


NA + NB= 196 N


NA = 0

NB = NB (300 m)

w1 = - (981 N) (0750 m) = - 736 N-m

sp = - (981 N) (250 m) = - 245 N-m

= 0


NB = 106 N

NA= 196 N ndash 106 N = 900 N


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT


momentum




(A) Down the slope

(B) Up the slope

(D) Into the page

(C) Out of the page

(E) Clockwise



(A) 0 rad

(B) 0700 rad

(C) 140 rad

(D) 280 rad

(E) 560 rad




(A) ML

(B) ML

2

(A) A

(B) B


kinetic energy


kinetic energy

(C) ML3

(D) ML

2

(E) LT

2



(A) = F r sin

(B) = F r cos

(C) = F r tan

(D) = F r cot

(E) = F r





LI = LF

LI = (12 m r2 ) + (12 m (2r)

2 ) (-)

LF = (12 m r2 ) F+ (12 m (2r)

2 ) F

(12 m r2 ) + (12 m (2r)

2 ) (-) = (12 m r

2 ) F+ (12 m (2r)

2 ) F

r2 + (2r)

2(-) = r

2 F+ (2r)

2 F

r2 + 4r

2(-) = r

2 F+ 4r

2 F

+ 4(-) = F+ 4F

F= - 35 I

F = - 0600 rads






TLx = 0

TLy = TL

TRx = 0

TRy = TR

wx = 0

wy = - m g

Fx = 0

Fy = 0


TL + TR = (100 kg)(981 ms2) = 981 N

TL= 0

TR= TR (125 m)

w= - (100 kg)(981 ms2)(100 m) = - 981 N-m

= 0

TR (125 m) ndash 981 N-m = 0

TR = 785 N

TL = 196 N


3 revmin during a

120 s interval


the interval


3 rads


3 rads

= + t

= (- ) t = 131 rads2


=

+ 2 ()

() = (-

) (2 = (469 rad)(1 rev2 rad) = 746 rev



I = frac12 m R2

I = frac12 (100 kg) (00500 m)2 = 00125 kg-m

2


2)(0314 rads)

2 = 616 J


2)(0471 rads)

2 = 1387 J

K = 771 J





= (119 N) (0350 m) sin(0o) = 0

= (146 N) (0350 m) sin(400o) = -328 N-m

= (850 N) (0350 m) sin(90o) = 298 N-m

= -0310 N-m


I = frac12 m R2

I = frac12 (500 kg) (0350 m)2 = 0306 kg-m

2


= I

I = -101 rads2


Clockwise


NAME____SOLUTION_________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

23

8 23 9

17

10

17

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




(4)


h All of the above


a Doubled

b Halved

(4)





a Potential energy

b Acceleration

(4)

c Kinetic energy

d Frequency


a ΣFx = 0

b ΣFx = 0 ΣΤ = 0

(4)

c ΣΤ = 0

d ΣFx = 0 ΣFy = 0


a Mass

b Moment of inertia

(4)

c Torque

d Angular velocity




(4)







Ii = 920 kgm2

If = 920 kgm2 + (750 kg)(300 m)

2 = 1595 kgm

2

i rads

Li = Lf

Ii i If f

f = Ii i If

f = (920 kgm2) 200 rads 1595 kgm

2) = 115 rads



2) (200 rads)

2 = 1840 J


2) (115 rads)

2 = 1055 J







(w)x = w cos(- 90o) = 0

(w)y = w sin(- 90o) = - 245 N

(FV)x = FV cos(90o) = 0




(T)x = T cos(145o)

(T)y = T sin(145o)

Fx = 0 FH + T cos(145o) = 0

Fy = 0 -245 N + FV + T sin(145o) = 0

w = - (245 N)(170 m) = - 4165 Nm

T= + (T)(1350 m) sin(35o)

FH = 0

FV = 0

= 0

-4165 Nm + (T)(1350 m) sin(35o) = 0 T = 538 N

FH = 441 N

FV = -636 N



TN2 = (4

2GMS )RN

3

TE2 = (4

2GMS )RE

3

TN2TE

2 = RN

3RE

3

TNTE = (RN3RE

3)

12

TNTE = ((450 x 109)3(150 x 10

8)

3)12

= 164

TN = 164 years


V = 2RT

V = 2(450 x 1012

m)(164x365x24x3600 s) = 5467 ms




Vmax = (km) A2

frac12

Vmax = [(120 Nm)(0250 kg)](0150 m)2

frac12 = 103 ms


x= 0


V = (km)(A2 ndash x

2)

frac12

V = [(120 Nm)(0250 kg)][(0150 m)2 - (00750 m)

2 ]

frac12 = 0900 ms


f = (12 (km)frac12 = (12 (120 Nm)(0250 kg)

frac12 = 110 Hz


a = Fm

F = k x

a = k xm

a = (120 Nm)(00750 m)(0250 kg) = 360 ms2


NAME_____SOLUTION________________________













shown



7 Start 1030 am

Stop 1145 am

PROBLEM

POINTS

CREDIT

1-5

20

6

20

7

20

8

20

9

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




b It increases

(4)

c It decreases




a 0857 rads2

b 117 rads2

(4)

c 300 rads2

d 350 rads2



a 110 ms2

b 087 ms2

(4)

c 055 ms2

d 023 ms2



a Nothing

b The Earth

(4)




length and time to

a ML2T

-2

b ML

(4)

c ML2

d ML-1

T-2



the scaffold

(20)





wpx = wp cos (-90o) = 0


2) = - 589 N

wsx = ws cos (-90o) = 0


2) = - 245 N



2) = - 392 N

Fx = 0 =gt 0 = 0


TR + TL= 873 N


TL = 0

TR = TR (400 m)

wp = - (589 N) (200 m) = - 1178 N-m

sp = - (245 N) (200 m) = - 490 N-m

wp = - (392 N) (100 m) = - 392 N-m

= 0


TR = 427 N

TL= 873 N ndash 417 N = 446 N






2 = 0400 kg m

2


2 = frac12 (500 kg)(0400 m)

2 + (100 kg)(0400 m)

2 = 0560 kg m

2

Ii i = If f





F = G m1 m2r2

F = G Ms MER2 = (6673 x 10

-11 Nm

2kg

2) (5500 kg) (600 x 10

24 kg )(726 x 10

6 m)

2 = 41779

N


T2 = (4

2GME )R

3

T2 GME = 4

2 R

3

R3 = T

2 GME (4

2)

R = (T2 GME (4

2))

13

R = ((6200 s)2 (6673 x 10

-11 Nm

2kg

2)(600 x 10

24 kg) (4

2))

13 = 726 x 10

6 m






= 0 + t

= ( - 0) t

= (175 rads) (100 s) = 0175 rads2



2 + m R

2 = frac12 (760 kg)(250 m)

2 + 2 (250 kg)(250 m)

2 = 2688 kg m

2


= I

= (2688 kgm2)(0175 rads

2) = 470 N-m


= F R

F = R = (470 N-m)(250 m) = 188 N


NAME______SOLUTION_______________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



a Child A

b Child B

(4)




a Mass

b Moment of inertia

(4)

c Angular velocity




a 14

b 12

(4)

c 2

d 4


a To your left

b To your right

(4)

c Forward

d Backward




(4)





a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)











2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55






(4)





a 45

b 85

(4)

c 2

d 4 (25 M (2R)2)(25 M R

2) = 4





= (500kg)(981 ms2)(180 m) = 883 N-m

= - (350kg)(981 ms2)(180 m) = - 618 N-m

C= - (250kg)(981 ms2)(x) = - (245N)(x)

= 0

883 N-m - 618 N-m ndash (245 N)(x) = 0

x = 108 m


Nx = 0

Ny = N

wAx = 0

wAy = - mA g

wBx = 0

wBy = - mB g

wCx = 0

wCy = - mC g

Fx = 0

Fy = 0







II I = IF F

II = 25 M0 R02

II = 25 (M02)(R0100)2

(25 M0 R02) I = (25 (M02)(R0100)

2) F

I = (12)(1100)2) F

I = (120000) F

F = (20000) I


rads

F = 484 x 10-2

rads



K = frac12 I 2

KI = frac12 (25 M0 R02I

2

KF = frac12 (25 (M02)(R0100)2 F

2

KF KI = [frac12 (25 (M02)(R0100)2 ) F

2] [frac12 (25 (M0)(R0)

2 ) I

2]

KF KI = [(12)(1100)2 ) F

2] [I

2] = (120000)(F

2 I

2) = 20000





TO2 = ((4

2)(GMS))RO

3

TO2 = ((4

2)(6673 x 10

-11 N-m

2kg

2)(570 x 10

26 kg)))(170000000 m)

3 = 506 x 10

9 s

2

TO = 711 x 104 s = 198 hours

TI2 = ((4

2)(GMS))RI

3

TI2 = ((4

2)(( 6673 x 10

-11 N-m

2kg

2)(570 x 10

26 kg)))(73000000 m)

3 = 400 x 10

8 s

2

TI = 200 x 104 s = 556 hours


NAME_______SOLUTION______________________













correct work shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL 100

PERCENTAGE



PARTIAL CREDIT



a 0330

b 100

(4)

c 300

d 900


a ΣFx = 0

b ΣFx = 0 ΣFy = 0 Σ = 0

(4)

c Σ = 0

d ΣFx = 0 ΣFy = 0






(4)







a 10 rads2

b 060 rads2

(4)

c 040 rads2

d 080 rads2



a Is maximum


(4)

c Is zero





(4)


d All of the above





I = frac12 MR2 = frac12 (480 kg)(00710 m)

2 = 00121 kgm

2


I = (120 N-m)(00121 kgm2) = 992 rads

2


2 =

2 + 2 )

= 10300 rpm = 1079 rads

= 0

)= - 2 (2 ) = (1079 rads)

2(-2 x 992 rads

2) = 5868 rad = 934 rev





Ii = frac12 MR2

If = frac12 MR2 + (112) ML

2 = frac12 MR

2 + (112) M(2R)

2 = frac12 MR

2 + (13) MR

2 = (56) MR

2

Iii = Iff

f = Iii If

f = (240 revs) (frac12 MR2) (

56 MR

2) = 144 revs




E = frac12 kA2 = frac12 (300 Nm) (0200 m)

2 = 600 J



2

m Vmax2 = kA

2

Vmax2 = (km)A

2



T = 2 (mk)frac12

T = 2 ((0150 kg)(300 Nm))frac12

T = 0140 s

F = 1T = 711 Hz



2))

frac12

V = +- ((300 Nm)(0150 kg)((0200 m)2 ndash (0100 m)

2))

frac12 = +- 775 ms

V = 775 ms







w1x = w1 cos (-90o) = 0

w1y = w1 sin (-90o) = - m1 g = - (100 kg)(981 ms

2) = - 981 N

w2x = w2 cos (-90o) = 0

w2y = w2 sin (-90o) = - m2 g = - (100 kg)(981 ms

2) = - 981 N

Fx = 0 =gt 0 = 0


NA + NB= 196 N


NA = 0

NB = NB (300 m)

w1 = - (981 N) (0750 m) = - 736 N-m

sp = - (981 N) (250 m) = - 245 N-m

= 0


NB = 106 N

NA= 196 N ndash 106 N = 900 N


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT


momentum




(A) Down the slope

(B) Up the slope

(D) Into the page

(C) Out of the page

(E) Clockwise



(A) 0 rad

(B) 0700 rad

(C) 140 rad

(D) 280 rad

(E) 560 rad




(A) ML

(B) ML

2

(A) A

(B) B


kinetic energy


kinetic energy

(C) ML3

(D) ML

2

(E) LT

2



(A) = F r sin

(B) = F r cos

(C) = F r tan

(D) = F r cot

(E) = F r





LI = LF

LI = (12 m r2 ) + (12 m (2r)

2 ) (-)

LF = (12 m r2 ) F+ (12 m (2r)

2 ) F

(12 m r2 ) + (12 m (2r)

2 ) (-) = (12 m r

2 ) F+ (12 m (2r)

2 ) F

r2 + (2r)

2(-) = r

2 F+ (2r)

2 F

r2 + 4r

2(-) = r

2 F+ 4r

2 F

+ 4(-) = F+ 4F

F= - 35 I

F = - 0600 rads






TLx = 0

TLy = TL

TRx = 0

TRy = TR

wx = 0

wy = - m g

Fx = 0

Fy = 0


TL + TR = (100 kg)(981 ms2) = 981 N

TL= 0

TR= TR (125 m)

w= - (100 kg)(981 ms2)(100 m) = - 981 N-m

= 0

TR (125 m) ndash 981 N-m = 0

TR = 785 N

TL = 196 N


3 revmin during a

120 s interval


the interval


3 rads


3 rads

= + t

= (- ) t = 131 rads2


=

+ 2 ()

() = (-

) (2 = (469 rad)(1 rev2 rad) = 746 rev



I = frac12 m R2

I = frac12 (100 kg) (00500 m)2 = 00125 kg-m

2


2)(0314 rads)

2 = 616 J


2)(0471 rads)

2 = 1387 J

K = 771 J





= (119 N) (0350 m) sin(0o) = 0

= (146 N) (0350 m) sin(400o) = -328 N-m

= (850 N) (0350 m) sin(90o) = 298 N-m

= -0310 N-m


I = frac12 m R2

I = frac12 (500 kg) (0350 m)2 = 0306 kg-m

2


= I

I = -101 rads2


Clockwise


NAME____SOLUTION_________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

23

8 23 9

17

10

17

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




(4)


h All of the above


a Doubled

b Halved

(4)





a Potential energy

b Acceleration

(4)

c Kinetic energy

d Frequency


a ΣFx = 0

b ΣFx = 0 ΣΤ = 0

(4)

c ΣΤ = 0

d ΣFx = 0 ΣFy = 0


a Mass

b Moment of inertia

(4)

c Torque

d Angular velocity




(4)







Ii = 920 kgm2

If = 920 kgm2 + (750 kg)(300 m)

2 = 1595 kgm

2

i rads

Li = Lf

Ii i If f

f = Ii i If

f = (920 kgm2) 200 rads 1595 kgm

2) = 115 rads



2) (200 rads)

2 = 1840 J


2) (115 rads)

2 = 1055 J







(w)x = w cos(- 90o) = 0

(w)y = w sin(- 90o) = - 245 N

(FV)x = FV cos(90o) = 0




(T)x = T cos(145o)

(T)y = T sin(145o)

Fx = 0 FH + T cos(145o) = 0

Fy = 0 -245 N + FV + T sin(145o) = 0

w = - (245 N)(170 m) = - 4165 Nm

T= + (T)(1350 m) sin(35o)

FH = 0

FV = 0

= 0

-4165 Nm + (T)(1350 m) sin(35o) = 0 T = 538 N

FH = 441 N

FV = -636 N



TN2 = (4

2GMS )RN

3

TE2 = (4

2GMS )RE

3

TN2TE

2 = RN

3RE

3

TNTE = (RN3RE

3)

12

TNTE = ((450 x 109)3(150 x 10

8)

3)12

= 164

TN = 164 years


V = 2RT

V = 2(450 x 1012

m)(164x365x24x3600 s) = 5467 ms




Vmax = (km) A2

frac12

Vmax = [(120 Nm)(0250 kg)](0150 m)2

frac12 = 103 ms


x= 0


V = (km)(A2 ndash x

2)

frac12

V = [(120 Nm)(0250 kg)][(0150 m)2 - (00750 m)

2 ]

frac12 = 0900 ms


f = (12 (km)frac12 = (12 (120 Nm)(0250 kg)

frac12 = 110 Hz


a = Fm

F = k x

a = k xm

a = (120 Nm)(00750 m)(0250 kg) = 360 ms2


NAME_____SOLUTION________________________













shown



7 Start 1030 am

Stop 1145 am

PROBLEM

POINTS

CREDIT

1-5

20

6

20

7

20

8

20

9

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




b It increases

(4)

c It decreases




a 0857 rads2

b 117 rads2

(4)

c 300 rads2

d 350 rads2



a 110 ms2

b 087 ms2

(4)

c 055 ms2

d 023 ms2



a Nothing

b The Earth

(4)




length and time to

a ML2T

-2

b ML

(4)

c ML2

d ML-1

T-2



the scaffold

(20)





wpx = wp cos (-90o) = 0


2) = - 589 N

wsx = ws cos (-90o) = 0


2) = - 245 N



2) = - 392 N

Fx = 0 =gt 0 = 0


TR + TL= 873 N


TL = 0

TR = TR (400 m)

wp = - (589 N) (200 m) = - 1178 N-m

sp = - (245 N) (200 m) = - 490 N-m

wp = - (392 N) (100 m) = - 392 N-m

= 0


TR = 427 N

TL= 873 N ndash 417 N = 446 N






2 = 0400 kg m

2


2 = frac12 (500 kg)(0400 m)

2 + (100 kg)(0400 m)

2 = 0560 kg m

2

Ii i = If f





F = G m1 m2r2

F = G Ms MER2 = (6673 x 10

-11 Nm

2kg

2) (5500 kg) (600 x 10

24 kg )(726 x 10

6 m)

2 = 41779

N


T2 = (4

2GME )R

3

T2 GME = 4

2 R

3

R3 = T

2 GME (4

2)

R = (T2 GME (4

2))

13

R = ((6200 s)2 (6673 x 10

-11 Nm

2kg

2)(600 x 10

24 kg) (4

2))

13 = 726 x 10

6 m






= 0 + t

= ( - 0) t

= (175 rads) (100 s) = 0175 rads2



2 + m R

2 = frac12 (760 kg)(250 m)

2 + 2 (250 kg)(250 m)

2 = 2688 kg m

2


= I

= (2688 kgm2)(0175 rads

2) = 470 N-m


= F R

F = R = (470 N-m)(250 m) = 188 N


NAME______SOLUTION_______________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



a Child A

b Child B

(4)




a Mass

b Moment of inertia

(4)

c Angular velocity




a 14

b 12

(4)

c 2

d 4


a To your left

b To your right

(4)

c Forward

d Backward




(4)





a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)











2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55


x = 108 m


Nx = 0

Ny = N

wAx = 0

wAy = - mA g

wBx = 0

wBy = - mB g

wCx = 0

wCy = - mC g

Fx = 0

Fy = 0







II I = IF F

II = 25 M0 R02

II = 25 (M02)(R0100)2

(25 M0 R02) I = (25 (M02)(R0100)

2) F

I = (12)(1100)2) F

I = (120000) F

F = (20000) I


rads

F = 484 x 10-2

rads



K = frac12 I 2

KI = frac12 (25 M0 R02I

2

KF = frac12 (25 (M02)(R0100)2 F

2

KF KI = [frac12 (25 (M02)(R0100)2 ) F

2] [frac12 (25 (M0)(R0)

2 ) I

2]

KF KI = [(12)(1100)2 ) F

2] [I

2] = (120000)(F

2 I

2) = 20000





TO2 = ((4

2)(GMS))RO

3

TO2 = ((4

2)(6673 x 10

-11 N-m

2kg

2)(570 x 10

26 kg)))(170000000 m)

3 = 506 x 10

9 s

2

TO = 711 x 104 s = 198 hours

TI2 = ((4

2)(GMS))RI

3

TI2 = ((4

2)(( 6673 x 10

-11 N-m

2kg

2)(570 x 10

26 kg)))(73000000 m)

3 = 400 x 10

8 s

2

TI = 200 x 104 s = 556 hours


NAME_______SOLUTION______________________













correct work shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL 100

PERCENTAGE



PARTIAL CREDIT



a 0330

b 100

(4)

c 300

d 900


a ΣFx = 0

b ΣFx = 0 ΣFy = 0 Σ = 0

(4)

c Σ = 0

d ΣFx = 0 ΣFy = 0






(4)







a 10 rads2

b 060 rads2

(4)

c 040 rads2

d 080 rads2



a Is maximum


(4)

c Is zero





(4)


d All of the above





I = frac12 MR2 = frac12 (480 kg)(00710 m)

2 = 00121 kgm

2


I = (120 N-m)(00121 kgm2) = 992 rads

2


2 =

2 + 2 )

= 10300 rpm = 1079 rads

= 0

)= - 2 (2 ) = (1079 rads)

2(-2 x 992 rads

2) = 5868 rad = 934 rev





Ii = frac12 MR2

If = frac12 MR2 + (112) ML

2 = frac12 MR

2 + (112) M(2R)

2 = frac12 MR

2 + (13) MR

2 = (56) MR

2

Iii = Iff

f = Iii If

f = (240 revs) (frac12 MR2) (

56 MR

2) = 144 revs




E = frac12 kA2 = frac12 (300 Nm) (0200 m)

2 = 600 J



2

m Vmax2 = kA

2

Vmax2 = (km)A

2



T = 2 (mk)frac12

T = 2 ((0150 kg)(300 Nm))frac12

T = 0140 s

F = 1T = 711 Hz



2))

frac12

V = +- ((300 Nm)(0150 kg)((0200 m)2 ndash (0100 m)

2))

frac12 = +- 775 ms

V = 775 ms







w1x = w1 cos (-90o) = 0

w1y = w1 sin (-90o) = - m1 g = - (100 kg)(981 ms

2) = - 981 N

w2x = w2 cos (-90o) = 0

w2y = w2 sin (-90o) = - m2 g = - (100 kg)(981 ms

2) = - 981 N

Fx = 0 =gt 0 = 0


NA + NB= 196 N


NA = 0

NB = NB (300 m)

w1 = - (981 N) (0750 m) = - 736 N-m

sp = - (981 N) (250 m) = - 245 N-m

= 0


NB = 106 N

NA= 196 N ndash 106 N = 900 N


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT


momentum




(A) Down the slope

(B) Up the slope

(D) Into the page

(C) Out of the page

(E) Clockwise



(A) 0 rad

(B) 0700 rad

(C) 140 rad

(D) 280 rad

(E) 560 rad




(A) ML

(B) ML

2

(A) A

(B) B


kinetic energy


kinetic energy

(C) ML3

(D) ML

2

(E) LT

2



(A) = F r sin

(B) = F r cos

(C) = F r tan

(D) = F r cot

(E) = F r





LI = LF

LI = (12 m r2 ) + (12 m (2r)

2 ) (-)

LF = (12 m r2 ) F+ (12 m (2r)

2 ) F

(12 m r2 ) + (12 m (2r)

2 ) (-) = (12 m r

2 ) F+ (12 m (2r)

2 ) F

r2 + (2r)

2(-) = r

2 F+ (2r)

2 F

r2 + 4r

2(-) = r

2 F+ 4r

2 F

+ 4(-) = F+ 4F

F= - 35 I

F = - 0600 rads






TLx = 0

TLy = TL

TRx = 0

TRy = TR

wx = 0

wy = - m g

Fx = 0

Fy = 0


TL + TR = (100 kg)(981 ms2) = 981 N

TL= 0

TR= TR (125 m)

w= - (100 kg)(981 ms2)(100 m) = - 981 N-m

= 0

TR (125 m) ndash 981 N-m = 0

TR = 785 N

TL = 196 N


3 revmin during a

120 s interval


the interval


3 rads


3 rads

= + t

= (- ) t = 131 rads2


=

+ 2 ()

() = (-

) (2 = (469 rad)(1 rev2 rad) = 746 rev



I = frac12 m R2

I = frac12 (100 kg) (00500 m)2 = 00125 kg-m

2


2)(0314 rads)

2 = 616 J


2)(0471 rads)

2 = 1387 J

K = 771 J





= (119 N) (0350 m) sin(0o) = 0

= (146 N) (0350 m) sin(400o) = -328 N-m

= (850 N) (0350 m) sin(90o) = 298 N-m

= -0310 N-m


I = frac12 m R2

I = frac12 (500 kg) (0350 m)2 = 0306 kg-m

2


= I

I = -101 rads2


Clockwise


NAME____SOLUTION_________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

23

8 23 9

17

10

17

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




(4)


h All of the above


a Doubled

b Halved

(4)





a Potential energy

b Acceleration

(4)

c Kinetic energy

d Frequency


a ΣFx = 0

b ΣFx = 0 ΣΤ = 0

(4)

c ΣΤ = 0

d ΣFx = 0 ΣFy = 0


a Mass

b Moment of inertia

(4)

c Torque

d Angular velocity




(4)







Ii = 920 kgm2

If = 920 kgm2 + (750 kg)(300 m)

2 = 1595 kgm

2

i rads

Li = Lf

Ii i If f

f = Ii i If

f = (920 kgm2) 200 rads 1595 kgm

2) = 115 rads



2) (200 rads)

2 = 1840 J


2) (115 rads)

2 = 1055 J







(w)x = w cos(- 90o) = 0

(w)y = w sin(- 90o) = - 245 N

(FV)x = FV cos(90o) = 0




(T)x = T cos(145o)

(T)y = T sin(145o)

Fx = 0 FH + T cos(145o) = 0

Fy = 0 -245 N + FV + T sin(145o) = 0

w = - (245 N)(170 m) = - 4165 Nm

T= + (T)(1350 m) sin(35o)

FH = 0

FV = 0

= 0

-4165 Nm + (T)(1350 m) sin(35o) = 0 T = 538 N

FH = 441 N

FV = -636 N



TN2 = (4

2GMS )RN

3

TE2 = (4

2GMS )RE

3

TN2TE

2 = RN

3RE

3

TNTE = (RN3RE

3)

12

TNTE = ((450 x 109)3(150 x 10

8)

3)12

= 164

TN = 164 years


V = 2RT

V = 2(450 x 1012

m)(164x365x24x3600 s) = 5467 ms




Vmax = (km) A2

frac12

Vmax = [(120 Nm)(0250 kg)](0150 m)2

frac12 = 103 ms


x= 0


V = (km)(A2 ndash x

2)

frac12

V = [(120 Nm)(0250 kg)][(0150 m)2 - (00750 m)

2 ]

frac12 = 0900 ms


f = (12 (km)frac12 = (12 (120 Nm)(0250 kg)

frac12 = 110 Hz


a = Fm

F = k x

a = k xm

a = (120 Nm)(00750 m)(0250 kg) = 360 ms2


NAME_____SOLUTION________________________













shown



7 Start 1030 am

Stop 1145 am

PROBLEM

POINTS

CREDIT

1-5

20

6

20

7

20

8

20

9

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




b It increases

(4)

c It decreases




a 0857 rads2

b 117 rads2

(4)

c 300 rads2

d 350 rads2



a 110 ms2

b 087 ms2

(4)

c 055 ms2

d 023 ms2



a Nothing

b The Earth

(4)




length and time to

a ML2T

-2

b ML

(4)

c ML2

d ML-1

T-2



the scaffold

(20)





wpx = wp cos (-90o) = 0


2) = - 589 N

wsx = ws cos (-90o) = 0


2) = - 245 N



2) = - 392 N

Fx = 0 =gt 0 = 0


TR + TL= 873 N


TL = 0

TR = TR (400 m)

wp = - (589 N) (200 m) = - 1178 N-m

sp = - (245 N) (200 m) = - 490 N-m

wp = - (392 N) (100 m) = - 392 N-m

= 0


TR = 427 N

TL= 873 N ndash 417 N = 446 N






2 = 0400 kg m

2


2 = frac12 (500 kg)(0400 m)

2 + (100 kg)(0400 m)

2 = 0560 kg m

2

Ii i = If f





F = G m1 m2r2

F = G Ms MER2 = (6673 x 10

-11 Nm

2kg

2) (5500 kg) (600 x 10

24 kg )(726 x 10

6 m)

2 = 41779

N


T2 = (4

2GME )R

3

T2 GME = 4

2 R

3

R3 = T

2 GME (4

2)

R = (T2 GME (4

2))

13

R = ((6200 s)2 (6673 x 10

-11 Nm

2kg

2)(600 x 10

24 kg) (4

2))

13 = 726 x 10

6 m






= 0 + t

= ( - 0) t

= (175 rads) (100 s) = 0175 rads2



2 + m R

2 = frac12 (760 kg)(250 m)

2 + 2 (250 kg)(250 m)

2 = 2688 kg m

2


= I

= (2688 kgm2)(0175 rads

2) = 470 N-m


= F R

F = R = (470 N-m)(250 m) = 188 N


NAME______SOLUTION_______________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



a Child A

b Child B

(4)




a Mass

b Moment of inertia

(4)

c Angular velocity




a 14

b 12

(4)

c 2

d 4


a To your left

b To your right

(4)

c Forward

d Backward




(4)





a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)











2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55


I = (12)(1100)2) F

I = (120000) F

F = (20000) I


rads

F = 484 x 10-2

rads



K = frac12 I 2

KI = frac12 (25 M0 R02I

2

KF = frac12 (25 (M02)(R0100)2 F

2

KF KI = [frac12 (25 (M02)(R0100)2 ) F

2] [frac12 (25 (M0)(R0)

2 ) I

2]

KF KI = [(12)(1100)2 ) F

2] [I

2] = (120000)(F

2 I

2) = 20000





TO2 = ((4

2)(GMS))RO

3

TO2 = ((4

2)(6673 x 10

-11 N-m

2kg

2)(570 x 10

26 kg)))(170000000 m)

3 = 506 x 10

9 s

2

TO = 711 x 104 s = 198 hours

TI2 = ((4

2)(GMS))RI

3

TI2 = ((4

2)(( 6673 x 10

-11 N-m

2kg

2)(570 x 10

26 kg)))(73000000 m)

3 = 400 x 10

8 s

2

TI = 200 x 104 s = 556 hours


NAME_______SOLUTION______________________













correct work shown



7 Start 600 pm

Stop 715 pm

PROBLEM

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1-6

20

7

20

8

20

9

20

10

20

TOTAL 100

PERCENTAGE



PARTIAL CREDIT



a 0330

b 100

(4)

c 300

d 900


a ΣFx = 0

b ΣFx = 0 ΣFy = 0 Σ = 0

(4)

c Σ = 0

d ΣFx = 0 ΣFy = 0






(4)







a 10 rads2

b 060 rads2

(4)

c 040 rads2

d 080 rads2



a Is maximum


(4)

c Is zero





(4)


d All of the above





I = frac12 MR2 = frac12 (480 kg)(00710 m)

2 = 00121 kgm

2


I = (120 N-m)(00121 kgm2) = 992 rads

2


2 =

2 + 2 )

= 10300 rpm = 1079 rads

= 0

)= - 2 (2 ) = (1079 rads)

2(-2 x 992 rads

2) = 5868 rad = 934 rev





Ii = frac12 MR2

If = frac12 MR2 + (112) ML

2 = frac12 MR

2 + (112) M(2R)

2 = frac12 MR

2 + (13) MR

2 = (56) MR

2

Iii = Iff

f = Iii If

f = (240 revs) (frac12 MR2) (

56 MR

2) = 144 revs




E = frac12 kA2 = frac12 (300 Nm) (0200 m)

2 = 600 J



2

m Vmax2 = kA

2

Vmax2 = (km)A

2



T = 2 (mk)frac12

T = 2 ((0150 kg)(300 Nm))frac12

T = 0140 s

F = 1T = 711 Hz



2))

frac12

V = +- ((300 Nm)(0150 kg)((0200 m)2 ndash (0100 m)

2))

frac12 = +- 775 ms

V = 775 ms







w1x = w1 cos (-90o) = 0

w1y = w1 sin (-90o) = - m1 g = - (100 kg)(981 ms

2) = - 981 N

w2x = w2 cos (-90o) = 0

w2y = w2 sin (-90o) = - m2 g = - (100 kg)(981 ms

2) = - 981 N

Fx = 0 =gt 0 = 0


NA + NB= 196 N


NA = 0

NB = NB (300 m)

w1 = - (981 N) (0750 m) = - 736 N-m

sp = - (981 N) (250 m) = - 245 N-m

= 0


NB = 106 N

NA= 196 N ndash 106 N = 900 N


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

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1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT


momentum




(A) Down the slope

(B) Up the slope

(D) Into the page

(C) Out of the page

(E) Clockwise



(A) 0 rad

(B) 0700 rad

(C) 140 rad

(D) 280 rad

(E) 560 rad




(A) ML

(B) ML

2

(A) A

(B) B


kinetic energy


kinetic energy

(C) ML3

(D) ML

2

(E) LT

2



(A) = F r sin

(B) = F r cos

(C) = F r tan

(D) = F r cot

(E) = F r





LI = LF

LI = (12 m r2 ) + (12 m (2r)

2 ) (-)

LF = (12 m r2 ) F+ (12 m (2r)

2 ) F

(12 m r2 ) + (12 m (2r)

2 ) (-) = (12 m r

2 ) F+ (12 m (2r)

2 ) F

r2 + (2r)

2(-) = r

2 F+ (2r)

2 F

r2 + 4r

2(-) = r

2 F+ 4r

2 F

+ 4(-) = F+ 4F

F= - 35 I

F = - 0600 rads






TLx = 0

TLy = TL

TRx = 0

TRy = TR

wx = 0

wy = - m g

Fx = 0

Fy = 0


TL + TR = (100 kg)(981 ms2) = 981 N

TL= 0

TR= TR (125 m)

w= - (100 kg)(981 ms2)(100 m) = - 981 N-m

= 0

TR (125 m) ndash 981 N-m = 0

TR = 785 N

TL = 196 N


3 revmin during a

120 s interval


the interval


3 rads


3 rads

= + t

= (- ) t = 131 rads2


=

+ 2 ()

() = (-

) (2 = (469 rad)(1 rev2 rad) = 746 rev



I = frac12 m R2

I = frac12 (100 kg) (00500 m)2 = 00125 kg-m

2


2)(0314 rads)

2 = 616 J


2)(0471 rads)

2 = 1387 J

K = 771 J





= (119 N) (0350 m) sin(0o) = 0

= (146 N) (0350 m) sin(400o) = -328 N-m

= (850 N) (0350 m) sin(90o) = 298 N-m

= -0310 N-m


I = frac12 m R2

I = frac12 (500 kg) (0350 m)2 = 0306 kg-m

2


= I

I = -101 rads2


Clockwise


NAME____SOLUTION_________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

23

8 23 9

17

10

17

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




(4)


h All of the above


a Doubled

b Halved

(4)





a Potential energy

b Acceleration

(4)

c Kinetic energy

d Frequency


a ΣFx = 0

b ΣFx = 0 ΣΤ = 0

(4)

c ΣΤ = 0

d ΣFx = 0 ΣFy = 0


a Mass

b Moment of inertia

(4)

c Torque

d Angular velocity




(4)







Ii = 920 kgm2

If = 920 kgm2 + (750 kg)(300 m)

2 = 1595 kgm

2

i rads

Li = Lf

Ii i If f

f = Ii i If

f = (920 kgm2) 200 rads 1595 kgm

2) = 115 rads



2) (200 rads)

2 = 1840 J


2) (115 rads)

2 = 1055 J







(w)x = w cos(- 90o) = 0

(w)y = w sin(- 90o) = - 245 N

(FV)x = FV cos(90o) = 0




(T)x = T cos(145o)

(T)y = T sin(145o)

Fx = 0 FH + T cos(145o) = 0

Fy = 0 -245 N + FV + T sin(145o) = 0

w = - (245 N)(170 m) = - 4165 Nm

T= + (T)(1350 m) sin(35o)

FH = 0

FV = 0

= 0

-4165 Nm + (T)(1350 m) sin(35o) = 0 T = 538 N

FH = 441 N

FV = -636 N



TN2 = (4

2GMS )RN

3

TE2 = (4

2GMS )RE

3

TN2TE

2 = RN

3RE

3

TNTE = (RN3RE

3)

12

TNTE = ((450 x 109)3(150 x 10

8)

3)12

= 164

TN = 164 years


V = 2RT

V = 2(450 x 1012

m)(164x365x24x3600 s) = 5467 ms




Vmax = (km) A2

frac12

Vmax = [(120 Nm)(0250 kg)](0150 m)2

frac12 = 103 ms


x= 0


V = (km)(A2 ndash x

2)

frac12

V = [(120 Nm)(0250 kg)][(0150 m)2 - (00750 m)

2 ]

frac12 = 0900 ms


f = (12 (km)frac12 = (12 (120 Nm)(0250 kg)

frac12 = 110 Hz


a = Fm

F = k x

a = k xm

a = (120 Nm)(00750 m)(0250 kg) = 360 ms2


NAME_____SOLUTION________________________













shown



7 Start 1030 am

Stop 1145 am

PROBLEM

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1-5

20

6

20

7

20

8

20

9

20

TOTAL

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PERCENTAGE



PARTIAL CREDIT




b It increases

(4)

c It decreases




a 0857 rads2

b 117 rads2

(4)

c 300 rads2

d 350 rads2



a 110 ms2

b 087 ms2

(4)

c 055 ms2

d 023 ms2



a Nothing

b The Earth

(4)




length and time to

a ML2T

-2

b ML

(4)

c ML2

d ML-1

T-2



the scaffold

(20)





wpx = wp cos (-90o) = 0


2) = - 589 N

wsx = ws cos (-90o) = 0


2) = - 245 N



2) = - 392 N

Fx = 0 =gt 0 = 0


TR + TL= 873 N


TL = 0

TR = TR (400 m)

wp = - (589 N) (200 m) = - 1178 N-m

sp = - (245 N) (200 m) = - 490 N-m

wp = - (392 N) (100 m) = - 392 N-m

= 0


TR = 427 N

TL= 873 N ndash 417 N = 446 N






2 = 0400 kg m

2


2 = frac12 (500 kg)(0400 m)

2 + (100 kg)(0400 m)

2 = 0560 kg m

2

Ii i = If f





F = G m1 m2r2

F = G Ms MER2 = (6673 x 10

-11 Nm

2kg

2) (5500 kg) (600 x 10

24 kg )(726 x 10

6 m)

2 = 41779

N


T2 = (4

2GME )R

3

T2 GME = 4

2 R

3

R3 = T

2 GME (4

2)

R = (T2 GME (4

2))

13

R = ((6200 s)2 (6673 x 10

-11 Nm

2kg

2)(600 x 10

24 kg) (4

2))

13 = 726 x 10

6 m






= 0 + t

= ( - 0) t

= (175 rads) (100 s) = 0175 rads2



2 + m R

2 = frac12 (760 kg)(250 m)

2 + 2 (250 kg)(250 m)

2 = 2688 kg m

2


= I

= (2688 kgm2)(0175 rads

2) = 470 N-m


= F R

F = R = (470 N-m)(250 m) = 188 N


NAME______SOLUTION_______________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

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1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



a Child A

b Child B

(4)




a Mass

b Moment of inertia

(4)

c Angular velocity




a 14

b 12

(4)

c 2

d 4


a To your left

b To your right

(4)

c Forward

d Backward




(4)





a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)











2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55


TI = 200 x 104 s = 556 hours


NAME_______SOLUTION______________________













correct work shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL 100

PERCENTAGE



PARTIAL CREDIT



a 0330

b 100

(4)

c 300

d 900


a ΣFx = 0

b ΣFx = 0 ΣFy = 0 Σ = 0

(4)

c Σ = 0

d ΣFx = 0 ΣFy = 0






(4)







a 10 rads2

b 060 rads2

(4)

c 040 rads2

d 080 rads2



a Is maximum


(4)

c Is zero





(4)


d All of the above





I = frac12 MR2 = frac12 (480 kg)(00710 m)

2 = 00121 kgm

2


I = (120 N-m)(00121 kgm2) = 992 rads

2


2 =

2 + 2 )

= 10300 rpm = 1079 rads

= 0

)= - 2 (2 ) = (1079 rads)

2(-2 x 992 rads

2) = 5868 rad = 934 rev





Ii = frac12 MR2

If = frac12 MR2 + (112) ML

2 = frac12 MR

2 + (112) M(2R)

2 = frac12 MR

2 + (13) MR

2 = (56) MR

2

Iii = Iff

f = Iii If

f = (240 revs) (frac12 MR2) (

56 MR

2) = 144 revs




E = frac12 kA2 = frac12 (300 Nm) (0200 m)

2 = 600 J



2

m Vmax2 = kA

2

Vmax2 = (km)A

2



T = 2 (mk)frac12

T = 2 ((0150 kg)(300 Nm))frac12

T = 0140 s

F = 1T = 711 Hz



2))

frac12

V = +- ((300 Nm)(0150 kg)((0200 m)2 ndash (0100 m)

2))

frac12 = +- 775 ms

V = 775 ms







w1x = w1 cos (-90o) = 0

w1y = w1 sin (-90o) = - m1 g = - (100 kg)(981 ms

2) = - 981 N

w2x = w2 cos (-90o) = 0

w2y = w2 sin (-90o) = - m2 g = - (100 kg)(981 ms

2) = - 981 N

Fx = 0 =gt 0 = 0


NA + NB= 196 N


NA = 0

NB = NB (300 m)

w1 = - (981 N) (0750 m) = - 736 N-m

sp = - (981 N) (250 m) = - 245 N-m

= 0


NB = 106 N

NA= 196 N ndash 106 N = 900 N


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT


momentum




(A) Down the slope

(B) Up the slope

(D) Into the page

(C) Out of the page

(E) Clockwise



(A) 0 rad

(B) 0700 rad

(C) 140 rad

(D) 280 rad

(E) 560 rad




(A) ML

(B) ML

2

(A) A

(B) B


kinetic energy


kinetic energy

(C) ML3

(D) ML

2

(E) LT

2



(A) = F r sin

(B) = F r cos

(C) = F r tan

(D) = F r cot

(E) = F r





LI = LF

LI = (12 m r2 ) + (12 m (2r)

2 ) (-)

LF = (12 m r2 ) F+ (12 m (2r)

2 ) F

(12 m r2 ) + (12 m (2r)

2 ) (-) = (12 m r

2 ) F+ (12 m (2r)

2 ) F

r2 + (2r)

2(-) = r

2 F+ (2r)

2 F

r2 + 4r

2(-) = r

2 F+ 4r

2 F

+ 4(-) = F+ 4F

F= - 35 I

F = - 0600 rads






TLx = 0

TLy = TL

TRx = 0

TRy = TR

wx = 0

wy = - m g

Fx = 0

Fy = 0


TL + TR = (100 kg)(981 ms2) = 981 N

TL= 0

TR= TR (125 m)

w= - (100 kg)(981 ms2)(100 m) = - 981 N-m

= 0

TR (125 m) ndash 981 N-m = 0

TR = 785 N

TL = 196 N


3 revmin during a

120 s interval


the interval


3 rads


3 rads

= + t

= (- ) t = 131 rads2


=

+ 2 ()

() = (-

) (2 = (469 rad)(1 rev2 rad) = 746 rev



I = frac12 m R2

I = frac12 (100 kg) (00500 m)2 = 00125 kg-m

2


2)(0314 rads)

2 = 616 J


2)(0471 rads)

2 = 1387 J

K = 771 J





= (119 N) (0350 m) sin(0o) = 0

= (146 N) (0350 m) sin(400o) = -328 N-m

= (850 N) (0350 m) sin(90o) = 298 N-m

= -0310 N-m


I = frac12 m R2

I = frac12 (500 kg) (0350 m)2 = 0306 kg-m

2


= I

I = -101 rads2


Clockwise


NAME____SOLUTION_________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

23

8 23 9

17

10

17

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




(4)


h All of the above


a Doubled

b Halved

(4)





a Potential energy

b Acceleration

(4)

c Kinetic energy

d Frequency


a ΣFx = 0

b ΣFx = 0 ΣΤ = 0

(4)

c ΣΤ = 0

d ΣFx = 0 ΣFy = 0


a Mass

b Moment of inertia

(4)

c Torque

d Angular velocity




(4)







Ii = 920 kgm2

If = 920 kgm2 + (750 kg)(300 m)

2 = 1595 kgm

2

i rads

Li = Lf

Ii i If f

f = Ii i If

f = (920 kgm2) 200 rads 1595 kgm

2) = 115 rads



2) (200 rads)

2 = 1840 J


2) (115 rads)

2 = 1055 J







(w)x = w cos(- 90o) = 0

(w)y = w sin(- 90o) = - 245 N

(FV)x = FV cos(90o) = 0




(T)x = T cos(145o)

(T)y = T sin(145o)

Fx = 0 FH + T cos(145o) = 0

Fy = 0 -245 N + FV + T sin(145o) = 0

w = - (245 N)(170 m) = - 4165 Nm

T= + (T)(1350 m) sin(35o)

FH = 0

FV = 0

= 0

-4165 Nm + (T)(1350 m) sin(35o) = 0 T = 538 N

FH = 441 N

FV = -636 N



TN2 = (4

2GMS )RN

3

TE2 = (4

2GMS )RE

3

TN2TE

2 = RN

3RE

3

TNTE = (RN3RE

3)

12

TNTE = ((450 x 109)3(150 x 10

8)

3)12

= 164

TN = 164 years


V = 2RT

V = 2(450 x 1012

m)(164x365x24x3600 s) = 5467 ms




Vmax = (km) A2

frac12

Vmax = [(120 Nm)(0250 kg)](0150 m)2

frac12 = 103 ms


x= 0


V = (km)(A2 ndash x

2)

frac12

V = [(120 Nm)(0250 kg)][(0150 m)2 - (00750 m)

2 ]

frac12 = 0900 ms


f = (12 (km)frac12 = (12 (120 Nm)(0250 kg)

frac12 = 110 Hz


a = Fm

F = k x

a = k xm

a = (120 Nm)(00750 m)(0250 kg) = 360 ms2


NAME_____SOLUTION________________________













shown



7 Start 1030 am

Stop 1145 am

PROBLEM

POINTS

CREDIT

1-5

20

6

20

7

20

8

20

9

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




b It increases

(4)

c It decreases




a 0857 rads2

b 117 rads2

(4)

c 300 rads2

d 350 rads2



a 110 ms2

b 087 ms2

(4)

c 055 ms2

d 023 ms2



a Nothing

b The Earth

(4)




length and time to

a ML2T

-2

b ML

(4)

c ML2

d ML-1

T-2



the scaffold

(20)





wpx = wp cos (-90o) = 0


2) = - 589 N

wsx = ws cos (-90o) = 0


2) = - 245 N



2) = - 392 N

Fx = 0 =gt 0 = 0


TR + TL= 873 N


TL = 0

TR = TR (400 m)

wp = - (589 N) (200 m) = - 1178 N-m

sp = - (245 N) (200 m) = - 490 N-m

wp = - (392 N) (100 m) = - 392 N-m

= 0


TR = 427 N

TL= 873 N ndash 417 N = 446 N






2 = 0400 kg m

2


2 = frac12 (500 kg)(0400 m)

2 + (100 kg)(0400 m)

2 = 0560 kg m

2

Ii i = If f





F = G m1 m2r2

F = G Ms MER2 = (6673 x 10

-11 Nm

2kg

2) (5500 kg) (600 x 10

24 kg )(726 x 10

6 m)

2 = 41779

N


T2 = (4

2GME )R

3

T2 GME = 4

2 R

3

R3 = T

2 GME (4

2)

R = (T2 GME (4

2))

13

R = ((6200 s)2 (6673 x 10

-11 Nm

2kg

2)(600 x 10

24 kg) (4

2))

13 = 726 x 10

6 m






= 0 + t

= ( - 0) t

= (175 rads) (100 s) = 0175 rads2



2 + m R

2 = frac12 (760 kg)(250 m)

2 + 2 (250 kg)(250 m)

2 = 2688 kg m

2


= I

= (2688 kgm2)(0175 rads

2) = 470 N-m


= F R

F = R = (470 N-m)(250 m) = 188 N


NAME______SOLUTION_______________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



a Child A

b Child B

(4)




a Mass

b Moment of inertia

(4)

c Angular velocity




a 14

b 12

(4)

c 2

d 4


a To your left

b To your right

(4)

c Forward

d Backward




(4)





a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)











2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55


TOTAL 100

PERCENTAGE



PARTIAL CREDIT



a 0330

b 100

(4)

c 300

d 900


a ΣFx = 0

b ΣFx = 0 ΣFy = 0 Σ = 0

(4)

c Σ = 0

d ΣFx = 0 ΣFy = 0






(4)







a 10 rads2

b 060 rads2

(4)

c 040 rads2

d 080 rads2



a Is maximum


(4)

c Is zero





(4)


d All of the above





I = frac12 MR2 = frac12 (480 kg)(00710 m)

2 = 00121 kgm

2


I = (120 N-m)(00121 kgm2) = 992 rads

2


2 =

2 + 2 )

= 10300 rpm = 1079 rads

= 0

)= - 2 (2 ) = (1079 rads)

2(-2 x 992 rads

2) = 5868 rad = 934 rev





Ii = frac12 MR2

If = frac12 MR2 + (112) ML

2 = frac12 MR

2 + (112) M(2R)

2 = frac12 MR

2 + (13) MR

2 = (56) MR

2

Iii = Iff

f = Iii If

f = (240 revs) (frac12 MR2) (

56 MR

2) = 144 revs




E = frac12 kA2 = frac12 (300 Nm) (0200 m)

2 = 600 J



2

m Vmax2 = kA

2

Vmax2 = (km)A

2



T = 2 (mk)frac12

T = 2 ((0150 kg)(300 Nm))frac12

T = 0140 s

F = 1T = 711 Hz



2))

frac12

V = +- ((300 Nm)(0150 kg)((0200 m)2 ndash (0100 m)

2))

frac12 = +- 775 ms

V = 775 ms







w1x = w1 cos (-90o) = 0

w1y = w1 sin (-90o) = - m1 g = - (100 kg)(981 ms

2) = - 981 N

w2x = w2 cos (-90o) = 0

w2y = w2 sin (-90o) = - m2 g = - (100 kg)(981 ms

2) = - 981 N

Fx = 0 =gt 0 = 0


NA + NB= 196 N


NA = 0

NB = NB (300 m)

w1 = - (981 N) (0750 m) = - 736 N-m

sp = - (981 N) (250 m) = - 245 N-m

= 0


NB = 106 N

NA= 196 N ndash 106 N = 900 N


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT


momentum




(A) Down the slope

(B) Up the slope

(D) Into the page

(C) Out of the page

(E) Clockwise



(A) 0 rad

(B) 0700 rad

(C) 140 rad

(D) 280 rad

(E) 560 rad




(A) ML

(B) ML

2

(A) A

(B) B


kinetic energy


kinetic energy

(C) ML3

(D) ML

2

(E) LT

2



(A) = F r sin

(B) = F r cos

(C) = F r tan

(D) = F r cot

(E) = F r





LI = LF

LI = (12 m r2 ) + (12 m (2r)

2 ) (-)

LF = (12 m r2 ) F+ (12 m (2r)

2 ) F

(12 m r2 ) + (12 m (2r)

2 ) (-) = (12 m r

2 ) F+ (12 m (2r)

2 ) F

r2 + (2r)

2(-) = r

2 F+ (2r)

2 F

r2 + 4r

2(-) = r

2 F+ 4r

2 F

+ 4(-) = F+ 4F

F= - 35 I

F = - 0600 rads






TLx = 0

TLy = TL

TRx = 0

TRy = TR

wx = 0

wy = - m g

Fx = 0

Fy = 0


TL + TR = (100 kg)(981 ms2) = 981 N

TL= 0

TR= TR (125 m)

w= - (100 kg)(981 ms2)(100 m) = - 981 N-m

= 0

TR (125 m) ndash 981 N-m = 0

TR = 785 N

TL = 196 N


3 revmin during a

120 s interval


the interval


3 rads


3 rads

= + t

= (- ) t = 131 rads2


=

+ 2 ()

() = (-

) (2 = (469 rad)(1 rev2 rad) = 746 rev



I = frac12 m R2

I = frac12 (100 kg) (00500 m)2 = 00125 kg-m

2


2)(0314 rads)

2 = 616 J


2)(0471 rads)

2 = 1387 J

K = 771 J





= (119 N) (0350 m) sin(0o) = 0

= (146 N) (0350 m) sin(400o) = -328 N-m

= (850 N) (0350 m) sin(90o) = 298 N-m

= -0310 N-m


I = frac12 m R2

I = frac12 (500 kg) (0350 m)2 = 0306 kg-m

2


= I

I = -101 rads2


Clockwise


NAME____SOLUTION_________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

23

8 23 9

17

10

17

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




(4)


h All of the above


a Doubled

b Halved

(4)





a Potential energy

b Acceleration

(4)

c Kinetic energy

d Frequency


a ΣFx = 0

b ΣFx = 0 ΣΤ = 0

(4)

c ΣΤ = 0

d ΣFx = 0 ΣFy = 0


a Mass

b Moment of inertia

(4)

c Torque

d Angular velocity




(4)







Ii = 920 kgm2

If = 920 kgm2 + (750 kg)(300 m)

2 = 1595 kgm

2

i rads

Li = Lf

Ii i If f

f = Ii i If

f = (920 kgm2) 200 rads 1595 kgm

2) = 115 rads



2) (200 rads)

2 = 1840 J


2) (115 rads)

2 = 1055 J







(w)x = w cos(- 90o) = 0

(w)y = w sin(- 90o) = - 245 N

(FV)x = FV cos(90o) = 0




(T)x = T cos(145o)

(T)y = T sin(145o)

Fx = 0 FH + T cos(145o) = 0

Fy = 0 -245 N + FV + T sin(145o) = 0

w = - (245 N)(170 m) = - 4165 Nm

T= + (T)(1350 m) sin(35o)

FH = 0

FV = 0

= 0

-4165 Nm + (T)(1350 m) sin(35o) = 0 T = 538 N

FH = 441 N

FV = -636 N



TN2 = (4

2GMS )RN

3

TE2 = (4

2GMS )RE

3

TN2TE

2 = RN

3RE

3

TNTE = (RN3RE

3)

12

TNTE = ((450 x 109)3(150 x 10

8)

3)12

= 164

TN = 164 years


V = 2RT

V = 2(450 x 1012

m)(164x365x24x3600 s) = 5467 ms




Vmax = (km) A2

frac12

Vmax = [(120 Nm)(0250 kg)](0150 m)2

frac12 = 103 ms


x= 0


V = (km)(A2 ndash x

2)

frac12

V = [(120 Nm)(0250 kg)][(0150 m)2 - (00750 m)

2 ]

frac12 = 0900 ms


f = (12 (km)frac12 = (12 (120 Nm)(0250 kg)

frac12 = 110 Hz


a = Fm

F = k x

a = k xm

a = (120 Nm)(00750 m)(0250 kg) = 360 ms2


NAME_____SOLUTION________________________













shown



7 Start 1030 am

Stop 1145 am

PROBLEM

POINTS

CREDIT

1-5

20

6

20

7

20

8

20

9

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




b It increases

(4)

c It decreases




a 0857 rads2

b 117 rads2

(4)

c 300 rads2

d 350 rads2



a 110 ms2

b 087 ms2

(4)

c 055 ms2

d 023 ms2



a Nothing

b The Earth

(4)




length and time to

a ML2T

-2

b ML

(4)

c ML2

d ML-1

T-2



the scaffold

(20)





wpx = wp cos (-90o) = 0


2) = - 589 N

wsx = ws cos (-90o) = 0


2) = - 245 N



2) = - 392 N

Fx = 0 =gt 0 = 0


TR + TL= 873 N


TL = 0

TR = TR (400 m)

wp = - (589 N) (200 m) = - 1178 N-m

sp = - (245 N) (200 m) = - 490 N-m

wp = - (392 N) (100 m) = - 392 N-m

= 0


TR = 427 N

TL= 873 N ndash 417 N = 446 N






2 = 0400 kg m

2


2 = frac12 (500 kg)(0400 m)

2 + (100 kg)(0400 m)

2 = 0560 kg m

2

Ii i = If f





F = G m1 m2r2

F = G Ms MER2 = (6673 x 10

-11 Nm

2kg

2) (5500 kg) (600 x 10

24 kg )(726 x 10

6 m)

2 = 41779

N


T2 = (4

2GME )R

3

T2 GME = 4

2 R

3

R3 = T

2 GME (4

2)

R = (T2 GME (4

2))

13

R = ((6200 s)2 (6673 x 10

-11 Nm

2kg

2)(600 x 10

24 kg) (4

2))

13 = 726 x 10

6 m






= 0 + t

= ( - 0) t

= (175 rads) (100 s) = 0175 rads2



2 + m R

2 = frac12 (760 kg)(250 m)

2 + 2 (250 kg)(250 m)

2 = 2688 kg m

2


= I

= (2688 kgm2)(0175 rads

2) = 470 N-m


= F R

F = R = (470 N-m)(250 m) = 188 N


NAME______SOLUTION_______________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



a Child A

b Child B

(4)




a Mass

b Moment of inertia

(4)

c Angular velocity




a 14

b 12

(4)

c 2

d 4


a To your left

b To your right

(4)

c Forward

d Backward




(4)





a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)











2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55






a 10 rads2

b 060 rads2

(4)

c 040 rads2

d 080 rads2



a Is maximum


(4)

c Is zero





(4)


d All of the above





I = frac12 MR2 = frac12 (480 kg)(00710 m)

2 = 00121 kgm

2


I = (120 N-m)(00121 kgm2) = 992 rads

2


2 =

2 + 2 )

= 10300 rpm = 1079 rads

= 0

)= - 2 (2 ) = (1079 rads)

2(-2 x 992 rads

2) = 5868 rad = 934 rev





Ii = frac12 MR2

If = frac12 MR2 + (112) ML

2 = frac12 MR

2 + (112) M(2R)

2 = frac12 MR

2 + (13) MR

2 = (56) MR

2

Iii = Iff

f = Iii If

f = (240 revs) (frac12 MR2) (

56 MR

2) = 144 revs




E = frac12 kA2 = frac12 (300 Nm) (0200 m)

2 = 600 J



2

m Vmax2 = kA

2

Vmax2 = (km)A

2



T = 2 (mk)frac12

T = 2 ((0150 kg)(300 Nm))frac12

T = 0140 s

F = 1T = 711 Hz



2))

frac12

V = +- ((300 Nm)(0150 kg)((0200 m)2 ndash (0100 m)

2))

frac12 = +- 775 ms

V = 775 ms







w1x = w1 cos (-90o) = 0

w1y = w1 sin (-90o) = - m1 g = - (100 kg)(981 ms

2) = - 981 N

w2x = w2 cos (-90o) = 0

w2y = w2 sin (-90o) = - m2 g = - (100 kg)(981 ms

2) = - 981 N

Fx = 0 =gt 0 = 0


NA + NB= 196 N


NA = 0

NB = NB (300 m)

w1 = - (981 N) (0750 m) = - 736 N-m

sp = - (981 N) (250 m) = - 245 N-m

= 0


NB = 106 N

NA= 196 N ndash 106 N = 900 N


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT


momentum




(A) Down the slope

(B) Up the slope

(D) Into the page

(C) Out of the page

(E) Clockwise



(A) 0 rad

(B) 0700 rad

(C) 140 rad

(D) 280 rad

(E) 560 rad




(A) ML

(B) ML

2

(A) A

(B) B


kinetic energy


kinetic energy

(C) ML3

(D) ML

2

(E) LT

2



(A) = F r sin

(B) = F r cos

(C) = F r tan

(D) = F r cot

(E) = F r





LI = LF

LI = (12 m r2 ) + (12 m (2r)

2 ) (-)

LF = (12 m r2 ) F+ (12 m (2r)

2 ) F

(12 m r2 ) + (12 m (2r)

2 ) (-) = (12 m r

2 ) F+ (12 m (2r)

2 ) F

r2 + (2r)

2(-) = r

2 F+ (2r)

2 F

r2 + 4r

2(-) = r

2 F+ 4r

2 F

+ 4(-) = F+ 4F

F= - 35 I

F = - 0600 rads






TLx = 0

TLy = TL

TRx = 0

TRy = TR

wx = 0

wy = - m g

Fx = 0

Fy = 0


TL + TR = (100 kg)(981 ms2) = 981 N

TL= 0

TR= TR (125 m)

w= - (100 kg)(981 ms2)(100 m) = - 981 N-m

= 0

TR (125 m) ndash 981 N-m = 0

TR = 785 N

TL = 196 N


3 revmin during a

120 s interval


the interval


3 rads


3 rads

= + t

= (- ) t = 131 rads2


=

+ 2 ()

() = (-

) (2 = (469 rad)(1 rev2 rad) = 746 rev



I = frac12 m R2

I = frac12 (100 kg) (00500 m)2 = 00125 kg-m

2


2)(0314 rads)

2 = 616 J


2)(0471 rads)

2 = 1387 J

K = 771 J





= (119 N) (0350 m) sin(0o) = 0

= (146 N) (0350 m) sin(400o) = -328 N-m

= (850 N) (0350 m) sin(90o) = 298 N-m

= -0310 N-m


I = frac12 m R2

I = frac12 (500 kg) (0350 m)2 = 0306 kg-m

2


= I

I = -101 rads2


Clockwise


NAME____SOLUTION_________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

23

8 23 9

17

10

17

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




(4)


h All of the above


a Doubled

b Halved

(4)





a Potential energy

b Acceleration

(4)

c Kinetic energy

d Frequency


a ΣFx = 0

b ΣFx = 0 ΣΤ = 0

(4)

c ΣΤ = 0

d ΣFx = 0 ΣFy = 0


a Mass

b Moment of inertia

(4)

c Torque

d Angular velocity




(4)







Ii = 920 kgm2

If = 920 kgm2 + (750 kg)(300 m)

2 = 1595 kgm

2

i rads

Li = Lf

Ii i If f

f = Ii i If

f = (920 kgm2) 200 rads 1595 kgm

2) = 115 rads



2) (200 rads)

2 = 1840 J


2) (115 rads)

2 = 1055 J







(w)x = w cos(- 90o) = 0

(w)y = w sin(- 90o) = - 245 N

(FV)x = FV cos(90o) = 0




(T)x = T cos(145o)

(T)y = T sin(145o)

Fx = 0 FH + T cos(145o) = 0

Fy = 0 -245 N + FV + T sin(145o) = 0

w = - (245 N)(170 m) = - 4165 Nm

T= + (T)(1350 m) sin(35o)

FH = 0

FV = 0

= 0

-4165 Nm + (T)(1350 m) sin(35o) = 0 T = 538 N

FH = 441 N

FV = -636 N



TN2 = (4

2GMS )RN

3

TE2 = (4

2GMS )RE

3

TN2TE

2 = RN

3RE

3

TNTE = (RN3RE

3)

12

TNTE = ((450 x 109)3(150 x 10

8)

3)12

= 164

TN = 164 years


V = 2RT

V = 2(450 x 1012

m)(164x365x24x3600 s) = 5467 ms




Vmax = (km) A2

frac12

Vmax = [(120 Nm)(0250 kg)](0150 m)2

frac12 = 103 ms


x= 0


V = (km)(A2 ndash x

2)

frac12

V = [(120 Nm)(0250 kg)][(0150 m)2 - (00750 m)

2 ]

frac12 = 0900 ms


f = (12 (km)frac12 = (12 (120 Nm)(0250 kg)

frac12 = 110 Hz


a = Fm

F = k x

a = k xm

a = (120 Nm)(00750 m)(0250 kg) = 360 ms2


NAME_____SOLUTION________________________













shown



7 Start 1030 am

Stop 1145 am

PROBLEM

POINTS

CREDIT

1-5

20

6

20

7

20

8

20

9

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




b It increases

(4)

c It decreases




a 0857 rads2

b 117 rads2

(4)

c 300 rads2

d 350 rads2



a 110 ms2

b 087 ms2

(4)

c 055 ms2

d 023 ms2



a Nothing

b The Earth

(4)




length and time to

a ML2T

-2

b ML

(4)

c ML2

d ML-1

T-2



the scaffold

(20)





wpx = wp cos (-90o) = 0


2) = - 589 N

wsx = ws cos (-90o) = 0


2) = - 245 N



2) = - 392 N

Fx = 0 =gt 0 = 0


TR + TL= 873 N


TL = 0

TR = TR (400 m)

wp = - (589 N) (200 m) = - 1178 N-m

sp = - (245 N) (200 m) = - 490 N-m

wp = - (392 N) (100 m) = - 392 N-m

= 0


TR = 427 N

TL= 873 N ndash 417 N = 446 N






2 = 0400 kg m

2


2 = frac12 (500 kg)(0400 m)

2 + (100 kg)(0400 m)

2 = 0560 kg m

2

Ii i = If f





F = G m1 m2r2

F = G Ms MER2 = (6673 x 10

-11 Nm

2kg

2) (5500 kg) (600 x 10

24 kg )(726 x 10

6 m)

2 = 41779

N


T2 = (4

2GME )R

3

T2 GME = 4

2 R

3

R3 = T

2 GME (4

2)

R = (T2 GME (4

2))

13

R = ((6200 s)2 (6673 x 10

-11 Nm

2kg

2)(600 x 10

24 kg) (4

2))

13 = 726 x 10

6 m






= 0 + t

= ( - 0) t

= (175 rads) (100 s) = 0175 rads2



2 + m R

2 = frac12 (760 kg)(250 m)

2 + 2 (250 kg)(250 m)

2 = 2688 kg m

2


= I

= (2688 kgm2)(0175 rads

2) = 470 N-m


= F R

F = R = (470 N-m)(250 m) = 188 N


NAME______SOLUTION_______________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



a Child A

b Child B

(4)




a Mass

b Moment of inertia

(4)

c Angular velocity




a 14

b 12

(4)

c 2

d 4


a To your left

b To your right

(4)

c Forward

d Backward




(4)





a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)











2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55






I = frac12 MR2 = frac12 (480 kg)(00710 m)

2 = 00121 kgm

2


I = (120 N-m)(00121 kgm2) = 992 rads

2


2 =

2 + 2 )

= 10300 rpm = 1079 rads

= 0

)= - 2 (2 ) = (1079 rads)

2(-2 x 992 rads

2) = 5868 rad = 934 rev





Ii = frac12 MR2

If = frac12 MR2 + (112) ML

2 = frac12 MR

2 + (112) M(2R)

2 = frac12 MR

2 + (13) MR

2 = (56) MR

2

Iii = Iff

f = Iii If

f = (240 revs) (frac12 MR2) (

56 MR

2) = 144 revs




E = frac12 kA2 = frac12 (300 Nm) (0200 m)

2 = 600 J



2

m Vmax2 = kA

2

Vmax2 = (km)A

2



T = 2 (mk)frac12

T = 2 ((0150 kg)(300 Nm))frac12

T = 0140 s

F = 1T = 711 Hz



2))

frac12

V = +- ((300 Nm)(0150 kg)((0200 m)2 ndash (0100 m)

2))

frac12 = +- 775 ms

V = 775 ms







w1x = w1 cos (-90o) = 0

w1y = w1 sin (-90o) = - m1 g = - (100 kg)(981 ms

2) = - 981 N

w2x = w2 cos (-90o) = 0

w2y = w2 sin (-90o) = - m2 g = - (100 kg)(981 ms

2) = - 981 N

Fx = 0 =gt 0 = 0


NA + NB= 196 N


NA = 0

NB = NB (300 m)

w1 = - (981 N) (0750 m) = - 736 N-m

sp = - (981 N) (250 m) = - 245 N-m

= 0


NB = 106 N

NA= 196 N ndash 106 N = 900 N


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT


momentum




(A) Down the slope

(B) Up the slope

(D) Into the page

(C) Out of the page

(E) Clockwise



(A) 0 rad

(B) 0700 rad

(C) 140 rad

(D) 280 rad

(E) 560 rad




(A) ML

(B) ML

2

(A) A

(B) B


kinetic energy


kinetic energy

(C) ML3

(D) ML

2

(E) LT

2



(A) = F r sin

(B) = F r cos

(C) = F r tan

(D) = F r cot

(E) = F r





LI = LF

LI = (12 m r2 ) + (12 m (2r)

2 ) (-)

LF = (12 m r2 ) F+ (12 m (2r)

2 ) F

(12 m r2 ) + (12 m (2r)

2 ) (-) = (12 m r

2 ) F+ (12 m (2r)

2 ) F

r2 + (2r)

2(-) = r

2 F+ (2r)

2 F

r2 + 4r

2(-) = r

2 F+ 4r

2 F

+ 4(-) = F+ 4F

F= - 35 I

F = - 0600 rads






TLx = 0

TLy = TL

TRx = 0

TRy = TR

wx = 0

wy = - m g

Fx = 0

Fy = 0


TL + TR = (100 kg)(981 ms2) = 981 N

TL= 0

TR= TR (125 m)

w= - (100 kg)(981 ms2)(100 m) = - 981 N-m

= 0

TR (125 m) ndash 981 N-m = 0

TR = 785 N

TL = 196 N


3 revmin during a

120 s interval


the interval


3 rads


3 rads

= + t

= (- ) t = 131 rads2


=

+ 2 ()

() = (-

) (2 = (469 rad)(1 rev2 rad) = 746 rev



I = frac12 m R2

I = frac12 (100 kg) (00500 m)2 = 00125 kg-m

2


2)(0314 rads)

2 = 616 J


2)(0471 rads)

2 = 1387 J

K = 771 J





= (119 N) (0350 m) sin(0o) = 0

= (146 N) (0350 m) sin(400o) = -328 N-m

= (850 N) (0350 m) sin(90o) = 298 N-m

= -0310 N-m


I = frac12 m R2

I = frac12 (500 kg) (0350 m)2 = 0306 kg-m

2


= I

I = -101 rads2


Clockwise


NAME____SOLUTION_________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

23

8 23 9

17

10

17

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




(4)


h All of the above


a Doubled

b Halved

(4)





a Potential energy

b Acceleration

(4)

c Kinetic energy

d Frequency


a ΣFx = 0

b ΣFx = 0 ΣΤ = 0

(4)

c ΣΤ = 0

d ΣFx = 0 ΣFy = 0


a Mass

b Moment of inertia

(4)

c Torque

d Angular velocity




(4)







Ii = 920 kgm2

If = 920 kgm2 + (750 kg)(300 m)

2 = 1595 kgm

2

i rads

Li = Lf

Ii i If f

f = Ii i If

f = (920 kgm2) 200 rads 1595 kgm

2) = 115 rads



2) (200 rads)

2 = 1840 J


2) (115 rads)

2 = 1055 J







(w)x = w cos(- 90o) = 0

(w)y = w sin(- 90o) = - 245 N

(FV)x = FV cos(90o) = 0




(T)x = T cos(145o)

(T)y = T sin(145o)

Fx = 0 FH + T cos(145o) = 0

Fy = 0 -245 N + FV + T sin(145o) = 0

w = - (245 N)(170 m) = - 4165 Nm

T= + (T)(1350 m) sin(35o)

FH = 0

FV = 0

= 0

-4165 Nm + (T)(1350 m) sin(35o) = 0 T = 538 N

FH = 441 N

FV = -636 N



TN2 = (4

2GMS )RN

3

TE2 = (4

2GMS )RE

3

TN2TE

2 = RN

3RE

3

TNTE = (RN3RE

3)

12

TNTE = ((450 x 109)3(150 x 10

8)

3)12

= 164

TN = 164 years


V = 2RT

V = 2(450 x 1012

m)(164x365x24x3600 s) = 5467 ms




Vmax = (km) A2

frac12

Vmax = [(120 Nm)(0250 kg)](0150 m)2

frac12 = 103 ms


x= 0


V = (km)(A2 ndash x

2)

frac12

V = [(120 Nm)(0250 kg)][(0150 m)2 - (00750 m)

2 ]

frac12 = 0900 ms


f = (12 (km)frac12 = (12 (120 Nm)(0250 kg)

frac12 = 110 Hz


a = Fm

F = k x

a = k xm

a = (120 Nm)(00750 m)(0250 kg) = 360 ms2


NAME_____SOLUTION________________________













shown



7 Start 1030 am

Stop 1145 am

PROBLEM

POINTS

CREDIT

1-5

20

6

20

7

20

8

20

9

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




b It increases

(4)

c It decreases




a 0857 rads2

b 117 rads2

(4)

c 300 rads2

d 350 rads2



a 110 ms2

b 087 ms2

(4)

c 055 ms2

d 023 ms2



a Nothing

b The Earth

(4)




length and time to

a ML2T

-2

b ML

(4)

c ML2

d ML-1

T-2



the scaffold

(20)





wpx = wp cos (-90o) = 0


2) = - 589 N

wsx = ws cos (-90o) = 0


2) = - 245 N



2) = - 392 N

Fx = 0 =gt 0 = 0


TR + TL= 873 N


TL = 0

TR = TR (400 m)

wp = - (589 N) (200 m) = - 1178 N-m

sp = - (245 N) (200 m) = - 490 N-m

wp = - (392 N) (100 m) = - 392 N-m

= 0


TR = 427 N

TL= 873 N ndash 417 N = 446 N






2 = 0400 kg m

2


2 = frac12 (500 kg)(0400 m)

2 + (100 kg)(0400 m)

2 = 0560 kg m

2

Ii i = If f





F = G m1 m2r2

F = G Ms MER2 = (6673 x 10

-11 Nm

2kg

2) (5500 kg) (600 x 10

24 kg )(726 x 10

6 m)

2 = 41779

N


T2 = (4

2GME )R

3

T2 GME = 4

2 R

3

R3 = T

2 GME (4

2)

R = (T2 GME (4

2))

13

R = ((6200 s)2 (6673 x 10

-11 Nm

2kg

2)(600 x 10

24 kg) (4

2))

13 = 726 x 10

6 m






= 0 + t

= ( - 0) t

= (175 rads) (100 s) = 0175 rads2



2 + m R

2 = frac12 (760 kg)(250 m)

2 + 2 (250 kg)(250 m)

2 = 2688 kg m

2


= I

= (2688 kgm2)(0175 rads

2) = 470 N-m


= F R

F = R = (470 N-m)(250 m) = 188 N


NAME______SOLUTION_______________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



a Child A

b Child B

(4)




a Mass

b Moment of inertia

(4)

c Angular velocity




a 14

b 12

(4)

c 2

d 4


a To your left

b To your right

(4)

c Forward

d Backward




(4)





a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)











2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55


If = frac12 MR2 + (112) ML

2 = frac12 MR

2 + (112) M(2R)

2 = frac12 MR

2 + (13) MR

2 = (56) MR

2

Iii = Iff

f = Iii If

f = (240 revs) (frac12 MR2) (

56 MR

2) = 144 revs




E = frac12 kA2 = frac12 (300 Nm) (0200 m)

2 = 600 J



2

m Vmax2 = kA

2

Vmax2 = (km)A

2



T = 2 (mk)frac12

T = 2 ((0150 kg)(300 Nm))frac12

T = 0140 s

F = 1T = 711 Hz



2))

frac12

V = +- ((300 Nm)(0150 kg)((0200 m)2 ndash (0100 m)

2))

frac12 = +- 775 ms

V = 775 ms







w1x = w1 cos (-90o) = 0

w1y = w1 sin (-90o) = - m1 g = - (100 kg)(981 ms

2) = - 981 N

w2x = w2 cos (-90o) = 0

w2y = w2 sin (-90o) = - m2 g = - (100 kg)(981 ms

2) = - 981 N

Fx = 0 =gt 0 = 0


NA + NB= 196 N


NA = 0

NB = NB (300 m)

w1 = - (981 N) (0750 m) = - 736 N-m

sp = - (981 N) (250 m) = - 245 N-m

= 0


NB = 106 N

NA= 196 N ndash 106 N = 900 N


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT


momentum




(A) Down the slope

(B) Up the slope

(D) Into the page

(C) Out of the page

(E) Clockwise



(A) 0 rad

(B) 0700 rad

(C) 140 rad

(D) 280 rad

(E) 560 rad




(A) ML

(B) ML

2

(A) A

(B) B


kinetic energy


kinetic energy

(C) ML3

(D) ML

2

(E) LT

2



(A) = F r sin

(B) = F r cos

(C) = F r tan

(D) = F r cot

(E) = F r





LI = LF

LI = (12 m r2 ) + (12 m (2r)

2 ) (-)

LF = (12 m r2 ) F+ (12 m (2r)

2 ) F

(12 m r2 ) + (12 m (2r)

2 ) (-) = (12 m r

2 ) F+ (12 m (2r)

2 ) F

r2 + (2r)

2(-) = r

2 F+ (2r)

2 F

r2 + 4r

2(-) = r

2 F+ 4r

2 F

+ 4(-) = F+ 4F

F= - 35 I

F = - 0600 rads






TLx = 0

TLy = TL

TRx = 0

TRy = TR

wx = 0

wy = - m g

Fx = 0

Fy = 0


TL + TR = (100 kg)(981 ms2) = 981 N

TL= 0

TR= TR (125 m)

w= - (100 kg)(981 ms2)(100 m) = - 981 N-m

= 0

TR (125 m) ndash 981 N-m = 0

TR = 785 N

TL = 196 N


3 revmin during a

120 s interval


the interval


3 rads


3 rads

= + t

= (- ) t = 131 rads2


=

+ 2 ()

() = (-

) (2 = (469 rad)(1 rev2 rad) = 746 rev



I = frac12 m R2

I = frac12 (100 kg) (00500 m)2 = 00125 kg-m

2


2)(0314 rads)

2 = 616 J


2)(0471 rads)

2 = 1387 J

K = 771 J





= (119 N) (0350 m) sin(0o) = 0

= (146 N) (0350 m) sin(400o) = -328 N-m

= (850 N) (0350 m) sin(90o) = 298 N-m

= -0310 N-m


I = frac12 m R2

I = frac12 (500 kg) (0350 m)2 = 0306 kg-m

2


= I

I = -101 rads2


Clockwise


NAME____SOLUTION_________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

23

8 23 9

17

10

17

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




(4)


h All of the above


a Doubled

b Halved

(4)





a Potential energy

b Acceleration

(4)

c Kinetic energy

d Frequency


a ΣFx = 0

b ΣFx = 0 ΣΤ = 0

(4)

c ΣΤ = 0

d ΣFx = 0 ΣFy = 0


a Mass

b Moment of inertia

(4)

c Torque

d Angular velocity




(4)







Ii = 920 kgm2

If = 920 kgm2 + (750 kg)(300 m)

2 = 1595 kgm

2

i rads

Li = Lf

Ii i If f

f = Ii i If

f = (920 kgm2) 200 rads 1595 kgm

2) = 115 rads



2) (200 rads)

2 = 1840 J


2) (115 rads)

2 = 1055 J







(w)x = w cos(- 90o) = 0

(w)y = w sin(- 90o) = - 245 N

(FV)x = FV cos(90o) = 0




(T)x = T cos(145o)

(T)y = T sin(145o)

Fx = 0 FH + T cos(145o) = 0

Fy = 0 -245 N + FV + T sin(145o) = 0

w = - (245 N)(170 m) = - 4165 Nm

T= + (T)(1350 m) sin(35o)

FH = 0

FV = 0

= 0

-4165 Nm + (T)(1350 m) sin(35o) = 0 T = 538 N

FH = 441 N

FV = -636 N



TN2 = (4

2GMS )RN

3

TE2 = (4

2GMS )RE

3

TN2TE

2 = RN

3RE

3

TNTE = (RN3RE

3)

12

TNTE = ((450 x 109)3(150 x 10

8)

3)12

= 164

TN = 164 years


V = 2RT

V = 2(450 x 1012

m)(164x365x24x3600 s) = 5467 ms




Vmax = (km) A2

frac12

Vmax = [(120 Nm)(0250 kg)](0150 m)2

frac12 = 103 ms


x= 0


V = (km)(A2 ndash x

2)

frac12

V = [(120 Nm)(0250 kg)][(0150 m)2 - (00750 m)

2 ]

frac12 = 0900 ms


f = (12 (km)frac12 = (12 (120 Nm)(0250 kg)

frac12 = 110 Hz


a = Fm

F = k x

a = k xm

a = (120 Nm)(00750 m)(0250 kg) = 360 ms2


NAME_____SOLUTION________________________













shown



7 Start 1030 am

Stop 1145 am

PROBLEM

POINTS

CREDIT

1-5

20

6

20

7

20

8

20

9

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




b It increases

(4)

c It decreases




a 0857 rads2

b 117 rads2

(4)

c 300 rads2

d 350 rads2



a 110 ms2

b 087 ms2

(4)

c 055 ms2

d 023 ms2



a Nothing

b The Earth

(4)




length and time to

a ML2T

-2

b ML

(4)

c ML2

d ML-1

T-2



the scaffold

(20)





wpx = wp cos (-90o) = 0


2) = - 589 N

wsx = ws cos (-90o) = 0


2) = - 245 N



2) = - 392 N

Fx = 0 =gt 0 = 0


TR + TL= 873 N


TL = 0

TR = TR (400 m)

wp = - (589 N) (200 m) = - 1178 N-m

sp = - (245 N) (200 m) = - 490 N-m

wp = - (392 N) (100 m) = - 392 N-m

= 0


TR = 427 N

TL= 873 N ndash 417 N = 446 N






2 = 0400 kg m

2


2 = frac12 (500 kg)(0400 m)

2 + (100 kg)(0400 m)

2 = 0560 kg m

2

Ii i = If f





F = G m1 m2r2

F = G Ms MER2 = (6673 x 10

-11 Nm

2kg

2) (5500 kg) (600 x 10

24 kg )(726 x 10

6 m)

2 = 41779

N


T2 = (4

2GME )R

3

T2 GME = 4

2 R

3

R3 = T

2 GME (4

2)

R = (T2 GME (4

2))

13

R = ((6200 s)2 (6673 x 10

-11 Nm

2kg

2)(600 x 10

24 kg) (4

2))

13 = 726 x 10

6 m






= 0 + t

= ( - 0) t

= (175 rads) (100 s) = 0175 rads2



2 + m R

2 = frac12 (760 kg)(250 m)

2 + 2 (250 kg)(250 m)

2 = 2688 kg m

2


= I

= (2688 kgm2)(0175 rads

2) = 470 N-m


= F R

F = R = (470 N-m)(250 m) = 188 N


NAME______SOLUTION_______________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



a Child A

b Child B

(4)




a Mass

b Moment of inertia

(4)

c Angular velocity




a 14

b 12

(4)

c 2

d 4


a To your left

b To your right

(4)

c Forward

d Backward




(4)





a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)











2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55


V = 775 ms







w1x = w1 cos (-90o) = 0

w1y = w1 sin (-90o) = - m1 g = - (100 kg)(981 ms

2) = - 981 N

w2x = w2 cos (-90o) = 0

w2y = w2 sin (-90o) = - m2 g = - (100 kg)(981 ms

2) = - 981 N

Fx = 0 =gt 0 = 0


NA + NB= 196 N


NA = 0

NB = NB (300 m)

w1 = - (981 N) (0750 m) = - 736 N-m

sp = - (981 N) (250 m) = - 245 N-m

= 0


NB = 106 N

NA= 196 N ndash 106 N = 900 N


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT


momentum




(A) Down the slope

(B) Up the slope

(D) Into the page

(C) Out of the page

(E) Clockwise



(A) 0 rad

(B) 0700 rad

(C) 140 rad

(D) 280 rad

(E) 560 rad




(A) ML

(B) ML

2

(A) A

(B) B


kinetic energy


kinetic energy

(C) ML3

(D) ML

2

(E) LT

2



(A) = F r sin

(B) = F r cos

(C) = F r tan

(D) = F r cot

(E) = F r





LI = LF

LI = (12 m r2 ) + (12 m (2r)

2 ) (-)

LF = (12 m r2 ) F+ (12 m (2r)

2 ) F

(12 m r2 ) + (12 m (2r)

2 ) (-) = (12 m r

2 ) F+ (12 m (2r)

2 ) F

r2 + (2r)

2(-) = r

2 F+ (2r)

2 F

r2 + 4r

2(-) = r

2 F+ 4r

2 F

+ 4(-) = F+ 4F

F= - 35 I

F = - 0600 rads






TLx = 0

TLy = TL

TRx = 0

TRy = TR

wx = 0

wy = - m g

Fx = 0

Fy = 0


TL + TR = (100 kg)(981 ms2) = 981 N

TL= 0

TR= TR (125 m)

w= - (100 kg)(981 ms2)(100 m) = - 981 N-m

= 0

TR (125 m) ndash 981 N-m = 0

TR = 785 N

TL = 196 N


3 revmin during a

120 s interval


the interval


3 rads


3 rads

= + t

= (- ) t = 131 rads2


=

+ 2 ()

() = (-

) (2 = (469 rad)(1 rev2 rad) = 746 rev



I = frac12 m R2

I = frac12 (100 kg) (00500 m)2 = 00125 kg-m

2


2)(0314 rads)

2 = 616 J


2)(0471 rads)

2 = 1387 J

K = 771 J





= (119 N) (0350 m) sin(0o) = 0

= (146 N) (0350 m) sin(400o) = -328 N-m

= (850 N) (0350 m) sin(90o) = 298 N-m

= -0310 N-m


I = frac12 m R2

I = frac12 (500 kg) (0350 m)2 = 0306 kg-m

2


= I

I = -101 rads2


Clockwise


NAME____SOLUTION_________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

23

8 23 9

17

10

17

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




(4)


h All of the above


a Doubled

b Halved

(4)





a Potential energy

b Acceleration

(4)

c Kinetic energy

d Frequency


a ΣFx = 0

b ΣFx = 0 ΣΤ = 0

(4)

c ΣΤ = 0

d ΣFx = 0 ΣFy = 0


a Mass

b Moment of inertia

(4)

c Torque

d Angular velocity




(4)







Ii = 920 kgm2

If = 920 kgm2 + (750 kg)(300 m)

2 = 1595 kgm

2

i rads

Li = Lf

Ii i If f

f = Ii i If

f = (920 kgm2) 200 rads 1595 kgm

2) = 115 rads



2) (200 rads)

2 = 1840 J


2) (115 rads)

2 = 1055 J







(w)x = w cos(- 90o) = 0

(w)y = w sin(- 90o) = - 245 N

(FV)x = FV cos(90o) = 0




(T)x = T cos(145o)

(T)y = T sin(145o)

Fx = 0 FH + T cos(145o) = 0

Fy = 0 -245 N + FV + T sin(145o) = 0

w = - (245 N)(170 m) = - 4165 Nm

T= + (T)(1350 m) sin(35o)

FH = 0

FV = 0

= 0

-4165 Nm + (T)(1350 m) sin(35o) = 0 T = 538 N

FH = 441 N

FV = -636 N



TN2 = (4

2GMS )RN

3

TE2 = (4

2GMS )RE

3

TN2TE

2 = RN

3RE

3

TNTE = (RN3RE

3)

12

TNTE = ((450 x 109)3(150 x 10

8)

3)12

= 164

TN = 164 years


V = 2RT

V = 2(450 x 1012

m)(164x365x24x3600 s) = 5467 ms




Vmax = (km) A2

frac12

Vmax = [(120 Nm)(0250 kg)](0150 m)2

frac12 = 103 ms


x= 0


V = (km)(A2 ndash x

2)

frac12

V = [(120 Nm)(0250 kg)][(0150 m)2 - (00750 m)

2 ]

frac12 = 0900 ms


f = (12 (km)frac12 = (12 (120 Nm)(0250 kg)

frac12 = 110 Hz


a = Fm

F = k x

a = k xm

a = (120 Nm)(00750 m)(0250 kg) = 360 ms2


NAME_____SOLUTION________________________













shown



7 Start 1030 am

Stop 1145 am

PROBLEM

POINTS

CREDIT

1-5

20

6

20

7

20

8

20

9

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




b It increases

(4)

c It decreases




a 0857 rads2

b 117 rads2

(4)

c 300 rads2

d 350 rads2



a 110 ms2

b 087 ms2

(4)

c 055 ms2

d 023 ms2



a Nothing

b The Earth

(4)




length and time to

a ML2T

-2

b ML

(4)

c ML2

d ML-1

T-2



the scaffold

(20)





wpx = wp cos (-90o) = 0


2) = - 589 N

wsx = ws cos (-90o) = 0


2) = - 245 N



2) = - 392 N

Fx = 0 =gt 0 = 0


TR + TL= 873 N


TL = 0

TR = TR (400 m)

wp = - (589 N) (200 m) = - 1178 N-m

sp = - (245 N) (200 m) = - 490 N-m

wp = - (392 N) (100 m) = - 392 N-m

= 0


TR = 427 N

TL= 873 N ndash 417 N = 446 N






2 = 0400 kg m

2


2 = frac12 (500 kg)(0400 m)

2 + (100 kg)(0400 m)

2 = 0560 kg m

2

Ii i = If f





F = G m1 m2r2

F = G Ms MER2 = (6673 x 10

-11 Nm

2kg

2) (5500 kg) (600 x 10

24 kg )(726 x 10

6 m)

2 = 41779

N


T2 = (4

2GME )R

3

T2 GME = 4

2 R

3

R3 = T

2 GME (4

2)

R = (T2 GME (4

2))

13

R = ((6200 s)2 (6673 x 10

-11 Nm

2kg

2)(600 x 10

24 kg) (4

2))

13 = 726 x 10

6 m






= 0 + t

= ( - 0) t

= (175 rads) (100 s) = 0175 rads2



2 + m R

2 = frac12 (760 kg)(250 m)

2 + 2 (250 kg)(250 m)

2 = 2688 kg m

2


= I

= (2688 kgm2)(0175 rads

2) = 470 N-m


= F R

F = R = (470 N-m)(250 m) = 188 N


NAME______SOLUTION_______________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



a Child A

b Child B

(4)




a Mass

b Moment of inertia

(4)

c Angular velocity




a 14

b 12

(4)

c 2

d 4


a To your left

b To your right

(4)

c Forward

d Backward




(4)





a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)











2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55


NA = 0

NB = NB (300 m)

w1 = - (981 N) (0750 m) = - 736 N-m

sp = - (981 N) (250 m) = - 245 N-m

= 0


NB = 106 N

NA= 196 N ndash 106 N = 900 N


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT


momentum




(A) Down the slope

(B) Up the slope

(D) Into the page

(C) Out of the page

(E) Clockwise



(A) 0 rad

(B) 0700 rad

(C) 140 rad

(D) 280 rad

(E) 560 rad




(A) ML

(B) ML

2

(A) A

(B) B


kinetic energy


kinetic energy

(C) ML3

(D) ML

2

(E) LT

2



(A) = F r sin

(B) = F r cos

(C) = F r tan

(D) = F r cot

(E) = F r





LI = LF

LI = (12 m r2 ) + (12 m (2r)

2 ) (-)

LF = (12 m r2 ) F+ (12 m (2r)

2 ) F

(12 m r2 ) + (12 m (2r)

2 ) (-) = (12 m r

2 ) F+ (12 m (2r)

2 ) F

r2 + (2r)

2(-) = r

2 F+ (2r)

2 F

r2 + 4r

2(-) = r

2 F+ 4r

2 F

+ 4(-) = F+ 4F

F= - 35 I

F = - 0600 rads






TLx = 0

TLy = TL

TRx = 0

TRy = TR

wx = 0

wy = - m g

Fx = 0

Fy = 0


TL + TR = (100 kg)(981 ms2) = 981 N

TL= 0

TR= TR (125 m)

w= - (100 kg)(981 ms2)(100 m) = - 981 N-m

= 0

TR (125 m) ndash 981 N-m = 0

TR = 785 N

TL = 196 N


3 revmin during a

120 s interval


the interval


3 rads


3 rads

= + t

= (- ) t = 131 rads2


=

+ 2 ()

() = (-

) (2 = (469 rad)(1 rev2 rad) = 746 rev



I = frac12 m R2

I = frac12 (100 kg) (00500 m)2 = 00125 kg-m

2


2)(0314 rads)

2 = 616 J


2)(0471 rads)

2 = 1387 J

K = 771 J





= (119 N) (0350 m) sin(0o) = 0

= (146 N) (0350 m) sin(400o) = -328 N-m

= (850 N) (0350 m) sin(90o) = 298 N-m

= -0310 N-m


I = frac12 m R2

I = frac12 (500 kg) (0350 m)2 = 0306 kg-m

2


= I

I = -101 rads2


Clockwise


NAME____SOLUTION_________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

23

8 23 9

17

10

17

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




(4)


h All of the above


a Doubled

b Halved

(4)





a Potential energy

b Acceleration

(4)

c Kinetic energy

d Frequency


a ΣFx = 0

b ΣFx = 0 ΣΤ = 0

(4)

c ΣΤ = 0

d ΣFx = 0 ΣFy = 0


a Mass

b Moment of inertia

(4)

c Torque

d Angular velocity




(4)







Ii = 920 kgm2

If = 920 kgm2 + (750 kg)(300 m)

2 = 1595 kgm

2

i rads

Li = Lf

Ii i If f

f = Ii i If

f = (920 kgm2) 200 rads 1595 kgm

2) = 115 rads



2) (200 rads)

2 = 1840 J


2) (115 rads)

2 = 1055 J







(w)x = w cos(- 90o) = 0

(w)y = w sin(- 90o) = - 245 N

(FV)x = FV cos(90o) = 0




(T)x = T cos(145o)

(T)y = T sin(145o)

Fx = 0 FH + T cos(145o) = 0

Fy = 0 -245 N + FV + T sin(145o) = 0

w = - (245 N)(170 m) = - 4165 Nm

T= + (T)(1350 m) sin(35o)

FH = 0

FV = 0

= 0

-4165 Nm + (T)(1350 m) sin(35o) = 0 T = 538 N

FH = 441 N

FV = -636 N



TN2 = (4

2GMS )RN

3

TE2 = (4

2GMS )RE

3

TN2TE

2 = RN

3RE

3

TNTE = (RN3RE

3)

12

TNTE = ((450 x 109)3(150 x 10

8)

3)12

= 164

TN = 164 years


V = 2RT

V = 2(450 x 1012

m)(164x365x24x3600 s) = 5467 ms




Vmax = (km) A2

frac12

Vmax = [(120 Nm)(0250 kg)](0150 m)2

frac12 = 103 ms


x= 0


V = (km)(A2 ndash x

2)

frac12

V = [(120 Nm)(0250 kg)][(0150 m)2 - (00750 m)

2 ]

frac12 = 0900 ms


f = (12 (km)frac12 = (12 (120 Nm)(0250 kg)

frac12 = 110 Hz


a = Fm

F = k x

a = k xm

a = (120 Nm)(00750 m)(0250 kg) = 360 ms2


NAME_____SOLUTION________________________













shown



7 Start 1030 am

Stop 1145 am

PROBLEM

POINTS

CREDIT

1-5

20

6

20

7

20

8

20

9

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




b It increases

(4)

c It decreases




a 0857 rads2

b 117 rads2

(4)

c 300 rads2

d 350 rads2



a 110 ms2

b 087 ms2

(4)

c 055 ms2

d 023 ms2



a Nothing

b The Earth

(4)




length and time to

a ML2T

-2

b ML

(4)

c ML2

d ML-1

T-2



the scaffold

(20)





wpx = wp cos (-90o) = 0


2) = - 589 N

wsx = ws cos (-90o) = 0


2) = - 245 N



2) = - 392 N

Fx = 0 =gt 0 = 0


TR + TL= 873 N


TL = 0

TR = TR (400 m)

wp = - (589 N) (200 m) = - 1178 N-m

sp = - (245 N) (200 m) = - 490 N-m

wp = - (392 N) (100 m) = - 392 N-m

= 0


TR = 427 N

TL= 873 N ndash 417 N = 446 N






2 = 0400 kg m

2


2 = frac12 (500 kg)(0400 m)

2 + (100 kg)(0400 m)

2 = 0560 kg m

2

Ii i = If f





F = G m1 m2r2

F = G Ms MER2 = (6673 x 10

-11 Nm

2kg

2) (5500 kg) (600 x 10

24 kg )(726 x 10

6 m)

2 = 41779

N


T2 = (4

2GME )R

3

T2 GME = 4

2 R

3

R3 = T

2 GME (4

2)

R = (T2 GME (4

2))

13

R = ((6200 s)2 (6673 x 10

-11 Nm

2kg

2)(600 x 10

24 kg) (4

2))

13 = 726 x 10

6 m






= 0 + t

= ( - 0) t

= (175 rads) (100 s) = 0175 rads2



2 + m R

2 = frac12 (760 kg)(250 m)

2 + 2 (250 kg)(250 m)

2 = 2688 kg m

2


= I

= (2688 kgm2)(0175 rads

2) = 470 N-m


= F R

F = R = (470 N-m)(250 m) = 188 N


NAME______SOLUTION_______________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



a Child A

b Child B

(4)




a Mass

b Moment of inertia

(4)

c Angular velocity




a 14

b 12

(4)

c 2

d 4


a To your left

b To your right

(4)

c Forward

d Backward




(4)





a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)











2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55


7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT


momentum




(A) Down the slope

(B) Up the slope

(D) Into the page

(C) Out of the page

(E) Clockwise



(A) 0 rad

(B) 0700 rad

(C) 140 rad

(D) 280 rad

(E) 560 rad




(A) ML

(B) ML

2

(A) A

(B) B


kinetic energy


kinetic energy

(C) ML3

(D) ML

2

(E) LT

2



(A) = F r sin

(B) = F r cos

(C) = F r tan

(D) = F r cot

(E) = F r





LI = LF

LI = (12 m r2 ) + (12 m (2r)

2 ) (-)

LF = (12 m r2 ) F+ (12 m (2r)

2 ) F

(12 m r2 ) + (12 m (2r)

2 ) (-) = (12 m r

2 ) F+ (12 m (2r)

2 ) F

r2 + (2r)

2(-) = r

2 F+ (2r)

2 F

r2 + 4r

2(-) = r

2 F+ 4r

2 F

+ 4(-) = F+ 4F

F= - 35 I

F = - 0600 rads






TLx = 0

TLy = TL

TRx = 0

TRy = TR

wx = 0

wy = - m g

Fx = 0

Fy = 0


TL + TR = (100 kg)(981 ms2) = 981 N

TL= 0

TR= TR (125 m)

w= - (100 kg)(981 ms2)(100 m) = - 981 N-m

= 0

TR (125 m) ndash 981 N-m = 0

TR = 785 N

TL = 196 N


3 revmin during a

120 s interval


the interval


3 rads


3 rads

= + t

= (- ) t = 131 rads2


=

+ 2 ()

() = (-

) (2 = (469 rad)(1 rev2 rad) = 746 rev



I = frac12 m R2

I = frac12 (100 kg) (00500 m)2 = 00125 kg-m

2


2)(0314 rads)

2 = 616 J


2)(0471 rads)

2 = 1387 J

K = 771 J





= (119 N) (0350 m) sin(0o) = 0

= (146 N) (0350 m) sin(400o) = -328 N-m

= (850 N) (0350 m) sin(90o) = 298 N-m

= -0310 N-m


I = frac12 m R2

I = frac12 (500 kg) (0350 m)2 = 0306 kg-m

2


= I

I = -101 rads2


Clockwise


NAME____SOLUTION_________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

23

8 23 9

17

10

17

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




(4)


h All of the above


a Doubled

b Halved

(4)





a Potential energy

b Acceleration

(4)

c Kinetic energy

d Frequency


a ΣFx = 0

b ΣFx = 0 ΣΤ = 0

(4)

c ΣΤ = 0

d ΣFx = 0 ΣFy = 0


a Mass

b Moment of inertia

(4)

c Torque

d Angular velocity




(4)







Ii = 920 kgm2

If = 920 kgm2 + (750 kg)(300 m)

2 = 1595 kgm

2

i rads

Li = Lf

Ii i If f

f = Ii i If

f = (920 kgm2) 200 rads 1595 kgm

2) = 115 rads



2) (200 rads)

2 = 1840 J


2) (115 rads)

2 = 1055 J







(w)x = w cos(- 90o) = 0

(w)y = w sin(- 90o) = - 245 N

(FV)x = FV cos(90o) = 0




(T)x = T cos(145o)

(T)y = T sin(145o)

Fx = 0 FH + T cos(145o) = 0

Fy = 0 -245 N + FV + T sin(145o) = 0

w = - (245 N)(170 m) = - 4165 Nm

T= + (T)(1350 m) sin(35o)

FH = 0

FV = 0

= 0

-4165 Nm + (T)(1350 m) sin(35o) = 0 T = 538 N

FH = 441 N

FV = -636 N



TN2 = (4

2GMS )RN

3

TE2 = (4

2GMS )RE

3

TN2TE

2 = RN

3RE

3

TNTE = (RN3RE

3)

12

TNTE = ((450 x 109)3(150 x 10

8)

3)12

= 164

TN = 164 years


V = 2RT

V = 2(450 x 1012

m)(164x365x24x3600 s) = 5467 ms




Vmax = (km) A2

frac12

Vmax = [(120 Nm)(0250 kg)](0150 m)2

frac12 = 103 ms


x= 0


V = (km)(A2 ndash x

2)

frac12

V = [(120 Nm)(0250 kg)][(0150 m)2 - (00750 m)

2 ]

frac12 = 0900 ms


f = (12 (km)frac12 = (12 (120 Nm)(0250 kg)

frac12 = 110 Hz


a = Fm

F = k x

a = k xm

a = (120 Nm)(00750 m)(0250 kg) = 360 ms2


NAME_____SOLUTION________________________













shown



7 Start 1030 am

Stop 1145 am

PROBLEM

POINTS

CREDIT

1-5

20

6

20

7

20

8

20

9

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




b It increases

(4)

c It decreases




a 0857 rads2

b 117 rads2

(4)

c 300 rads2

d 350 rads2



a 110 ms2

b 087 ms2

(4)

c 055 ms2

d 023 ms2



a Nothing

b The Earth

(4)




length and time to

a ML2T

-2

b ML

(4)

c ML2

d ML-1

T-2



the scaffold

(20)





wpx = wp cos (-90o) = 0


2) = - 589 N

wsx = ws cos (-90o) = 0


2) = - 245 N



2) = - 392 N

Fx = 0 =gt 0 = 0


TR + TL= 873 N


TL = 0

TR = TR (400 m)

wp = - (589 N) (200 m) = - 1178 N-m

sp = - (245 N) (200 m) = - 490 N-m

wp = - (392 N) (100 m) = - 392 N-m

= 0


TR = 427 N

TL= 873 N ndash 417 N = 446 N






2 = 0400 kg m

2


2 = frac12 (500 kg)(0400 m)

2 + (100 kg)(0400 m)

2 = 0560 kg m

2

Ii i = If f





F = G m1 m2r2

F = G Ms MER2 = (6673 x 10

-11 Nm

2kg

2) (5500 kg) (600 x 10

24 kg )(726 x 10

6 m)

2 = 41779

N


T2 = (4

2GME )R

3

T2 GME = 4

2 R

3

R3 = T

2 GME (4

2)

R = (T2 GME (4

2))

13

R = ((6200 s)2 (6673 x 10

-11 Nm

2kg

2)(600 x 10

24 kg) (4

2))

13 = 726 x 10

6 m






= 0 + t

= ( - 0) t

= (175 rads) (100 s) = 0175 rads2



2 + m R

2 = frac12 (760 kg)(250 m)

2 + 2 (250 kg)(250 m)

2 = 2688 kg m

2


= I

= (2688 kgm2)(0175 rads

2) = 470 N-m


= F R

F = R = (470 N-m)(250 m) = 188 N


NAME______SOLUTION_______________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



a Child A

b Child B

(4)




a Mass

b Moment of inertia

(4)

c Angular velocity




a 14

b 12

(4)

c 2

d 4


a To your left

b To your right

(4)

c Forward

d Backward




(4)





a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)











2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55




(A) 0 rad

(B) 0700 rad

(C) 140 rad

(D) 280 rad

(E) 560 rad




(A) ML

(B) ML

2

(A) A

(B) B


kinetic energy


kinetic energy

(C) ML3

(D) ML

2

(E) LT

2



(A) = F r sin

(B) = F r cos

(C) = F r tan

(D) = F r cot

(E) = F r





LI = LF

LI = (12 m r2 ) + (12 m (2r)

2 ) (-)

LF = (12 m r2 ) F+ (12 m (2r)

2 ) F

(12 m r2 ) + (12 m (2r)

2 ) (-) = (12 m r

2 ) F+ (12 m (2r)

2 ) F

r2 + (2r)

2(-) = r

2 F+ (2r)

2 F

r2 + 4r

2(-) = r

2 F+ 4r

2 F

+ 4(-) = F+ 4F

F= - 35 I

F = - 0600 rads






TLx = 0

TLy = TL

TRx = 0

TRy = TR

wx = 0

wy = - m g

Fx = 0

Fy = 0


TL + TR = (100 kg)(981 ms2) = 981 N

TL= 0

TR= TR (125 m)

w= - (100 kg)(981 ms2)(100 m) = - 981 N-m

= 0

TR (125 m) ndash 981 N-m = 0

TR = 785 N

TL = 196 N


3 revmin during a

120 s interval


the interval


3 rads


3 rads

= + t

= (- ) t = 131 rads2


=

+ 2 ()

() = (-

) (2 = (469 rad)(1 rev2 rad) = 746 rev



I = frac12 m R2

I = frac12 (100 kg) (00500 m)2 = 00125 kg-m

2


2)(0314 rads)

2 = 616 J


2)(0471 rads)

2 = 1387 J

K = 771 J





= (119 N) (0350 m) sin(0o) = 0

= (146 N) (0350 m) sin(400o) = -328 N-m

= (850 N) (0350 m) sin(90o) = 298 N-m

= -0310 N-m


I = frac12 m R2

I = frac12 (500 kg) (0350 m)2 = 0306 kg-m

2


= I

I = -101 rads2


Clockwise


NAME____SOLUTION_________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

23

8 23 9

17

10

17

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




(4)


h All of the above


a Doubled

b Halved

(4)





a Potential energy

b Acceleration

(4)

c Kinetic energy

d Frequency


a ΣFx = 0

b ΣFx = 0 ΣΤ = 0

(4)

c ΣΤ = 0

d ΣFx = 0 ΣFy = 0


a Mass

b Moment of inertia

(4)

c Torque

d Angular velocity




(4)







Ii = 920 kgm2

If = 920 kgm2 + (750 kg)(300 m)

2 = 1595 kgm

2

i rads

Li = Lf

Ii i If f

f = Ii i If

f = (920 kgm2) 200 rads 1595 kgm

2) = 115 rads



2) (200 rads)

2 = 1840 J


2) (115 rads)

2 = 1055 J







(w)x = w cos(- 90o) = 0

(w)y = w sin(- 90o) = - 245 N

(FV)x = FV cos(90o) = 0




(T)x = T cos(145o)

(T)y = T sin(145o)

Fx = 0 FH + T cos(145o) = 0

Fy = 0 -245 N + FV + T sin(145o) = 0

w = - (245 N)(170 m) = - 4165 Nm

T= + (T)(1350 m) sin(35o)

FH = 0

FV = 0

= 0

-4165 Nm + (T)(1350 m) sin(35o) = 0 T = 538 N

FH = 441 N

FV = -636 N



TN2 = (4

2GMS )RN

3

TE2 = (4

2GMS )RE

3

TN2TE

2 = RN

3RE

3

TNTE = (RN3RE

3)

12

TNTE = ((450 x 109)3(150 x 10

8)

3)12

= 164

TN = 164 years


V = 2RT

V = 2(450 x 1012

m)(164x365x24x3600 s) = 5467 ms




Vmax = (km) A2

frac12

Vmax = [(120 Nm)(0250 kg)](0150 m)2

frac12 = 103 ms


x= 0


V = (km)(A2 ndash x

2)

frac12

V = [(120 Nm)(0250 kg)][(0150 m)2 - (00750 m)

2 ]

frac12 = 0900 ms


f = (12 (km)frac12 = (12 (120 Nm)(0250 kg)

frac12 = 110 Hz


a = Fm

F = k x

a = k xm

a = (120 Nm)(00750 m)(0250 kg) = 360 ms2


NAME_____SOLUTION________________________













shown



7 Start 1030 am

Stop 1145 am

PROBLEM

POINTS

CREDIT

1-5

20

6

20

7

20

8

20

9

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




b It increases

(4)

c It decreases




a 0857 rads2

b 117 rads2

(4)

c 300 rads2

d 350 rads2



a 110 ms2

b 087 ms2

(4)

c 055 ms2

d 023 ms2



a Nothing

b The Earth

(4)




length and time to

a ML2T

-2

b ML

(4)

c ML2

d ML-1

T-2



the scaffold

(20)





wpx = wp cos (-90o) = 0


2) = - 589 N

wsx = ws cos (-90o) = 0


2) = - 245 N



2) = - 392 N

Fx = 0 =gt 0 = 0


TR + TL= 873 N


TL = 0

TR = TR (400 m)

wp = - (589 N) (200 m) = - 1178 N-m

sp = - (245 N) (200 m) = - 490 N-m

wp = - (392 N) (100 m) = - 392 N-m

= 0


TR = 427 N

TL= 873 N ndash 417 N = 446 N






2 = 0400 kg m

2


2 = frac12 (500 kg)(0400 m)

2 + (100 kg)(0400 m)

2 = 0560 kg m

2

Ii i = If f





F = G m1 m2r2

F = G Ms MER2 = (6673 x 10

-11 Nm

2kg

2) (5500 kg) (600 x 10

24 kg )(726 x 10

6 m)

2 = 41779

N


T2 = (4

2GME )R

3

T2 GME = 4

2 R

3

R3 = T

2 GME (4

2)

R = (T2 GME (4

2))

13

R = ((6200 s)2 (6673 x 10

-11 Nm

2kg

2)(600 x 10

24 kg) (4

2))

13 = 726 x 10

6 m






= 0 + t

= ( - 0) t

= (175 rads) (100 s) = 0175 rads2



2 + m R

2 = frac12 (760 kg)(250 m)

2 + 2 (250 kg)(250 m)

2 = 2688 kg m

2


= I

= (2688 kgm2)(0175 rads

2) = 470 N-m


= F R

F = R = (470 N-m)(250 m) = 188 N


NAME______SOLUTION_______________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



a Child A

b Child B

(4)




a Mass

b Moment of inertia

(4)

c Angular velocity




a 14

b 12

(4)

c 2

d 4


a To your left

b To your right

(4)

c Forward

d Backward




(4)





a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)











2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55


(C) ML3

(D) ML

2

(E) LT

2



(A) = F r sin

(B) = F r cos

(C) = F r tan

(D) = F r cot

(E) = F r





LI = LF

LI = (12 m r2 ) + (12 m (2r)

2 ) (-)

LF = (12 m r2 ) F+ (12 m (2r)

2 ) F

(12 m r2 ) + (12 m (2r)

2 ) (-) = (12 m r

2 ) F+ (12 m (2r)

2 ) F

r2 + (2r)

2(-) = r

2 F+ (2r)

2 F

r2 + 4r

2(-) = r

2 F+ 4r

2 F

+ 4(-) = F+ 4F

F= - 35 I

F = - 0600 rads






TLx = 0

TLy = TL

TRx = 0

TRy = TR

wx = 0

wy = - m g

Fx = 0

Fy = 0


TL + TR = (100 kg)(981 ms2) = 981 N

TL= 0

TR= TR (125 m)

w= - (100 kg)(981 ms2)(100 m) = - 981 N-m

= 0

TR (125 m) ndash 981 N-m = 0

TR = 785 N

TL = 196 N


3 revmin during a

120 s interval


the interval


3 rads


3 rads

= + t

= (- ) t = 131 rads2


=

+ 2 ()

() = (-

) (2 = (469 rad)(1 rev2 rad) = 746 rev



I = frac12 m R2

I = frac12 (100 kg) (00500 m)2 = 00125 kg-m

2


2)(0314 rads)

2 = 616 J


2)(0471 rads)

2 = 1387 J

K = 771 J





= (119 N) (0350 m) sin(0o) = 0

= (146 N) (0350 m) sin(400o) = -328 N-m

= (850 N) (0350 m) sin(90o) = 298 N-m

= -0310 N-m


I = frac12 m R2

I = frac12 (500 kg) (0350 m)2 = 0306 kg-m

2


= I

I = -101 rads2


Clockwise


NAME____SOLUTION_________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

23

8 23 9

17

10

17

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




(4)


h All of the above


a Doubled

b Halved

(4)





a Potential energy

b Acceleration

(4)

c Kinetic energy

d Frequency


a ΣFx = 0

b ΣFx = 0 ΣΤ = 0

(4)

c ΣΤ = 0

d ΣFx = 0 ΣFy = 0


a Mass

b Moment of inertia

(4)

c Torque

d Angular velocity




(4)







Ii = 920 kgm2

If = 920 kgm2 + (750 kg)(300 m)

2 = 1595 kgm

2

i rads

Li = Lf

Ii i If f

f = Ii i If

f = (920 kgm2) 200 rads 1595 kgm

2) = 115 rads



2) (200 rads)

2 = 1840 J


2) (115 rads)

2 = 1055 J







(w)x = w cos(- 90o) = 0

(w)y = w sin(- 90o) = - 245 N

(FV)x = FV cos(90o) = 0




(T)x = T cos(145o)

(T)y = T sin(145o)

Fx = 0 FH + T cos(145o) = 0

Fy = 0 -245 N + FV + T sin(145o) = 0

w = - (245 N)(170 m) = - 4165 Nm

T= + (T)(1350 m) sin(35o)

FH = 0

FV = 0

= 0

-4165 Nm + (T)(1350 m) sin(35o) = 0 T = 538 N

FH = 441 N

FV = -636 N



TN2 = (4

2GMS )RN

3

TE2 = (4

2GMS )RE

3

TN2TE

2 = RN

3RE

3

TNTE = (RN3RE

3)

12

TNTE = ((450 x 109)3(150 x 10

8)

3)12

= 164

TN = 164 years


V = 2RT

V = 2(450 x 1012

m)(164x365x24x3600 s) = 5467 ms




Vmax = (km) A2

frac12

Vmax = [(120 Nm)(0250 kg)](0150 m)2

frac12 = 103 ms


x= 0


V = (km)(A2 ndash x

2)

frac12

V = [(120 Nm)(0250 kg)][(0150 m)2 - (00750 m)

2 ]

frac12 = 0900 ms


f = (12 (km)frac12 = (12 (120 Nm)(0250 kg)

frac12 = 110 Hz


a = Fm

F = k x

a = k xm

a = (120 Nm)(00750 m)(0250 kg) = 360 ms2


NAME_____SOLUTION________________________













shown



7 Start 1030 am

Stop 1145 am

PROBLEM

POINTS

CREDIT

1-5

20

6

20

7

20

8

20

9

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




b It increases

(4)

c It decreases




a 0857 rads2

b 117 rads2

(4)

c 300 rads2

d 350 rads2



a 110 ms2

b 087 ms2

(4)

c 055 ms2

d 023 ms2



a Nothing

b The Earth

(4)




length and time to

a ML2T

-2

b ML

(4)

c ML2

d ML-1

T-2



the scaffold

(20)





wpx = wp cos (-90o) = 0


2) = - 589 N

wsx = ws cos (-90o) = 0


2) = - 245 N



2) = - 392 N

Fx = 0 =gt 0 = 0


TR + TL= 873 N


TL = 0

TR = TR (400 m)

wp = - (589 N) (200 m) = - 1178 N-m

sp = - (245 N) (200 m) = - 490 N-m

wp = - (392 N) (100 m) = - 392 N-m

= 0


TR = 427 N

TL= 873 N ndash 417 N = 446 N






2 = 0400 kg m

2


2 = frac12 (500 kg)(0400 m)

2 + (100 kg)(0400 m)

2 = 0560 kg m

2

Ii i = If f





F = G m1 m2r2

F = G Ms MER2 = (6673 x 10

-11 Nm

2kg

2) (5500 kg) (600 x 10

24 kg )(726 x 10

6 m)

2 = 41779

N


T2 = (4

2GME )R

3

T2 GME = 4

2 R

3

R3 = T

2 GME (4

2)

R = (T2 GME (4

2))

13

R = ((6200 s)2 (6673 x 10

-11 Nm

2kg

2)(600 x 10

24 kg) (4

2))

13 = 726 x 10

6 m






= 0 + t

= ( - 0) t

= (175 rads) (100 s) = 0175 rads2



2 + m R

2 = frac12 (760 kg)(250 m)

2 + 2 (250 kg)(250 m)

2 = 2688 kg m

2


= I

= (2688 kgm2)(0175 rads

2) = 470 N-m


= F R

F = R = (470 N-m)(250 m) = 188 N


NAME______SOLUTION_______________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



a Child A

b Child B

(4)




a Mass

b Moment of inertia

(4)

c Angular velocity




a 14

b 12

(4)

c 2

d 4


a To your left

b To your right

(4)

c Forward

d Backward




(4)





a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)











2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55


F = - 0600 rads






TLx = 0

TLy = TL

TRx = 0

TRy = TR

wx = 0

wy = - m g

Fx = 0

Fy = 0


TL + TR = (100 kg)(981 ms2) = 981 N

TL= 0

TR= TR (125 m)

w= - (100 kg)(981 ms2)(100 m) = - 981 N-m

= 0

TR (125 m) ndash 981 N-m = 0

TR = 785 N

TL = 196 N


3 revmin during a

120 s interval


the interval


3 rads


3 rads

= + t

= (- ) t = 131 rads2


=

+ 2 ()

() = (-

) (2 = (469 rad)(1 rev2 rad) = 746 rev



I = frac12 m R2

I = frac12 (100 kg) (00500 m)2 = 00125 kg-m

2


2)(0314 rads)

2 = 616 J


2)(0471 rads)

2 = 1387 J

K = 771 J





= (119 N) (0350 m) sin(0o) = 0

= (146 N) (0350 m) sin(400o) = -328 N-m

= (850 N) (0350 m) sin(90o) = 298 N-m

= -0310 N-m


I = frac12 m R2

I = frac12 (500 kg) (0350 m)2 = 0306 kg-m

2


= I

I = -101 rads2


Clockwise


NAME____SOLUTION_________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

23

8 23 9

17

10

17

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




(4)


h All of the above


a Doubled

b Halved

(4)





a Potential energy

b Acceleration

(4)

c Kinetic energy

d Frequency


a ΣFx = 0

b ΣFx = 0 ΣΤ = 0

(4)

c ΣΤ = 0

d ΣFx = 0 ΣFy = 0


a Mass

b Moment of inertia

(4)

c Torque

d Angular velocity




(4)







Ii = 920 kgm2

If = 920 kgm2 + (750 kg)(300 m)

2 = 1595 kgm

2

i rads

Li = Lf

Ii i If f

f = Ii i If

f = (920 kgm2) 200 rads 1595 kgm

2) = 115 rads



2) (200 rads)

2 = 1840 J


2) (115 rads)

2 = 1055 J







(w)x = w cos(- 90o) = 0

(w)y = w sin(- 90o) = - 245 N

(FV)x = FV cos(90o) = 0




(T)x = T cos(145o)

(T)y = T sin(145o)

Fx = 0 FH + T cos(145o) = 0

Fy = 0 -245 N + FV + T sin(145o) = 0

w = - (245 N)(170 m) = - 4165 Nm

T= + (T)(1350 m) sin(35o)

FH = 0

FV = 0

= 0

-4165 Nm + (T)(1350 m) sin(35o) = 0 T = 538 N

FH = 441 N

FV = -636 N



TN2 = (4

2GMS )RN

3

TE2 = (4

2GMS )RE

3

TN2TE

2 = RN

3RE

3

TNTE = (RN3RE

3)

12

TNTE = ((450 x 109)3(150 x 10

8)

3)12

= 164

TN = 164 years


V = 2RT

V = 2(450 x 1012

m)(164x365x24x3600 s) = 5467 ms




Vmax = (km) A2

frac12

Vmax = [(120 Nm)(0250 kg)](0150 m)2

frac12 = 103 ms


x= 0


V = (km)(A2 ndash x

2)

frac12

V = [(120 Nm)(0250 kg)][(0150 m)2 - (00750 m)

2 ]

frac12 = 0900 ms


f = (12 (km)frac12 = (12 (120 Nm)(0250 kg)

frac12 = 110 Hz


a = Fm

F = k x

a = k xm

a = (120 Nm)(00750 m)(0250 kg) = 360 ms2


NAME_____SOLUTION________________________













shown



7 Start 1030 am

Stop 1145 am

PROBLEM

POINTS

CREDIT

1-5

20

6

20

7

20

8

20

9

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




b It increases

(4)

c It decreases




a 0857 rads2

b 117 rads2

(4)

c 300 rads2

d 350 rads2



a 110 ms2

b 087 ms2

(4)

c 055 ms2

d 023 ms2



a Nothing

b The Earth

(4)




length and time to

a ML2T

-2

b ML

(4)

c ML2

d ML-1

T-2



the scaffold

(20)





wpx = wp cos (-90o) = 0


2) = - 589 N

wsx = ws cos (-90o) = 0


2) = - 245 N



2) = - 392 N

Fx = 0 =gt 0 = 0


TR + TL= 873 N


TL = 0

TR = TR (400 m)

wp = - (589 N) (200 m) = - 1178 N-m

sp = - (245 N) (200 m) = - 490 N-m

wp = - (392 N) (100 m) = - 392 N-m

= 0


TR = 427 N

TL= 873 N ndash 417 N = 446 N






2 = 0400 kg m

2


2 = frac12 (500 kg)(0400 m)

2 + (100 kg)(0400 m)

2 = 0560 kg m

2

Ii i = If f





F = G m1 m2r2

F = G Ms MER2 = (6673 x 10

-11 Nm

2kg

2) (5500 kg) (600 x 10

24 kg )(726 x 10

6 m)

2 = 41779

N


T2 = (4

2GME )R

3

T2 GME = 4

2 R

3

R3 = T

2 GME (4

2)

R = (T2 GME (4

2))

13

R = ((6200 s)2 (6673 x 10

-11 Nm

2kg

2)(600 x 10

24 kg) (4

2))

13 = 726 x 10

6 m






= 0 + t

= ( - 0) t

= (175 rads) (100 s) = 0175 rads2



2 + m R

2 = frac12 (760 kg)(250 m)

2 + 2 (250 kg)(250 m)

2 = 2688 kg m

2


= I

= (2688 kgm2)(0175 rads

2) = 470 N-m


= F R

F = R = (470 N-m)(250 m) = 188 N


NAME______SOLUTION_______________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



a Child A

b Child B

(4)




a Mass

b Moment of inertia

(4)

c Angular velocity




a 14

b 12

(4)

c 2

d 4


a To your left

b To your right

(4)

c Forward

d Backward




(4)





a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)











2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55


Fy = 0


TL + TR = (100 kg)(981 ms2) = 981 N

TL= 0

TR= TR (125 m)

w= - (100 kg)(981 ms2)(100 m) = - 981 N-m

= 0

TR (125 m) ndash 981 N-m = 0

TR = 785 N

TL = 196 N


3 revmin during a

120 s interval


the interval


3 rads


3 rads

= + t

= (- ) t = 131 rads2


=

+ 2 ()

() = (-

) (2 = (469 rad)(1 rev2 rad) = 746 rev



I = frac12 m R2

I = frac12 (100 kg) (00500 m)2 = 00125 kg-m

2


2)(0314 rads)

2 = 616 J


2)(0471 rads)

2 = 1387 J

K = 771 J





= (119 N) (0350 m) sin(0o) = 0

= (146 N) (0350 m) sin(400o) = -328 N-m

= (850 N) (0350 m) sin(90o) = 298 N-m

= -0310 N-m


I = frac12 m R2

I = frac12 (500 kg) (0350 m)2 = 0306 kg-m

2


= I

I = -101 rads2


Clockwise


NAME____SOLUTION_________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

23

8 23 9

17

10

17

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




(4)


h All of the above


a Doubled

b Halved

(4)





a Potential energy

b Acceleration

(4)

c Kinetic energy

d Frequency


a ΣFx = 0

b ΣFx = 0 ΣΤ = 0

(4)

c ΣΤ = 0

d ΣFx = 0 ΣFy = 0


a Mass

b Moment of inertia

(4)

c Torque

d Angular velocity




(4)







Ii = 920 kgm2

If = 920 kgm2 + (750 kg)(300 m)

2 = 1595 kgm

2

i rads

Li = Lf

Ii i If f

f = Ii i If

f = (920 kgm2) 200 rads 1595 kgm

2) = 115 rads



2) (200 rads)

2 = 1840 J


2) (115 rads)

2 = 1055 J







(w)x = w cos(- 90o) = 0

(w)y = w sin(- 90o) = - 245 N

(FV)x = FV cos(90o) = 0




(T)x = T cos(145o)

(T)y = T sin(145o)

Fx = 0 FH + T cos(145o) = 0

Fy = 0 -245 N + FV + T sin(145o) = 0

w = - (245 N)(170 m) = - 4165 Nm

T= + (T)(1350 m) sin(35o)

FH = 0

FV = 0

= 0

-4165 Nm + (T)(1350 m) sin(35o) = 0 T = 538 N

FH = 441 N

FV = -636 N



TN2 = (4

2GMS )RN

3

TE2 = (4

2GMS )RE

3

TN2TE

2 = RN

3RE

3

TNTE = (RN3RE

3)

12

TNTE = ((450 x 109)3(150 x 10

8)

3)12

= 164

TN = 164 years


V = 2RT

V = 2(450 x 1012

m)(164x365x24x3600 s) = 5467 ms




Vmax = (km) A2

frac12

Vmax = [(120 Nm)(0250 kg)](0150 m)2

frac12 = 103 ms


x= 0


V = (km)(A2 ndash x

2)

frac12

V = [(120 Nm)(0250 kg)][(0150 m)2 - (00750 m)

2 ]

frac12 = 0900 ms


f = (12 (km)frac12 = (12 (120 Nm)(0250 kg)

frac12 = 110 Hz


a = Fm

F = k x

a = k xm

a = (120 Nm)(00750 m)(0250 kg) = 360 ms2


NAME_____SOLUTION________________________













shown



7 Start 1030 am

Stop 1145 am

PROBLEM

POINTS

CREDIT

1-5

20

6

20

7

20

8

20

9

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




b It increases

(4)

c It decreases




a 0857 rads2

b 117 rads2

(4)

c 300 rads2

d 350 rads2



a 110 ms2

b 087 ms2

(4)

c 055 ms2

d 023 ms2



a Nothing

b The Earth

(4)




length and time to

a ML2T

-2

b ML

(4)

c ML2

d ML-1

T-2



the scaffold

(20)





wpx = wp cos (-90o) = 0


2) = - 589 N

wsx = ws cos (-90o) = 0


2) = - 245 N



2) = - 392 N

Fx = 0 =gt 0 = 0


TR + TL= 873 N


TL = 0

TR = TR (400 m)

wp = - (589 N) (200 m) = - 1178 N-m

sp = - (245 N) (200 m) = - 490 N-m

wp = - (392 N) (100 m) = - 392 N-m

= 0


TR = 427 N

TL= 873 N ndash 417 N = 446 N






2 = 0400 kg m

2


2 = frac12 (500 kg)(0400 m)

2 + (100 kg)(0400 m)

2 = 0560 kg m

2

Ii i = If f





F = G m1 m2r2

F = G Ms MER2 = (6673 x 10

-11 Nm

2kg

2) (5500 kg) (600 x 10

24 kg )(726 x 10

6 m)

2 = 41779

N


T2 = (4

2GME )R

3

T2 GME = 4

2 R

3

R3 = T

2 GME (4

2)

R = (T2 GME (4

2))

13

R = ((6200 s)2 (6673 x 10

-11 Nm

2kg

2)(600 x 10

24 kg) (4

2))

13 = 726 x 10

6 m






= 0 + t

= ( - 0) t

= (175 rads) (100 s) = 0175 rads2



2 + m R

2 = frac12 (760 kg)(250 m)

2 + 2 (250 kg)(250 m)

2 = 2688 kg m

2


= I

= (2688 kgm2)(0175 rads

2) = 470 N-m


= F R

F = R = (470 N-m)(250 m) = 188 N


NAME______SOLUTION_______________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



a Child A

b Child B

(4)




a Mass

b Moment of inertia

(4)

c Angular velocity




a 14

b 12

(4)

c 2

d 4


a To your left

b To your right

(4)

c Forward

d Backward




(4)





a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)











2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55


I = frac12 m R2

I = frac12 (100 kg) (00500 m)2 = 00125 kg-m

2


2)(0314 rads)

2 = 616 J


2)(0471 rads)

2 = 1387 J

K = 771 J





= (119 N) (0350 m) sin(0o) = 0

= (146 N) (0350 m) sin(400o) = -328 N-m

= (850 N) (0350 m) sin(90o) = 298 N-m

= -0310 N-m


I = frac12 m R2

I = frac12 (500 kg) (0350 m)2 = 0306 kg-m

2


= I

I = -101 rads2


Clockwise


NAME____SOLUTION_________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

23

8 23 9

17

10

17

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




(4)


h All of the above


a Doubled

b Halved

(4)





a Potential energy

b Acceleration

(4)

c Kinetic energy

d Frequency


a ΣFx = 0

b ΣFx = 0 ΣΤ = 0

(4)

c ΣΤ = 0

d ΣFx = 0 ΣFy = 0


a Mass

b Moment of inertia

(4)

c Torque

d Angular velocity




(4)







Ii = 920 kgm2

If = 920 kgm2 + (750 kg)(300 m)

2 = 1595 kgm

2

i rads

Li = Lf

Ii i If f

f = Ii i If

f = (920 kgm2) 200 rads 1595 kgm

2) = 115 rads



2) (200 rads)

2 = 1840 J


2) (115 rads)

2 = 1055 J







(w)x = w cos(- 90o) = 0

(w)y = w sin(- 90o) = - 245 N

(FV)x = FV cos(90o) = 0




(T)x = T cos(145o)

(T)y = T sin(145o)

Fx = 0 FH + T cos(145o) = 0

Fy = 0 -245 N + FV + T sin(145o) = 0

w = - (245 N)(170 m) = - 4165 Nm

T= + (T)(1350 m) sin(35o)

FH = 0

FV = 0

= 0

-4165 Nm + (T)(1350 m) sin(35o) = 0 T = 538 N

FH = 441 N

FV = -636 N



TN2 = (4

2GMS )RN

3

TE2 = (4

2GMS )RE

3

TN2TE

2 = RN

3RE

3

TNTE = (RN3RE

3)

12

TNTE = ((450 x 109)3(150 x 10

8)

3)12

= 164

TN = 164 years


V = 2RT

V = 2(450 x 1012

m)(164x365x24x3600 s) = 5467 ms




Vmax = (km) A2

frac12

Vmax = [(120 Nm)(0250 kg)](0150 m)2

frac12 = 103 ms


x= 0


V = (km)(A2 ndash x

2)

frac12

V = [(120 Nm)(0250 kg)][(0150 m)2 - (00750 m)

2 ]

frac12 = 0900 ms


f = (12 (km)frac12 = (12 (120 Nm)(0250 kg)

frac12 = 110 Hz


a = Fm

F = k x

a = k xm

a = (120 Nm)(00750 m)(0250 kg) = 360 ms2


NAME_____SOLUTION________________________













shown



7 Start 1030 am

Stop 1145 am

PROBLEM

POINTS

CREDIT

1-5

20

6

20

7

20

8

20

9

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




b It increases

(4)

c It decreases




a 0857 rads2

b 117 rads2

(4)

c 300 rads2

d 350 rads2



a 110 ms2

b 087 ms2

(4)

c 055 ms2

d 023 ms2



a Nothing

b The Earth

(4)




length and time to

a ML2T

-2

b ML

(4)

c ML2

d ML-1

T-2



the scaffold

(20)





wpx = wp cos (-90o) = 0


2) = - 589 N

wsx = ws cos (-90o) = 0


2) = - 245 N



2) = - 392 N

Fx = 0 =gt 0 = 0


TR + TL= 873 N


TL = 0

TR = TR (400 m)

wp = - (589 N) (200 m) = - 1178 N-m

sp = - (245 N) (200 m) = - 490 N-m

wp = - (392 N) (100 m) = - 392 N-m

= 0


TR = 427 N

TL= 873 N ndash 417 N = 446 N






2 = 0400 kg m

2


2 = frac12 (500 kg)(0400 m)

2 + (100 kg)(0400 m)

2 = 0560 kg m

2

Ii i = If f





F = G m1 m2r2

F = G Ms MER2 = (6673 x 10

-11 Nm

2kg

2) (5500 kg) (600 x 10

24 kg )(726 x 10

6 m)

2 = 41779

N


T2 = (4

2GME )R

3

T2 GME = 4

2 R

3

R3 = T

2 GME (4

2)

R = (T2 GME (4

2))

13

R = ((6200 s)2 (6673 x 10

-11 Nm

2kg

2)(600 x 10

24 kg) (4

2))

13 = 726 x 10

6 m






= 0 + t

= ( - 0) t

= (175 rads) (100 s) = 0175 rads2



2 + m R

2 = frac12 (760 kg)(250 m)

2 + 2 (250 kg)(250 m)

2 = 2688 kg m

2


= I

= (2688 kgm2)(0175 rads

2) = 470 N-m


= F R

F = R = (470 N-m)(250 m) = 188 N


NAME______SOLUTION_______________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



a Child A

b Child B

(4)




a Mass

b Moment of inertia

(4)

c Angular velocity




a 14

b 12

(4)

c 2

d 4


a To your left

b To your right

(4)

c Forward

d Backward




(4)





a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)











2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55



Clockwise


NAME____SOLUTION_________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

23

8 23 9

17

10

17

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




(4)


h All of the above


a Doubled

b Halved

(4)





a Potential energy

b Acceleration

(4)

c Kinetic energy

d Frequency


a ΣFx = 0

b ΣFx = 0 ΣΤ = 0

(4)

c ΣΤ = 0

d ΣFx = 0 ΣFy = 0


a Mass

b Moment of inertia

(4)

c Torque

d Angular velocity




(4)







Ii = 920 kgm2

If = 920 kgm2 + (750 kg)(300 m)

2 = 1595 kgm

2

i rads

Li = Lf

Ii i If f

f = Ii i If

f = (920 kgm2) 200 rads 1595 kgm

2) = 115 rads



2) (200 rads)

2 = 1840 J


2) (115 rads)

2 = 1055 J







(w)x = w cos(- 90o) = 0

(w)y = w sin(- 90o) = - 245 N

(FV)x = FV cos(90o) = 0




(T)x = T cos(145o)

(T)y = T sin(145o)

Fx = 0 FH + T cos(145o) = 0

Fy = 0 -245 N + FV + T sin(145o) = 0

w = - (245 N)(170 m) = - 4165 Nm

T= + (T)(1350 m) sin(35o)

FH = 0

FV = 0

= 0

-4165 Nm + (T)(1350 m) sin(35o) = 0 T = 538 N

FH = 441 N

FV = -636 N



TN2 = (4

2GMS )RN

3

TE2 = (4

2GMS )RE

3

TN2TE

2 = RN

3RE

3

TNTE = (RN3RE

3)

12

TNTE = ((450 x 109)3(150 x 10

8)

3)12

= 164

TN = 164 years


V = 2RT

V = 2(450 x 1012

m)(164x365x24x3600 s) = 5467 ms




Vmax = (km) A2

frac12

Vmax = [(120 Nm)(0250 kg)](0150 m)2

frac12 = 103 ms


x= 0


V = (km)(A2 ndash x

2)

frac12

V = [(120 Nm)(0250 kg)][(0150 m)2 - (00750 m)

2 ]

frac12 = 0900 ms


f = (12 (km)frac12 = (12 (120 Nm)(0250 kg)

frac12 = 110 Hz


a = Fm

F = k x

a = k xm

a = (120 Nm)(00750 m)(0250 kg) = 360 ms2


NAME_____SOLUTION________________________













shown



7 Start 1030 am

Stop 1145 am

PROBLEM

POINTS

CREDIT

1-5

20

6

20

7

20

8

20

9

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




b It increases

(4)

c It decreases




a 0857 rads2

b 117 rads2

(4)

c 300 rads2

d 350 rads2



a 110 ms2

b 087 ms2

(4)

c 055 ms2

d 023 ms2



a Nothing

b The Earth

(4)




length and time to

a ML2T

-2

b ML

(4)

c ML2

d ML-1

T-2



the scaffold

(20)





wpx = wp cos (-90o) = 0


2) = - 589 N

wsx = ws cos (-90o) = 0


2) = - 245 N



2) = - 392 N

Fx = 0 =gt 0 = 0


TR + TL= 873 N


TL = 0

TR = TR (400 m)

wp = - (589 N) (200 m) = - 1178 N-m

sp = - (245 N) (200 m) = - 490 N-m

wp = - (392 N) (100 m) = - 392 N-m

= 0


TR = 427 N

TL= 873 N ndash 417 N = 446 N






2 = 0400 kg m

2


2 = frac12 (500 kg)(0400 m)

2 + (100 kg)(0400 m)

2 = 0560 kg m

2

Ii i = If f





F = G m1 m2r2

F = G Ms MER2 = (6673 x 10

-11 Nm

2kg

2) (5500 kg) (600 x 10

24 kg )(726 x 10

6 m)

2 = 41779

N


T2 = (4

2GME )R

3

T2 GME = 4

2 R

3

R3 = T

2 GME (4

2)

R = (T2 GME (4

2))

13

R = ((6200 s)2 (6673 x 10

-11 Nm

2kg

2)(600 x 10

24 kg) (4

2))

13 = 726 x 10

6 m






= 0 + t

= ( - 0) t

= (175 rads) (100 s) = 0175 rads2



2 + m R

2 = frac12 (760 kg)(250 m)

2 + 2 (250 kg)(250 m)

2 = 2688 kg m

2


= I

= (2688 kgm2)(0175 rads

2) = 470 N-m


= F R

F = R = (470 N-m)(250 m) = 188 N


NAME______SOLUTION_______________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



a Child A

b Child B

(4)




a Mass

b Moment of inertia

(4)

c Angular velocity




a 14

b 12

(4)

c 2

d 4


a To your left

b To your right

(4)

c Forward

d Backward




(4)





a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)











2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55


8 23 9

17

10

17

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




(4)


h All of the above


a Doubled

b Halved

(4)





a Potential energy

b Acceleration

(4)

c Kinetic energy

d Frequency


a ΣFx = 0

b ΣFx = 0 ΣΤ = 0

(4)

c ΣΤ = 0

d ΣFx = 0 ΣFy = 0


a Mass

b Moment of inertia

(4)

c Torque

d Angular velocity




(4)







Ii = 920 kgm2

If = 920 kgm2 + (750 kg)(300 m)

2 = 1595 kgm

2

i rads

Li = Lf

Ii i If f

f = Ii i If

f = (920 kgm2) 200 rads 1595 kgm

2) = 115 rads



2) (200 rads)

2 = 1840 J


2) (115 rads)

2 = 1055 J







(w)x = w cos(- 90o) = 0

(w)y = w sin(- 90o) = - 245 N

(FV)x = FV cos(90o) = 0




(T)x = T cos(145o)

(T)y = T sin(145o)

Fx = 0 FH + T cos(145o) = 0

Fy = 0 -245 N + FV + T sin(145o) = 0

w = - (245 N)(170 m) = - 4165 Nm

T= + (T)(1350 m) sin(35o)

FH = 0

FV = 0

= 0

-4165 Nm + (T)(1350 m) sin(35o) = 0 T = 538 N

FH = 441 N

FV = -636 N



TN2 = (4

2GMS )RN

3

TE2 = (4

2GMS )RE

3

TN2TE

2 = RN

3RE

3

TNTE = (RN3RE

3)

12

TNTE = ((450 x 109)3(150 x 10

8)

3)12

= 164

TN = 164 years


V = 2RT

V = 2(450 x 1012

m)(164x365x24x3600 s) = 5467 ms




Vmax = (km) A2

frac12

Vmax = [(120 Nm)(0250 kg)](0150 m)2

frac12 = 103 ms


x= 0


V = (km)(A2 ndash x

2)

frac12

V = [(120 Nm)(0250 kg)][(0150 m)2 - (00750 m)

2 ]

frac12 = 0900 ms


f = (12 (km)frac12 = (12 (120 Nm)(0250 kg)

frac12 = 110 Hz


a = Fm

F = k x

a = k xm

a = (120 Nm)(00750 m)(0250 kg) = 360 ms2


NAME_____SOLUTION________________________













shown



7 Start 1030 am

Stop 1145 am

PROBLEM

POINTS

CREDIT

1-5

20

6

20

7

20

8

20

9

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




b It increases

(4)

c It decreases




a 0857 rads2

b 117 rads2

(4)

c 300 rads2

d 350 rads2



a 110 ms2

b 087 ms2

(4)

c 055 ms2

d 023 ms2



a Nothing

b The Earth

(4)




length and time to

a ML2T

-2

b ML

(4)

c ML2

d ML-1

T-2



the scaffold

(20)





wpx = wp cos (-90o) = 0


2) = - 589 N

wsx = ws cos (-90o) = 0


2) = - 245 N



2) = - 392 N

Fx = 0 =gt 0 = 0


TR + TL= 873 N


TL = 0

TR = TR (400 m)

wp = - (589 N) (200 m) = - 1178 N-m

sp = - (245 N) (200 m) = - 490 N-m

wp = - (392 N) (100 m) = - 392 N-m

= 0


TR = 427 N

TL= 873 N ndash 417 N = 446 N






2 = 0400 kg m

2


2 = frac12 (500 kg)(0400 m)

2 + (100 kg)(0400 m)

2 = 0560 kg m

2

Ii i = If f





F = G m1 m2r2

F = G Ms MER2 = (6673 x 10

-11 Nm

2kg

2) (5500 kg) (600 x 10

24 kg )(726 x 10

6 m)

2 = 41779

N


T2 = (4

2GME )R

3

T2 GME = 4

2 R

3

R3 = T

2 GME (4

2)

R = (T2 GME (4

2))

13

R = ((6200 s)2 (6673 x 10

-11 Nm

2kg

2)(600 x 10

24 kg) (4

2))

13 = 726 x 10

6 m






= 0 + t

= ( - 0) t

= (175 rads) (100 s) = 0175 rads2



2 + m R

2 = frac12 (760 kg)(250 m)

2 + 2 (250 kg)(250 m)

2 = 2688 kg m

2


= I

= (2688 kgm2)(0175 rads

2) = 470 N-m


= F R

F = R = (470 N-m)(250 m) = 188 N


NAME______SOLUTION_______________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



a Child A

b Child B

(4)




a Mass

b Moment of inertia

(4)

c Angular velocity




a 14

b 12

(4)

c 2

d 4


a To your left

b To your right

(4)

c Forward

d Backward




(4)





a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)











2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55




PARTIAL CREDIT




(4)


h All of the above


a Doubled

b Halved

(4)





a Potential energy

b Acceleration

(4)

c Kinetic energy

d Frequency


a ΣFx = 0

b ΣFx = 0 ΣΤ = 0

(4)

c ΣΤ = 0

d ΣFx = 0 ΣFy = 0


a Mass

b Moment of inertia

(4)

c Torque

d Angular velocity




(4)







Ii = 920 kgm2

If = 920 kgm2 + (750 kg)(300 m)

2 = 1595 kgm

2

i rads

Li = Lf

Ii i If f

f = Ii i If

f = (920 kgm2) 200 rads 1595 kgm

2) = 115 rads



2) (200 rads)

2 = 1840 J


2) (115 rads)

2 = 1055 J







(w)x = w cos(- 90o) = 0

(w)y = w sin(- 90o) = - 245 N

(FV)x = FV cos(90o) = 0




(T)x = T cos(145o)

(T)y = T sin(145o)

Fx = 0 FH + T cos(145o) = 0

Fy = 0 -245 N + FV + T sin(145o) = 0

w = - (245 N)(170 m) = - 4165 Nm

T= + (T)(1350 m) sin(35o)

FH = 0

FV = 0

= 0

-4165 Nm + (T)(1350 m) sin(35o) = 0 T = 538 N

FH = 441 N

FV = -636 N



TN2 = (4

2GMS )RN

3

TE2 = (4

2GMS )RE

3

TN2TE

2 = RN

3RE

3

TNTE = (RN3RE

3)

12

TNTE = ((450 x 109)3(150 x 10

8)

3)12

= 164

TN = 164 years


V = 2RT

V = 2(450 x 1012

m)(164x365x24x3600 s) = 5467 ms




Vmax = (km) A2

frac12

Vmax = [(120 Nm)(0250 kg)](0150 m)2

frac12 = 103 ms


x= 0


V = (km)(A2 ndash x

2)

frac12

V = [(120 Nm)(0250 kg)][(0150 m)2 - (00750 m)

2 ]

frac12 = 0900 ms


f = (12 (km)frac12 = (12 (120 Nm)(0250 kg)

frac12 = 110 Hz


a = Fm

F = k x

a = k xm

a = (120 Nm)(00750 m)(0250 kg) = 360 ms2


NAME_____SOLUTION________________________













shown



7 Start 1030 am

Stop 1145 am

PROBLEM

POINTS

CREDIT

1-5

20

6

20

7

20

8

20

9

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




b It increases

(4)

c It decreases




a 0857 rads2

b 117 rads2

(4)

c 300 rads2

d 350 rads2



a 110 ms2

b 087 ms2

(4)

c 055 ms2

d 023 ms2



a Nothing

b The Earth

(4)




length and time to

a ML2T

-2

b ML

(4)

c ML2

d ML-1

T-2



the scaffold

(20)





wpx = wp cos (-90o) = 0


2) = - 589 N

wsx = ws cos (-90o) = 0


2) = - 245 N



2) = - 392 N

Fx = 0 =gt 0 = 0


TR + TL= 873 N


TL = 0

TR = TR (400 m)

wp = - (589 N) (200 m) = - 1178 N-m

sp = - (245 N) (200 m) = - 490 N-m

wp = - (392 N) (100 m) = - 392 N-m

= 0


TR = 427 N

TL= 873 N ndash 417 N = 446 N






2 = 0400 kg m

2


2 = frac12 (500 kg)(0400 m)

2 + (100 kg)(0400 m)

2 = 0560 kg m

2

Ii i = If f





F = G m1 m2r2

F = G Ms MER2 = (6673 x 10

-11 Nm

2kg

2) (5500 kg) (600 x 10

24 kg )(726 x 10

6 m)

2 = 41779

N


T2 = (4

2GME )R

3

T2 GME = 4

2 R

3

R3 = T

2 GME (4

2)

R = (T2 GME (4

2))

13

R = ((6200 s)2 (6673 x 10

-11 Nm

2kg

2)(600 x 10

24 kg) (4

2))

13 = 726 x 10

6 m






= 0 + t

= ( - 0) t

= (175 rads) (100 s) = 0175 rads2



2 + m R

2 = frac12 (760 kg)(250 m)

2 + 2 (250 kg)(250 m)

2 = 2688 kg m

2


= I

= (2688 kgm2)(0175 rads

2) = 470 N-m


= F R

F = R = (470 N-m)(250 m) = 188 N


NAME______SOLUTION_______________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



a Child A

b Child B

(4)




a Mass

b Moment of inertia

(4)

c Angular velocity




a 14

b 12

(4)

c 2

d 4


a To your left

b To your right

(4)

c Forward

d Backward




(4)





a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)











2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55



a ΣFx = 0

b ΣFx = 0 ΣΤ = 0

(4)

c ΣΤ = 0

d ΣFx = 0 ΣFy = 0


a Mass

b Moment of inertia

(4)

c Torque

d Angular velocity




(4)







Ii = 920 kgm2

If = 920 kgm2 + (750 kg)(300 m)

2 = 1595 kgm

2

i rads

Li = Lf

Ii i If f

f = Ii i If

f = (920 kgm2) 200 rads 1595 kgm

2) = 115 rads



2) (200 rads)

2 = 1840 J


2) (115 rads)

2 = 1055 J







(w)x = w cos(- 90o) = 0

(w)y = w sin(- 90o) = - 245 N

(FV)x = FV cos(90o) = 0




(T)x = T cos(145o)

(T)y = T sin(145o)

Fx = 0 FH + T cos(145o) = 0

Fy = 0 -245 N + FV + T sin(145o) = 0

w = - (245 N)(170 m) = - 4165 Nm

T= + (T)(1350 m) sin(35o)

FH = 0

FV = 0

= 0

-4165 Nm + (T)(1350 m) sin(35o) = 0 T = 538 N

FH = 441 N

FV = -636 N



TN2 = (4

2GMS )RN

3

TE2 = (4

2GMS )RE

3

TN2TE

2 = RN

3RE

3

TNTE = (RN3RE

3)

12

TNTE = ((450 x 109)3(150 x 10

8)

3)12

= 164

TN = 164 years


V = 2RT

V = 2(450 x 1012

m)(164x365x24x3600 s) = 5467 ms




Vmax = (km) A2

frac12

Vmax = [(120 Nm)(0250 kg)](0150 m)2

frac12 = 103 ms


x= 0


V = (km)(A2 ndash x

2)

frac12

V = [(120 Nm)(0250 kg)][(0150 m)2 - (00750 m)

2 ]

frac12 = 0900 ms


f = (12 (km)frac12 = (12 (120 Nm)(0250 kg)

frac12 = 110 Hz


a = Fm

F = k x

a = k xm

a = (120 Nm)(00750 m)(0250 kg) = 360 ms2


NAME_____SOLUTION________________________













shown



7 Start 1030 am

Stop 1145 am

PROBLEM

POINTS

CREDIT

1-5

20

6

20

7

20

8

20

9

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




b It increases

(4)

c It decreases




a 0857 rads2

b 117 rads2

(4)

c 300 rads2

d 350 rads2



a 110 ms2

b 087 ms2

(4)

c 055 ms2

d 023 ms2



a Nothing

b The Earth

(4)




length and time to

a ML2T

-2

b ML

(4)

c ML2

d ML-1

T-2



the scaffold

(20)





wpx = wp cos (-90o) = 0


2) = - 589 N

wsx = ws cos (-90o) = 0


2) = - 245 N



2) = - 392 N

Fx = 0 =gt 0 = 0


TR + TL= 873 N


TL = 0

TR = TR (400 m)

wp = - (589 N) (200 m) = - 1178 N-m

sp = - (245 N) (200 m) = - 490 N-m

wp = - (392 N) (100 m) = - 392 N-m

= 0


TR = 427 N

TL= 873 N ndash 417 N = 446 N






2 = 0400 kg m

2


2 = frac12 (500 kg)(0400 m)

2 + (100 kg)(0400 m)

2 = 0560 kg m

2

Ii i = If f





F = G m1 m2r2

F = G Ms MER2 = (6673 x 10

-11 Nm

2kg

2) (5500 kg) (600 x 10

24 kg )(726 x 10

6 m)

2 = 41779

N


T2 = (4

2GME )R

3

T2 GME = 4

2 R

3

R3 = T

2 GME (4

2)

R = (T2 GME (4

2))

13

R = ((6200 s)2 (6673 x 10

-11 Nm

2kg

2)(600 x 10

24 kg) (4

2))

13 = 726 x 10

6 m






= 0 + t

= ( - 0) t

= (175 rads) (100 s) = 0175 rads2



2 + m R

2 = frac12 (760 kg)(250 m)

2 + 2 (250 kg)(250 m)

2 = 2688 kg m

2


= I

= (2688 kgm2)(0175 rads

2) = 470 N-m


= F R

F = R = (470 N-m)(250 m) = 188 N


NAME______SOLUTION_______________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



a Child A

b Child B

(4)




a Mass

b Moment of inertia

(4)

c Angular velocity




a 14

b 12

(4)

c 2

d 4


a To your left

b To your right

(4)

c Forward

d Backward




(4)





a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)











2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55


i rads

Li = Lf

Ii i If f

f = Ii i If

f = (920 kgm2) 200 rads 1595 kgm

2) = 115 rads



2) (200 rads)

2 = 1840 J


2) (115 rads)

2 = 1055 J







(w)x = w cos(- 90o) = 0

(w)y = w sin(- 90o) = - 245 N

(FV)x = FV cos(90o) = 0




(T)x = T cos(145o)

(T)y = T sin(145o)

Fx = 0 FH + T cos(145o) = 0

Fy = 0 -245 N + FV + T sin(145o) = 0

w = - (245 N)(170 m) = - 4165 Nm

T= + (T)(1350 m) sin(35o)

FH = 0

FV = 0

= 0

-4165 Nm + (T)(1350 m) sin(35o) = 0 T = 538 N

FH = 441 N

FV = -636 N



TN2 = (4

2GMS )RN

3

TE2 = (4

2GMS )RE

3

TN2TE

2 = RN

3RE

3

TNTE = (RN3RE

3)

12

TNTE = ((450 x 109)3(150 x 10

8)

3)12

= 164

TN = 164 years


V = 2RT

V = 2(450 x 1012

m)(164x365x24x3600 s) = 5467 ms




Vmax = (km) A2

frac12

Vmax = [(120 Nm)(0250 kg)](0150 m)2

frac12 = 103 ms


x= 0


V = (km)(A2 ndash x

2)

frac12

V = [(120 Nm)(0250 kg)][(0150 m)2 - (00750 m)

2 ]

frac12 = 0900 ms


f = (12 (km)frac12 = (12 (120 Nm)(0250 kg)

frac12 = 110 Hz


a = Fm

F = k x

a = k xm

a = (120 Nm)(00750 m)(0250 kg) = 360 ms2


NAME_____SOLUTION________________________













shown



7 Start 1030 am

Stop 1145 am

PROBLEM

POINTS

CREDIT

1-5

20

6

20

7

20

8

20

9

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




b It increases

(4)

c It decreases




a 0857 rads2

b 117 rads2

(4)

c 300 rads2

d 350 rads2



a 110 ms2

b 087 ms2

(4)

c 055 ms2

d 023 ms2



a Nothing

b The Earth

(4)




length and time to

a ML2T

-2

b ML

(4)

c ML2

d ML-1

T-2



the scaffold

(20)





wpx = wp cos (-90o) = 0


2) = - 589 N

wsx = ws cos (-90o) = 0


2) = - 245 N



2) = - 392 N

Fx = 0 =gt 0 = 0


TR + TL= 873 N


TL = 0

TR = TR (400 m)

wp = - (589 N) (200 m) = - 1178 N-m

sp = - (245 N) (200 m) = - 490 N-m

wp = - (392 N) (100 m) = - 392 N-m

= 0


TR = 427 N

TL= 873 N ndash 417 N = 446 N






2 = 0400 kg m

2


2 = frac12 (500 kg)(0400 m)

2 + (100 kg)(0400 m)

2 = 0560 kg m

2

Ii i = If f





F = G m1 m2r2

F = G Ms MER2 = (6673 x 10

-11 Nm

2kg

2) (5500 kg) (600 x 10

24 kg )(726 x 10

6 m)

2 = 41779

N


T2 = (4

2GME )R

3

T2 GME = 4

2 R

3

R3 = T

2 GME (4

2)

R = (T2 GME (4

2))

13

R = ((6200 s)2 (6673 x 10

-11 Nm

2kg

2)(600 x 10

24 kg) (4

2))

13 = 726 x 10

6 m






= 0 + t

= ( - 0) t

= (175 rads) (100 s) = 0175 rads2



2 + m R

2 = frac12 (760 kg)(250 m)

2 + 2 (250 kg)(250 m)

2 = 2688 kg m

2


= I

= (2688 kgm2)(0175 rads

2) = 470 N-m


= F R

F = R = (470 N-m)(250 m) = 188 N


NAME______SOLUTION_______________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



a Child A

b Child B

(4)




a Mass

b Moment of inertia

(4)

c Angular velocity




a 14

b 12

(4)

c 2

d 4


a To your left

b To your right

(4)

c Forward

d Backward




(4)





a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)











2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55



(T)x = T cos(145o)

(T)y = T sin(145o)

Fx = 0 FH + T cos(145o) = 0

Fy = 0 -245 N + FV + T sin(145o) = 0

w = - (245 N)(170 m) = - 4165 Nm

T= + (T)(1350 m) sin(35o)

FH = 0

FV = 0

= 0

-4165 Nm + (T)(1350 m) sin(35o) = 0 T = 538 N

FH = 441 N

FV = -636 N



TN2 = (4

2GMS )RN

3

TE2 = (4

2GMS )RE

3

TN2TE

2 = RN

3RE

3

TNTE = (RN3RE

3)

12

TNTE = ((450 x 109)3(150 x 10

8)

3)12

= 164

TN = 164 years


V = 2RT

V = 2(450 x 1012

m)(164x365x24x3600 s) = 5467 ms




Vmax = (km) A2

frac12

Vmax = [(120 Nm)(0250 kg)](0150 m)2

frac12 = 103 ms


x= 0


V = (km)(A2 ndash x

2)

frac12

V = [(120 Nm)(0250 kg)][(0150 m)2 - (00750 m)

2 ]

frac12 = 0900 ms


f = (12 (km)frac12 = (12 (120 Nm)(0250 kg)

frac12 = 110 Hz


a = Fm

F = k x

a = k xm

a = (120 Nm)(00750 m)(0250 kg) = 360 ms2


NAME_____SOLUTION________________________













shown



7 Start 1030 am

Stop 1145 am

PROBLEM

POINTS

CREDIT

1-5

20

6

20

7

20

8

20

9

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




b It increases

(4)

c It decreases




a 0857 rads2

b 117 rads2

(4)

c 300 rads2

d 350 rads2



a 110 ms2

b 087 ms2

(4)

c 055 ms2

d 023 ms2



a Nothing

b The Earth

(4)




length and time to

a ML2T

-2

b ML

(4)

c ML2

d ML-1

T-2



the scaffold

(20)





wpx = wp cos (-90o) = 0


2) = - 589 N

wsx = ws cos (-90o) = 0


2) = - 245 N



2) = - 392 N

Fx = 0 =gt 0 = 0


TR + TL= 873 N


TL = 0

TR = TR (400 m)

wp = - (589 N) (200 m) = - 1178 N-m

sp = - (245 N) (200 m) = - 490 N-m

wp = - (392 N) (100 m) = - 392 N-m

= 0


TR = 427 N

TL= 873 N ndash 417 N = 446 N






2 = 0400 kg m

2


2 = frac12 (500 kg)(0400 m)

2 + (100 kg)(0400 m)

2 = 0560 kg m

2

Ii i = If f





F = G m1 m2r2

F = G Ms MER2 = (6673 x 10

-11 Nm

2kg

2) (5500 kg) (600 x 10

24 kg )(726 x 10

6 m)

2 = 41779

N


T2 = (4

2GME )R

3

T2 GME = 4

2 R

3

R3 = T

2 GME (4

2)

R = (T2 GME (4

2))

13

R = ((6200 s)2 (6673 x 10

-11 Nm

2kg

2)(600 x 10

24 kg) (4

2))

13 = 726 x 10

6 m






= 0 + t

= ( - 0) t

= (175 rads) (100 s) = 0175 rads2



2 + m R

2 = frac12 (760 kg)(250 m)

2 + 2 (250 kg)(250 m)

2 = 2688 kg m

2


= I

= (2688 kgm2)(0175 rads

2) = 470 N-m


= F R

F = R = (470 N-m)(250 m) = 188 N


NAME______SOLUTION_______________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



a Child A

b Child B

(4)




a Mass

b Moment of inertia

(4)

c Angular velocity




a 14

b 12

(4)

c 2

d 4


a To your left

b To your right

(4)

c Forward

d Backward




(4)





a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)











2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55


V = 2RT

V = 2(450 x 1012

m)(164x365x24x3600 s) = 5467 ms




Vmax = (km) A2

frac12

Vmax = [(120 Nm)(0250 kg)](0150 m)2

frac12 = 103 ms


x= 0


V = (km)(A2 ndash x

2)

frac12

V = [(120 Nm)(0250 kg)][(0150 m)2 - (00750 m)

2 ]

frac12 = 0900 ms


f = (12 (km)frac12 = (12 (120 Nm)(0250 kg)

frac12 = 110 Hz


a = Fm

F = k x

a = k xm

a = (120 Nm)(00750 m)(0250 kg) = 360 ms2


NAME_____SOLUTION________________________













shown



7 Start 1030 am

Stop 1145 am

PROBLEM

POINTS

CREDIT

1-5

20

6

20

7

20

8

20

9

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




b It increases

(4)

c It decreases




a 0857 rads2

b 117 rads2

(4)

c 300 rads2

d 350 rads2



a 110 ms2

b 087 ms2

(4)

c 055 ms2

d 023 ms2



a Nothing

b The Earth

(4)




length and time to

a ML2T

-2

b ML

(4)

c ML2

d ML-1

T-2



the scaffold

(20)





wpx = wp cos (-90o) = 0


2) = - 589 N

wsx = ws cos (-90o) = 0


2) = - 245 N



2) = - 392 N

Fx = 0 =gt 0 = 0


TR + TL= 873 N


TL = 0

TR = TR (400 m)

wp = - (589 N) (200 m) = - 1178 N-m

sp = - (245 N) (200 m) = - 490 N-m

wp = - (392 N) (100 m) = - 392 N-m

= 0


TR = 427 N

TL= 873 N ndash 417 N = 446 N






2 = 0400 kg m

2


2 = frac12 (500 kg)(0400 m)

2 + (100 kg)(0400 m)

2 = 0560 kg m

2

Ii i = If f





F = G m1 m2r2

F = G Ms MER2 = (6673 x 10

-11 Nm

2kg

2) (5500 kg) (600 x 10

24 kg )(726 x 10

6 m)

2 = 41779

N


T2 = (4

2GME )R

3

T2 GME = 4

2 R

3

R3 = T

2 GME (4

2)

R = (T2 GME (4

2))

13

R = ((6200 s)2 (6673 x 10

-11 Nm

2kg

2)(600 x 10

24 kg) (4

2))

13 = 726 x 10

6 m






= 0 + t

= ( - 0) t

= (175 rads) (100 s) = 0175 rads2



2 + m R

2 = frac12 (760 kg)(250 m)

2 + 2 (250 kg)(250 m)

2 = 2688 kg m

2


= I

= (2688 kgm2)(0175 rads

2) = 470 N-m


= F R

F = R = (470 N-m)(250 m) = 188 N


NAME______SOLUTION_______________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



a Child A

b Child B

(4)




a Mass

b Moment of inertia

(4)

c Angular velocity




a 14

b 12

(4)

c 2

d 4


a To your left

b To your right

(4)

c Forward

d Backward




(4)





a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)











2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55














shown



7 Start 1030 am

Stop 1145 am

PROBLEM

POINTS

CREDIT

1-5

20

6

20

7

20

8

20

9

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT




b It increases

(4)

c It decreases




a 0857 rads2

b 117 rads2

(4)

c 300 rads2

d 350 rads2



a 110 ms2

b 087 ms2

(4)

c 055 ms2

d 023 ms2



a Nothing

b The Earth

(4)




length and time to

a ML2T

-2

b ML

(4)

c ML2

d ML-1

T-2



the scaffold

(20)





wpx = wp cos (-90o) = 0


2) = - 589 N

wsx = ws cos (-90o) = 0


2) = - 245 N



2) = - 392 N

Fx = 0 =gt 0 = 0


TR + TL= 873 N


TL = 0

TR = TR (400 m)

wp = - (589 N) (200 m) = - 1178 N-m

sp = - (245 N) (200 m) = - 490 N-m

wp = - (392 N) (100 m) = - 392 N-m

= 0


TR = 427 N

TL= 873 N ndash 417 N = 446 N






2 = 0400 kg m

2


2 = frac12 (500 kg)(0400 m)

2 + (100 kg)(0400 m)

2 = 0560 kg m

2

Ii i = If f





F = G m1 m2r2

F = G Ms MER2 = (6673 x 10

-11 Nm

2kg

2) (5500 kg) (600 x 10

24 kg )(726 x 10

6 m)

2 = 41779

N


T2 = (4

2GME )R

3

T2 GME = 4

2 R

3

R3 = T

2 GME (4

2)

R = (T2 GME (4

2))

13

R = ((6200 s)2 (6673 x 10

-11 Nm

2kg

2)(600 x 10

24 kg) (4

2))

13 = 726 x 10

6 m






= 0 + t

= ( - 0) t

= (175 rads) (100 s) = 0175 rads2



2 + m R

2 = frac12 (760 kg)(250 m)

2 + 2 (250 kg)(250 m)

2 = 2688 kg m

2


= I

= (2688 kgm2)(0175 rads

2) = 470 N-m


= F R

F = R = (470 N-m)(250 m) = 188 N


NAME______SOLUTION_______________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



a Child A

b Child B

(4)




a Mass

b Moment of inertia

(4)

c Angular velocity




a 14

b 12

(4)

c 2

d 4


a To your left

b To your right

(4)

c Forward

d Backward




(4)





a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)











2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55


PARTIAL CREDIT




b It increases

(4)

c It decreases




a 0857 rads2

b 117 rads2

(4)

c 300 rads2

d 350 rads2



a 110 ms2

b 087 ms2

(4)

c 055 ms2

d 023 ms2



a Nothing

b The Earth

(4)




length and time to

a ML2T

-2

b ML

(4)

c ML2

d ML-1

T-2



the scaffold

(20)





wpx = wp cos (-90o) = 0


2) = - 589 N

wsx = ws cos (-90o) = 0


2) = - 245 N



2) = - 392 N

Fx = 0 =gt 0 = 0


TR + TL= 873 N


TL = 0

TR = TR (400 m)

wp = - (589 N) (200 m) = - 1178 N-m

sp = - (245 N) (200 m) = - 490 N-m

wp = - (392 N) (100 m) = - 392 N-m

= 0


TR = 427 N

TL= 873 N ndash 417 N = 446 N






2 = 0400 kg m

2


2 = frac12 (500 kg)(0400 m)

2 + (100 kg)(0400 m)

2 = 0560 kg m

2

Ii i = If f





F = G m1 m2r2

F = G Ms MER2 = (6673 x 10

-11 Nm

2kg

2) (5500 kg) (600 x 10

24 kg )(726 x 10

6 m)

2 = 41779

N


T2 = (4

2GME )R

3

T2 GME = 4

2 R

3

R3 = T

2 GME (4

2)

R = (T2 GME (4

2))

13

R = ((6200 s)2 (6673 x 10

-11 Nm

2kg

2)(600 x 10

24 kg) (4

2))

13 = 726 x 10

6 m






= 0 + t

= ( - 0) t

= (175 rads) (100 s) = 0175 rads2



2 + m R

2 = frac12 (760 kg)(250 m)

2 + 2 (250 kg)(250 m)

2 = 2688 kg m

2


= I

= (2688 kgm2)(0175 rads

2) = 470 N-m


= F R

F = R = (470 N-m)(250 m) = 188 N


NAME______SOLUTION_______________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



a Child A

b Child B

(4)




a Mass

b Moment of inertia

(4)

c Angular velocity




a 14

b 12

(4)

c 2

d 4


a To your left

b To your right

(4)

c Forward

d Backward




(4)





a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)











2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55





length and time to

a ML2T

-2

b ML

(4)

c ML2

d ML-1

T-2



the scaffold

(20)





wpx = wp cos (-90o) = 0


2) = - 589 N

wsx = ws cos (-90o) = 0


2) = - 245 N



2) = - 392 N

Fx = 0 =gt 0 = 0


TR + TL= 873 N


TL = 0

TR = TR (400 m)

wp = - (589 N) (200 m) = - 1178 N-m

sp = - (245 N) (200 m) = - 490 N-m

wp = - (392 N) (100 m) = - 392 N-m

= 0


TR = 427 N

TL= 873 N ndash 417 N = 446 N






2 = 0400 kg m

2


2 = frac12 (500 kg)(0400 m)

2 + (100 kg)(0400 m)

2 = 0560 kg m

2

Ii i = If f





F = G m1 m2r2

F = G Ms MER2 = (6673 x 10

-11 Nm

2kg

2) (5500 kg) (600 x 10

24 kg )(726 x 10

6 m)

2 = 41779

N


T2 = (4

2GME )R

3

T2 GME = 4

2 R

3

R3 = T

2 GME (4

2)

R = (T2 GME (4

2))

13

R = ((6200 s)2 (6673 x 10

-11 Nm

2kg

2)(600 x 10

24 kg) (4

2))

13 = 726 x 10

6 m






= 0 + t

= ( - 0) t

= (175 rads) (100 s) = 0175 rads2



2 + m R

2 = frac12 (760 kg)(250 m)

2 + 2 (250 kg)(250 m)

2 = 2688 kg m

2


= I

= (2688 kgm2)(0175 rads

2) = 470 N-m


= F R

F = R = (470 N-m)(250 m) = 188 N


NAME______SOLUTION_______________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



a Child A

b Child B

(4)




a Mass

b Moment of inertia

(4)

c Angular velocity




a 14

b 12

(4)

c 2

d 4


a To your left

b To your right

(4)

c Forward

d Backward




(4)





a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)











2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55



2) = - 589 N

wsx = ws cos (-90o) = 0


2) = - 245 N



2) = - 392 N

Fx = 0 =gt 0 = 0


TR + TL= 873 N


TL = 0

TR = TR (400 m)

wp = - (589 N) (200 m) = - 1178 N-m

sp = - (245 N) (200 m) = - 490 N-m

wp = - (392 N) (100 m) = - 392 N-m

= 0


TR = 427 N

TL= 873 N ndash 417 N = 446 N






2 = 0400 kg m

2


2 = frac12 (500 kg)(0400 m)

2 + (100 kg)(0400 m)

2 = 0560 kg m

2

Ii i = If f





F = G m1 m2r2

F = G Ms MER2 = (6673 x 10

-11 Nm

2kg

2) (5500 kg) (600 x 10

24 kg )(726 x 10

6 m)

2 = 41779

N


T2 = (4

2GME )R

3

T2 GME = 4

2 R

3

R3 = T

2 GME (4

2)

R = (T2 GME (4

2))

13

R = ((6200 s)2 (6673 x 10

-11 Nm

2kg

2)(600 x 10

24 kg) (4

2))

13 = 726 x 10

6 m






= 0 + t

= ( - 0) t

= (175 rads) (100 s) = 0175 rads2



2 + m R

2 = frac12 (760 kg)(250 m)

2 + 2 (250 kg)(250 m)

2 = 2688 kg m

2


= I

= (2688 kgm2)(0175 rads

2) = 470 N-m


= F R

F = R = (470 N-m)(250 m) = 188 N


NAME______SOLUTION_______________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



a Child A

b Child B

(4)




a Mass

b Moment of inertia

(4)

c Angular velocity




a 14

b 12

(4)

c 2

d 4


a To your left

b To your right

(4)

c Forward

d Backward




(4)





a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)











2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55



2 = 0400 kg m

2


2 = frac12 (500 kg)(0400 m)

2 + (100 kg)(0400 m)

2 = 0560 kg m

2

Ii i = If f





F = G m1 m2r2

F = G Ms MER2 = (6673 x 10

-11 Nm

2kg

2) (5500 kg) (600 x 10

24 kg )(726 x 10

6 m)

2 = 41779

N


T2 = (4

2GME )R

3

T2 GME = 4

2 R

3

R3 = T

2 GME (4

2)

R = (T2 GME (4

2))

13

R = ((6200 s)2 (6673 x 10

-11 Nm

2kg

2)(600 x 10

24 kg) (4

2))

13 = 726 x 10

6 m






= 0 + t

= ( - 0) t

= (175 rads) (100 s) = 0175 rads2



2 + m R

2 = frac12 (760 kg)(250 m)

2 + 2 (250 kg)(250 m)

2 = 2688 kg m

2


= I

= (2688 kgm2)(0175 rads

2) = 470 N-m


= F R

F = R = (470 N-m)(250 m) = 188 N


NAME______SOLUTION_______________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



a Child A

b Child B

(4)




a Mass

b Moment of inertia

(4)

c Angular velocity




a 14

b 12

(4)

c 2

d 4


a To your left

b To your right

(4)

c Forward

d Backward




(4)





a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)











2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55


= ( - 0) t

= (175 rads) (100 s) = 0175 rads2



2 + m R

2 = frac12 (760 kg)(250 m)

2 + 2 (250 kg)(250 m)

2 = 2688 kg m

2


= I

= (2688 kgm2)(0175 rads

2) = 470 N-m


= F R

F = R = (470 N-m)(250 m) = 188 N


NAME______SOLUTION_______________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



a Child A

b Child B

(4)




a Mass

b Moment of inertia

(4)

c Angular velocity




a 14

b 12

(4)

c 2

d 4


a To your left

b To your right

(4)

c Forward

d Backward




(4)





a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)











2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55



shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



a Child A

b Child B

(4)




a Mass

b Moment of inertia

(4)

c Angular velocity




a 14

b 12

(4)

c 2

d 4


a To your left

b To your right

(4)

c Forward

d Backward




(4)





a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)











2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55


a Mass

b Moment of inertia

(4)

c Angular velocity




a 14

b 12

(4)

c 2

d 4


a To your left

b To your right

(4)

c Forward

d Backward




(4)





a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)











2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55






a (GM2)(R

2)

b (GM2)(2R

2)

(4)

c (GM2)(4R

2)

d (2GM2)(R

2)











2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55







2) = - 491 N



2) = - 392 N

Fx = 0 =gt 0 = 0


NR + NS= 442 N


NR = 0

NS = NS (200 m)

w board = - (491 N) (100 m) = - 491 N-m

w block = - (392 N) (150 m) = - 588 N-m

= 0


NS = 319 N

NR= 442 N ndash 319 N = 123 N



0 =


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55


= 0 + t

= ( - 0) t

= (524 rads) (550 s) = 952 rads2


=0 +0 t +t2 = frac12 (524 rads

2)(550 s)

2 = 793 rad = 126 rev



Vt = R = (523 rads)(00600 m) = 314 ms




Ei = Ef



2 + frac12 I f

2

yi = 0

Vi = 200 ms

i = Vi R

yf = h

Vf = 0

f = 0


2 = m g yf

m Vi 2 + I i

2 = 2 m g yf

m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55


m Vi 2 + (25 m R

2)(Vi R)

2 = 2 m g yf

Vi 2 + (25) Vi

2 = 2 g yf

(710) Vi2 = g yf

yf = 0285 m








I = 500 kg-m2 +(250 kg)(100 m)

2 + (250 kg)(200 m)

2 = 625 kg-m

2




= (300 N)(300 m) + (200 N)(300 m) ndash (150 N)(300 m) = 1050 N-m


I = (1050 N-m)(625 kg-m2) = 168 rads

2


NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55



NAME___SOLUTION__________________________













shown



7 Start 600 pm

Stop 715 pm

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55


PERCENTAGE



PARTIAL CREDIT



(A) Velocity

(B) Acceleration

(C) Net force




(A) Impulse

(B) Momentum

(C) Force

(D) Energy




(A) 14

(B) 12

(C) 2

(D) 4

(E) 8



(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55




(A) Child A

(B) Child B





















t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55





t



)

) =

)

) = (156 rads)2(2 x 260 rads

2) = 468 rad





Vt = r = (156 rads)(0300 m) = 468 ms

at = r = (260 rads2)(0300 m) = 0780 ms

2






NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55




NLx = NL cos (900o) = 0

NLy = NL sin (900o) = NL = 290 N

NRx = NR cos (900o) = 0

NRy = NR sin (900o) = NR = 122 N

wx = m g cos (-900o) = 0

wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55


wy = m g sin (-900o) = - m g

Fx = 0

Fy = 0

290 N + 122 N - mg = 0

mg = 412 N

m = 420 kg


NL = 0

NR = (122 N) (250 m) = 305 N-m

w = -w (x) = -(412 N) (x)

= 0

305 N-m ndash (412 N) (x) = 0

x = 0740 m




= (2 rad)(500 s) = 126 rads



2(200 m) = 318 ms

2




T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55





T = m acp = (250 kg)(319 ms2) = 794 N








r1 = r2 = r3 = ((00900 m)2 + (00900 m)

2)12

2 = 0127 m

1 = F1 r1 sin(1) = (180 N)(0127 m) sin (450o) = 162 Nm

2 = - F2 r2 sin(2) = - (260 N)(0127 m) sin (450o) = -233 Nm

3 = F3 r3 sin(3) = (140 N)(0127 m) sin (900o) = 178 Nm

= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55


= 162 Nm -233 Nm + 178 Nm = 107 Nm


NAME_____SOLUTION________________________















7 Start 1130 am

Stop 1250 pm

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55


PERCENTAGE



PARTIAL CREDIT



e Potential energy

(5)

f Acceleration

g Kinetic energy

h Frequency





(5)






a MLT -2

b ML2T

-1

(5)

c ML2T

-3

d ML3T



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55



Fall 2005 47



a 20 days

b 80 days

(5)

c 160 days

d 320 days




(5)





a The boy

b The girl

(5)







Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55



Fall 2005 48


N1x = N1 cos (-90o) = 0

N1y = N1 sin (-90o) = - N1 = -828 N

N2x = N2 cos (90o) = 0

N2y = N2 sin (90o) = N2

wx = w cos (-90o) = 0

wy = w sin (-90o) = - w = - m g = - (640 kg)(980 ms

2) = - 627 N

Fx = 0 =gt 0 = 0

Fy = 0 =gt (-828 N) + N2 ndash (627 N) = 0

N2 = 1455 N

N2 = N2 (110 m) = (1455 N)(110 m) = 1601 Nm

N1 = 0

w = - (627 N) (x)

= 0

- (627 N) x + 1601 N m = 0

x = 255 m


Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55



Fall 2005 49


N2 = 1455 N







2 = 0100 kg m

2


2 = frac12 (500 kg)(0200 m)

2 + frac12 (310 kg)(00800 m)

2 = 010992 kg m

2

Ii i = If f






Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55



Fall 2005 50


released


A = 0160 m


Vmax = (km)12

A = ((800 Nm)(0500 kg)) 12

(0160 m) = 202 ms



2 = 102 J


f = (1(2)(km)12

= 201 Hz


T = 1f = 0497 s



center



2 = frac12 (100 kg)(100 m)

2 + 4 (0250 kg)(500 m)

2 = 750 kg m

2


Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55



Fall 2005 51



1 = F1 d1 = (400N)(100m) = 400 Nm

2 = - F2 d2 = - (200N)(500m) = -100 Nm

= 400 Nm ndash 100 Nm = 300 Nm


= I

= I = (300 Nm) (750 kgm2) = 400 rads

2




2 = frac12 (750 kgm

2)((400 rads

2)(100 s))

2 = 60000 J


Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55



Fall 2005 52


Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55



Fall 2005 53


Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55



Fall 2005 54





2 +

frac12 Ii

2 = 0 + frac12 mVi

2 +

frac12 Ii

2 = frac12 mVi

2 +

frac12 (25mR

2)i

2 = frac12

mVi2

+ frac12 (25)mVi

2


+ frac12 If


2 +

frac12 (25mR

2)f


2 +

frac12 (25)mVf2

Ei = Ei

Vi2

+ (25)Vi

2 = 2 g yf + Vf

2 +

(25)Vf

2

(75)Vi2

= 2 g yf + (75)Vf2

Vi2

= (57)(2 g yf)+ Vf2

Vf2

= - (57)(2 g yf)+ Vi2

Vf = (- (57)(2 g yf)+ Vi

2)

12

Vf = (- (57)(2 (981 ms

2) (0530 m))+ (285 ms)

2)

12

Vf = 0834 ms


Fall 2005 55



Fall 2005 55


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