physics 201: lecture 23, pg 1 lecture 23 l exam 3: 2103 chamberlin hall, b102 van vleck & quiet...

Physics 201: Lecture 23, Pg 1

Lecture 23

Exam 3: 2103 Chamberlin Hall, B102 Van Vleck & quiet room CH Sections 604 605 606 609 610 611 (TA: Moriah T., Abdallah

C., Tamara S.) VV Sections 602 603 607 608 612 (TA: Eric P., Dan C., Zhe D.) Format: Closed book, three 8 x11” sheets, hand written Electronics: Any calculator is okay but no web/cell access Quiet room: Test anxiety, special accommodations, etc.

Chapters Covered Chapter 9: Linear momentum and collision (not 9.9) Chapter 10: Rotation about fixed axis and rolling Chapter 11: Angular Momentum (not 11.5) Chapter 12: Static equilibrium and elasticity


Basic Concepts and Quantities

Momentum, Angular Momentum Linear Momentum, Impulse Angular Momentum (magnitude and direction)

Torque Collisions: Elastic & Inelastic Center of Mass Rotational Motion (1-axis) Angular displacement (Δ) / Velocity(), Acceleration (). Moments of Inertia Rotational Kinetic Energy Conservation Laws: Energy, Momentum, Angular Momentum Static Equilibrium Elasticity: Young’s, Shear, Bulk Modulus


Chapter 9 : Momentum and Momentum Conservation

Ix =

Ix


Chapter 9


Center of Mass

Chapter 9

M

rmrmrm

m

rmr N

ii

N

iii

CM

332211

1

1

M

vmvmvm

m

vmv N

ii

N

iii

CM

332211

1

1


Chapter 10

At a point a distance R away from the axis of rotation, the tangential motion:

s (arc) = R vT (tangential) = R

aT = R

R

v = R

s

)(2

21

rad/s)in elocity (angular v

)rad/sin accelation(angular constant

22

2

2

ifif

iif

if

tt

t

rrv

a

ra

rv

c

t

22


Chapter 10


Chapter 11

dtLd


Chapter 12


Approach to Statics:

In general, we can use the two equations

to solve any statics problems.

When choosing axes about which to calculate torque, choose one that makes the problem easy....

However if there is acceleration, then restrict rotation axis to center of mass (as well as for translation).

0F

0


Young’s modulus: measures the resistance of a solid to a change in its length.

Changes in length: Young’s modulus

L0 L

Felasticity in length

0//

strain tensilestress tensileY

LLAF

A


Bulk modulus: measures the resistance of solids or liquids to changes in their volume.

Changes in volume: Bulk Modulus

V0

Vi + V

F

Volume elasticity

iΔV/VF/AB

PressureF/AP


Changes in shape: Shear Modulus

hx

AF

/

/

strainshear

stressshear S

Applying a force perpendicular to a surface


Comparison Kinematics

Angular Linear

constant

0 t

200 2

1tt

constanta

at 0vv

221

00 v attxx

22

02

)( 021

AVE

ax2vv 2

02

)vv(v 021

AVE


Comparison: Dynamics

Angular Linear

K 1

2I 2

K 1

2mv2

I = i mi ri2 m

F = a mr x F = I

L = r x = I p = mv

EXT dLdt

FEXT dp

dt

W = W = F •x

K = WNET K = WNET


Momentum and collisions

Remember vector components

A 5 kg cart rolling without friction to the right at 10 m/s collides and sticks to a 5 kg motionless block on a 30° frictionless incline.

How far along the incline do the joined blocks slide?


Momentum and collisions Remember vector components

A 5 kg cart rolling without friction to the right at 10 m/s collides and sticks to a 5 kg motionless block on a 30° frictionless incline.

Momentum parallel to incline is conserved Normal force (by ground on cart) is to the incline

mvi cos + m 0 = 2m vf vf = vi cos / 2 = 4.4 m/s

Now use work-energy

2mgh + 0 = ½ 2m v2f d = h / sin = v2

f / (2g sin )

mvi

mvi cos


Example problem: Going in circles B A 2.0 kg disk tied to a long string undergoes circular motion on

a frictionless horizontal table top. The string passes through a hole and then hangs vertically. The disk starts out at 5.0 rev/sec 0.50 m away from hole. If you pull slowly down on the string so that the final radius is 0.25 m, what is the final angular velocity?

No external torque so angular momentum is conserved. Ib b = Ia a

I = mR2

(0.50)2 b = (0.25)2 a

4 b = a = 20 rev/sec ri

i


Spinning ball on incline

A solid disk of mass m and radius R is spinning with angular velocity . It is positioned so that it can either move directly up or down an incline of angle (but it is not rolling motion). The coefficient of kinetic friction is . At what angle will the disk’s position on the incline not change?


Spinning ball on incline A solid disk of mass m and radius R is spinning with

angular velocity . It is positioned so that it can either move directly up or down an incline of angle (but it is not rolling motion). The coefficient of kinetic friction is .

At what angle will the disk’s position on the incline not change?

0 F

mg

N

fsin0 mgfFx

cos0 mgNFy

cosmgNf

tan


Spinning ball on incline A solid disk of mass m and radius r is spinning with angular

velocity . It is positioned so that it can either move directly up or down an incline of angle (but it is not rolling motion). The coefficient of kinetic friction is . While spinning the disk’s position will not change.

How long will it be before it starts to roll? This will occur only when =0.

CMdisk about 0 z

mg

N

f90sinsin|||| rffrz

cosmgrrfI rg /cos2

rgttf /cos20


Example Problem A 3.0 kg mass is attached to a light, rigid rod 1.5 m long. The

rod is vertical and is anchored to the ground through a frictionless pivot. It sits perfectly balanced at an unstable equilibrium. A 500 gm bullet is shot horizontally at 100 m/s through the mass. The force versus time plot is shown.

How fast is the bullet going when

it leaves? What is the tension in the rod just

after the bullet exits?

Method: Impulse

I = area under curve

dttFI xx )(

dttI x ])15.0(10225[ 24




How fast is the bullet going when

it leaves?

dttI x ])15.0(10225[ 24

3.00.0

34 |3/)15.0(10225 ttI x

Ns 45xI

Ns)451000.5(

Ns 45

if PP

m/s 10 Ns 5 ff vP




What is the tension in the rod just

after the bullet exits?

m/s 15345

Ns 45

/ v

P

f

f

Tmgr

vmmaF cy

2

]N 305.1/)15(3[ 2 T

N 420N 30)(450 T


Statics: Example

A sign of mass M is hung 1 m from the end of a 4 m long beam (mass m) as shown in the diagram. The beam is hinged at the wall. What is the tension in the wire in terms of m, M, g and any other given quantity?

wire

1 m

SIGN


Statics: Example

mgMg

T

Fx

30°

Process: Make a FBD and note known / unknown forces.

Chose axis of rotation at support because Fx & Fy are not known

Fy

2 m

3 m

F = 0 0 = Fx – T cos 30°

0 = Fy + T sin 30° - mg - Mg

z-dir: z = 0 = -mg 2r – Mg 3r + T sin 30° 4r (r = 1m)

The torque equation get us where we need to go, T

T = (2m + 3M) g / 2

X


Center of Mass Example: Astronauts & Rope

Two astronauts are initially at rest in outer space and 20 meters apart. The one on the right has 1.5 times the mass of the other (as shown). The 1.5 m astronaut wants to get back to the ship but his jet pack is broken. There happens to be a rope connected between the two. The heavier astronaut starts pulling in the rope.

(1) Does he/she get back to the ship ?

(2) Does he/she meet the other astronaut ?

M = 1.5mm


Example: Astronauts & Rope(1) There is no external force so if the larger astronaut

pulls on the rope he will create an impulse that accelerates him/her to the left and the small astronaut to the right. The larger one’s velocity will be less than the smaller one’s so he/she doesn’t let go of the rope they will either collide (elastically or inelastically) and thus never make it.

M = 1.5mm


Example: Astronauts & Rope

(2) However if the larger astronaut lets go of the rope he will get to the ship. (Too bad for the smaller astronaut!)

In all cases the center of mass will remain fixed because they are initially at rest and there is no external force.

To find the position where they meet all we need do is find the Center of Mass

M = 1.5mm


Forces and rigid body rotation

To change the angular velocity of a rotating object, a force must be applied

How effective an applied force is at changing the rotation depends on several factorsThe magnitude of the forceWhere, relative to the axis of rotation the force is appliedThe direction of the force

F F

Which applied force will cause the wheel to spin the fastest?

A B C


Leverage

The same concept applies to leveragethe lever undergoes rigid body rotation about a pivot

point:

F

F

F

Which applied force provides the greatest lift ?

AC

B


More on torques

You need to change the tire on you car. You use a tire wrench which allows you to apply a pair of forces.

(A) What is the torque produced by a tire wrench of length L, given an applied couple of magnitude F, acting on a lug nut (point F) as shown in the figure?

(B) Assume the lug nut is stuck What is the torque acting on the wheel, if the lug nut is a distance r from the center?

Image courtesy John Wiley & Sons, Inc.


Wheel wrench

F = (L/2) F + (L/2) F = LF

F = r F sin + r F sin ( )= = LF

3. F = L F + 0 F

Notice the torque is the same everywhere.


For Thursday

Chapter 13 (Newton’s Law of Gravitation)


Momentum & Impulse

A. 0

B. ½ v

C. 2 v

D. 4 v

A rubber ball collides head on (i.e., velocities are opposite) with a clay ball of the same mass. The balls have the same speed, v, before the collision, and stick together after the collision. What is their speed immediately after the collision?


Momentum & Impulse

A rubber ball collides head on with a clay ball of the same mass. The balls have the same speed, v, before the collision, and stick together after the collision. What is their speed after the collision?

(a) 0

(b) ½ v

(c) 2 v

(d) 4 v


Momentum, Work and Energy

A 0.40 kg block is pushed up against a spring (with spring constant 270 N/m ) on a frictionless surface so that the spring is compressed 0.20 m. When the block is released, it slides across the surface and collides with the 0.60 kg bob of a pendulum. The bob is made of clay and the block sticks to it. The length of the pendulum is 0.80 m. (See the diagram.)

To what maximum height above the surface will the ball/block assembly rise after the collision? (g=9.8 m/s2)

A. 2.2 cmB. 4.4 cmC. 11. cmD. 22 cmE. 44 cmF. 55 cm


Momentum, Work and Energy

A 0.40 kg block is pushed up against a spring (with spring constant 270 N/m ) on a frictionless surface so that the spring is compressed 0.20 m. When the block is released, it slides across the surface and collides with the 0.60 kg bob of a pendulum. The bob is made of clay and the block sticks to it. The length of the pendulum is .80 m. (See the diagram.)

To what maximum height above the surface will the ball/block assembly rise after the collision?

A. 2.2 cmB. 4.4 cmC. 11. cmD. 22 cmE. 44 cmF. 55 cm


Momentum, Work and Energy ( Now with friction)

A 0.40 kg block is pushed up against a spring (with spring constant 270 N/m ) on a surface.

If mstatic = 0.54, how far can the spring be compressed and the block remain stationary (i.e., maximum static friction)?

F = 0 = k u - f = k u - N

u = mg/k = 0.54 (0.40x10 N) / 270 N/m= 0.0080 m


Momentum, Work and Energy ( Now with friction)

A 0.40 kg block is pushed up against a spring (with spring constant 270 N/m ) on a surface. The spring is compressed 0.20 m

If mkinetic = 0.50 and the block is 9.8 m away from the unstretched spring, how high with the clay/block pair rise?

Emech (at collision) = Uspring + Wfriction = ½ k u2 - mg d 1/2 m v2 = 135(0.04)-0.50(0.40x10.)10.=(540-20) J=520 J v2 = 1040/0.40 m2/s2 Now the collision (cons. of momentum) and the swing.


Momentum and Impulse

Henri Lamothe holds the world record for the highest shallow dive. He belly-flopped from a platform 12.0 m high into a tank of water just 30.0 cm deep! Assuming that he had a mass of 50.0 kg and that he stopped just as he reached the bottom of the tank, what is the magnitude of the impulse imparted to him while in the tank of water (in units of kg m/s)?

(a) 121

(b) 286

(c) 490

(d) 623

(e) 767



Henri Lamothe holds the world record for the highest shallow dive. He belly-flopped from a platform 12.0 m high into a tank of water just 30.0 cm deep! Assuming that he had a mass of 50.0 kg and that he stopped just as he reached the bottom of the tank, what is the magnitude of the impulse imparted to him while in the tank of water (in units of kg m/s)?

(a) 121

(b) 286

(c) 490

(d) 623

(e) 767p = sqrt(2x9.8x12.3)x50


Momentum & Impulse

Suppose that in the previous problem, the positively charged particle is a proton and the negatively charged particle, an electron. (The mass of a proton is approximately 1,840 times the mass of an electron.) Suppose that they are released from rest simultaneously. If, after a certain time, the change in momentum of the proton is p, what is the magnitude of the change in momentum of the electron?

(a) p / 1840

(b) p

(c) 1840 p


Conservation of Momentum

A woman is skating to the right with a speed of 12.0 m/s when she is hit in the stomach by a giant snowball moving to the left. The mass of the snowball is 2.00 kg, its speed is 25.0 m/s and it sticks to the woman's stomach. If the mass of the woman is 60.0 kg, what is her speed after the collision?

(a) 10.8 m/s

(b) 11.2 m/s

(c) 12.4 m/s

(d) 12.8 m/s



A woman is skating to the right with a speed of 12.0 m/s when she is hit in the stomach by a giant snowball moving to the left. The mass of the snowball is 2.00 kg, its speed is 25.0 m/s and it sticks to the woman's stomach. If the mass of the woman is 60.0 kg, what is her speed after the collision?

(a) 10.8 m/s

(b) 11.2 m/s

(c) 12.4 m/s

(d) 12.8 m/s



Sean is carrying 24 bottles of beer when he slips in a large frictionless puddle. He slides forwards with a speed of 2.50 m/s towards a very steep cliff. The only way for Sean to stop before he reaches the edge of the cliff is to throw the bottles forward at 20.0 m/s (relative to the ground). If the mass of each bottle is 500 g, and Sean's mass is 72 kg, what is the minimum number of bottles that he needs to throw?

(a) 18 (b) 20 (c) 21 (d) 24 (e) more than 24



A stunt man jumps from the roof of a tall building, but no injury occurs because the person lands on a large, air-filled bag. Which one of the following statements best describes why no injury occurs?

(a) The bag provides the necessary force to stop the person.

(b) The bag reduces the impulse to the person.

(c) The bag reduces the change in momentum.

(d) The bag decreases the amount of time during which the momentum is changing and reduces the average force on the person.

(e) The bag increases the amount of time during which the momentum is changing and reduces the average force on the person.



Two blocks of mass m1 = M and m2 = 2M are both sliding towards you on a frictionless surface. The linear momentum of block 1 is half the linear momentum of block 2. You apply the same constant force to both objects in order to bring them to rest. What is the ratio of the two stopping distances d2/d1?

(a) 1/ 2 (b) 1/ 2½

(c) 1 (d) 2½ (e) 2 (f) Cannot be determined without knowing the masses of

the objects and their velocities.


Momentum and Impulse In a table-top shuffleboard game, a heavy moving puck collides with a

lighter stationary puck as shown. The incident puck is deflected through an angle of 20° and both pucks are eventually brought to rest by friction with the table. The impulse approximation is valid (i.e.,the time of the collision is short relative to the time of motion so that momentum is conserved).

Which of the following statements is correct?A. The collision must be inelastic because the pucks have different

masses.B. The collision must be inelastic because there is friction between the

pucks and the surface.C. The collision must be elastic because the pucks bounce off each other.D. The collision must be elastic because, in the impulse approximation,momentum is conserved.E. There is not enough information given to decide whether the collision iselastic or inelastic.


Bill (mass M) is climbing a ladder (length L, mass m) that leans against a smooth wall (no friction between wall and ladder). A frictional force F between the ladder and the floor keeps it from slipping. The angle between the ladder and the wall is .

What is the magnitude of F as a function of Bill’s distance up the ladder?

Bill

Lm

F

Exercise: Ladder against smooth wall


Ladder against smooth wall...

Consider all of the forces acting. In addition to gravity and friction, there will be normal forces Nf and Nw by the floor and wall respectively on the ladder.

First sketch the FBD

Mgd

L/2

F

mg

Nw

Nf

y

x

Again use the fact that FNET = 0 in both x and y directions:

x: Nw = F

y: Nf = Mg + mg


Ladder against smooth wall...

Since we are not interested in Nw, calculate torques about an axis through the top end of the ladder, in the z direction.

Mg

L/2

m

F

mg

Nw

Nf

y

x Substituting: Nf = Mg + mg and

solve for F :

torque axis

d

cos

0 sin sin sin)(sin2

fNLFLMgdLmgL

Mm

MgF2L

d tan


Example: Ladder against smooth wall

For a given coefficient of static friction s,the maximum force of friction F that can beprovided is sNf = s g(M + m).

The ladder will slip if F exceedsthis value.

m

F

dCautionary note:Cautionary note:

(1) Brace the bottom of ladders!

(2) Don’t make too big!

We have just calculated that

Mm

MgF2L

d tan

physics 201: lecture 23, pg 1 lecture 23 l exam 3: 2103 chamberlin hall, b102 van vleck & quiet...

Documents