ME 274 – Summer 2020 SOLUTION Examination No. 1 (AM)

July 2, 2020

INSTRUCTIONS Begin each problem on a separate sheet. All problems are of equal value and will be graded on the basis of 20 points maximum for each problem. Please remember that in order for you to obtain maximum credit for a problem, the solution must be clearly presented, i. e.:

• coordinate systems used must be clearly identified. • where ever appropriate, free body diagrams must be drawn. These should be

drawn separately from the given figures. • units must be clearly stated as part of the answer. • vectors must be clearly identified with proper vector notation (e.g.,

�

v ,

�

v or

�

v) If the solution does not follow a logical thought process, it will be assumed in error.

ME 274 – Summer 2020 SOLUTION Examination No. 1 (AM) PROBLEM NO. 1 – 20 points

Given: Particle P travels within the x-y plane along a path given by x = y2 − 3 , and

with y t( ) = t2 −1 where x and y are given in feet, and t is given in seconds. Find: For the position of P corresponding to t = 2 seconds :

a) Determine the velocity and acceleration vectors for point P. Write your answers using the Cartesian unit vectors i and j .

b) Make a sketch in the xy-plane of the velocity and acceleration vectors for P.

c) Determine the rate of change of speed of P. d) Determine the radius of curvature for the path of P.

SOLUTION Using the given information:

y = t2 −1= 3 ft!y = 2t = 4 ft / s

!!y = 2 ft / s2

!x = 2y!y = 2( ) 3( ) 4( ) = 24 ft / s

!!x = 2 !y2 + y!!y( ) = 2( ) 4( )2 + 3( ) 2( )⎡⎣⎢

⎤⎦⎥= 44 ft / s2

Therefore:

!v = "xi + "yj = 24i + 4 j( ) ft / s

!a = ""xi + ""yj = 44 i + 2 j( ) ft / s2

From these, we have:

!v = "a i et =

"a i"vv

⎛⎝⎜

⎞⎠⎟= 44 i + 2 j( ) i 24i + 4 j

242 + 42

⎛

⎝⎜

⎞

⎠⎟ =

44( ) 24( ) + 2( ) 4( )592

= 26637

ft / s2

!a 2 = "v2 + v2

ρ⎛

⎝⎜

⎞

⎠⎟

2

⇒ ρ = v2

!a 2 − "v2= 592

1940− 7075637

⎛⎝⎜

⎞⎠⎟

= 37 372

ft

4

3

A B

L

vA

L

L

D

x

y

See practice problem from 6/22 tutorial session.

ME 274 – Summer SOLUTION Examination No. 1 (AM) PROBLEM NO. 2 – 20 points Given: End A of the L-shaped bar AD is constrained to move along a straight, inclined

path with a constant speed of vA . In addition, point B on the bar is constrained to move along a straight, horizontal path.

Find: At the position shown bar section AB is horizontal and section CD is vertical.

At this position: a) Determine the angular velocity and angular acceleration of bar AD. Write

your answers as vectors. b) Determine the acceleration of point D on bar AD. Write your answer as a

vector. Leave your answers in terms of, at most, L and vA .

SOLUTION

!vB = !vA +!ω × !rA/ B

vBi = vA cosθ i + sinθ j( ) + ω ABk( )× Li( )= vAcosθ( ) i + vAsinθ + Lω AB( ) j

Therefore, balancing coefficients of the above vector equation:

j : 0 = vAsinθ + Lω AB ⇒

!ω AB = −

vAL

sinθ k

For acceleration:

!aB = !aA +!α × !rA/ B −ω AB

2 !rA/ B

aBi =!0+ α ABk( )× Li( )−ω AB

2 Li( )= −Lω AB

2( ) i + Lα AB( ) j

Therefore, balancing coefficients of the above vector equation: j : 0 = Lα AB ⇒

!α AB =

!0

To find the acceleration of point D:

!aD = !aA +!α AB × !rD/ A −ω AB

2 !rD/ A

=!0+!0−ω AB

2 2Li + Lj( )= −2Lω AB

2( ) i + −Lω AB2( ) j

= −vA

2sin2θL

2i + j( )

Chapter 2: Planar Rigid Body Kinematics Homework

Homework 2.A.8

Given: End A of bar AB is constrained to move along a straight inclined surface. End B of thebar is constrained to move along a straight horizontal surface. End A moves in the direction shownwith a speed of vA and acceleration aA. At the position shown, bar AB is horizontal.

Find: For this position:

(a) Determine the angular velocity of bar AB and the velocity of end B. Write your answers asvectors.

(b) Determine the angular acceleration of bar AB and the acceleration of end B. Write your answersas vectors.

vA,aA no slip

O O

R

A

no slip

B

vA,aA

L A B

!

Use the following parameters in your analysis: ✓ = 36.87�, L = 3 ft, vA = 20 ft/s and aA = 10ft/s2.

2-10 Freeform c�2018

ME 274 – Summer 2020 SOLUTION Examination No. 1 (AM) PROBLEM NO. 3 – 20 points TOTAL PART A – 4 points You are a passenger in automobile P. Your forward motion, in terms of the polar coordinates shown in the figure, is described by the following velocity and acceleration vectors when θ = 225° :

!v = −60eR −80eθ( ) ft / s!a = −10eθ( ) ft / s2

Choose the response below that most accurately describes the motion of you in the automobile:

a) You are turning left.

b) You are traveling on a straight road.

c) You are turning right. d) More information is needed in order to determine the characteristics of the

roadway path. SOLUTION

P

eR

R

O

eθ

θ i

j

P

O x

y

P eR

O

eθ

i

j

P

O x

y

6

8

!v

!a

et

en

vB B

sA

h

sB

A A

ME 274 – Summer 2020 SOLUTION Examination No. 1 (AM) PROBLEM NO. 3 (continued) PART B – 4 points Blocks A and B are connected by an inextensible cable. B is moving to the left with a speed of vB . Let vA represent the speed of block A.

Choose the response below that most accurately describes the motion of A:

a) vA < vB / 2

b) vA = vB / 2

c) vB / 2 < vA < vB

d) vA = vB

e) vA > vB

SOLUTION

L = length of cable = 2sA + sB2 + h2 + constant = constant ⇒

dLdt

= 0 = 2 !sA + 12

2 !sBsB

sB2 + h2

⇒ vA = 12

sB

sB2 + h2

vB < 12

vB

See practice problem from 6/19 tutorial session.

A

B

O

D E

ωOA

H

ICAB

ICDE

a

b

c

d e

f

g

A

B

D

E

H ωHK

A

B

D

E

H ωHK

K

ω AB = CCWω AB = 0ω AB = CW

ω BE = CCWω BE = 0ω BE = CW

ωDH = CCW

ωDH = 0ωDH = CW

ω AB >ω BE

ω AB =ω BE

ω AB <ω BE

ω BE >ωDH

ω BE =ωDH

ω BE <ωDH

ωHK >ωDH

ωHK =ωDH

ωHK <ωDH

A

B

D

E

H ωHK

A

B

D

E

H ωHK

K

ω AB = CCWω AB = 0ω AB = CW

ω BE = CCWω BE = 0ω BE = CW

ωDH = CCW

ωDH = 0ωDH = CW

ω AB >ω BE

ω AB =ω BE

ω AB <ω BE

ω BE >ωDH

ω BE =ωDH

ω BE <ωDH

ωHK >ωDH

ωHK =ωDH

ωHK <ωDH

A

B

D

E

H ωHK

A

B

D

E

H ωHK

K

ω AB = CCWω AB = 0ω AB = CW

ω BE = CCWω BE = 0ω BE = CW

ωDH = CCW

ωDH = 0ωDH = CW

ω AB >ω BE

ω AB =ω BE

ω AB <ω BE

ω BE >ωDH

ω BE =ωDH

ω BE <ωDH

ωHK >ωDH

ωHK =ωDH

ωHK <ωDH

ME 274 – Summer 2020 Name Quiz 2 Problem 1 Consider the mechanism below. Link HK is rotating in the CCW sense with a rate of ωHK . For this problem:

a) Locate the instant centers for links AB, BE, DH and HK. b) Let ω AB , ω BE , ωDH , ωHK represent the angular speeds of links AB, BE, DH and HK,

respectively. Circle the correct answers below in regard to the relative sizes of these angular speeds (three answers):

c) Circle the correct answers below regarding the direction of rotation for links AB, BE and DH (three answers).

See Quiz #2.

ME 274 – Summer 2020 SOLUTION Examination No. 1 (AM) PROBLEM NO. 3 (continued) PART C – 4 points Consider the scaled drawing of a mechanism shown below, with the locations for the instant centers for links AB and DE are shown in the figure. Link OA is rotating CCW with an angular speed of ωOA . Let ω AB ,ω BH andωDE represent the angular speeds of

links AB, BH and DE, respectively, and let vD and vE represent the speeds of points D and E, respectively. Choose the correct responses in Parts C1, C2, C3 and C4 below.

vB = aω AB = b+ c( )ω BH ⇒ ω BH = a

b+ cω AB >ω AB

vA = fω AB = f + g( )ωOA ⇒ ω AB = f + g

fωOA >ωOA

vD = cω BH = c + d( )ωDE ⇒ ω BH = c + d

cωDE >ωDE

vD = c + d( )ωDE and vE = eωDE ⇒

vDvE

= c + de

>1

Part C2 a)

b)

c)

Part C3

a)

b)

c)

Part C4

a)

b)

c)

Part C1 a)

b)

c)

ME 274 – Summer 2020 SOLUTION Examination No. 1 (AM) PROBLEM NO. 3 (continued) PART D – 4 points Consider the linkage made up of links OA and AB, with the angular orientation of these links given by the angles θ1 and θ2 . (Note that these angles are both measured from fixed, horizontal lines, as shown in the figure.) The constant rotation rates for these two links are given by

!θ1 and !θ2 . The following moving reference frame kinematics equation

is to be used to describe the acceleration of point B:

!aB = !aA + !aB/ A( )rel

+!α × !rB/ A + 2

!ω × !vB/ A( )rel

+!ω ×

!ω × !rB/ A( )

Using an observer attached to link AB, fill in the following terms below for this equation:

!ω = "θ2k

!α =!0

!vB/ A( )rel=!0

!aB/ A( )rel=!0

A

B

O θ1

θ2

L

L

x

y

A

B

O θ1

θ2

L

L

x

y

ME 274 – Summer 2013 Name Examination No. 1 PROBLEM NO. 3 (continued) PART B – 4 points

The rotation rates ω1 = 2rad / sec and ω2 = 4rad / sec are both constant. Determine the angular velocity and angular acceleration vectors for the disk.

Moving reference frame kinematics III-26 ME274

Example III.16

X

Y

P

x

L

!1

R

y

A

!2

O

ME 274 – Summer 2020 Name Quiz 3 Arm OA rotates about the fixed Y-axis with a constant rate of

ω1. The disk rotates with respect to arm OA with a constant

rate of ω2 . It is desired to find the acceleration of point P using the moving reference frame acceleration equation:

!aP = !aA + !aP/ A( )rel

+!α × !rP/ A + 2

!ω × !vP/ A( )rel

+!ω ×

!ω × !rP/ A( )

at the instant shown when the xyz-axes are aligned with the XYZ-axes, as shown in the figure.

a) Using an observer attached to arm OA, fill in the expressions for the following, and then calculate the acceleration of P using the above equation:

!ω =

!α =

!vP/ A( )rel=

!aP/ A( )rel=

!aA =

b) Using an observer attached to the disk, fill in the expressions for the following, and then calculate the acceleration of P using the above equation:

!ω =

!α =

!vP/ A( )rel=

!aP/ A( )rel=

!aA =

ME 274 – Summer 2013 Name Examination No. 1 PROBLEM NO. 3 (continued) PART B – 4 points

The rotation rates ω1 = 2rad / sec and ω2 = 4rad / sec are both constant. Determine the angular velocity and angular acceleration vectors for the disk.

Moving reference frame kinematics III-26 ME274

Example III.16

X

Y

P

x

L

!1

R

y

A

!2

O

ME 274 – Summer 2020 Name Quiz 3 Arm OA rotates about the fixed Y-axis with a constant rate of

ω1. The disk rotates with respect to arm OA with a constant

rate of ω2 . It is desired to find the acceleration of point P using the moving reference frame acceleration equation:

!aP = !aA + !aP/ A( )rel

+!α × !rP/ A + 2

!ω × !vP/ A( )rel

+!ω ×

!ω × !rP/ A( )

at the instant shown when the xyz-axes are aligned with the XYZ-axes, as shown in the figure.

a) Using an observer attached to arm OA, fill in the expressions for the following, and then calculate the acceleration of P using the above equation:

!ω =

!α =

!vP/ A( )rel=

!aP/ A( )rel=

!aA =

b) Using an observer attached to the disk, fill in the expressions for the following, and then calculate the acceleration of P using the above equation:

!ω =

!α =

!vP/ A( )rel=

!aP/ A( )rel=

!aA =

See Quiz #3.

ME 274 – Summer 2020 SOLUTION Examination No. 1 (AM) PROBLEM NO. 3 (continued) PART E – 4 points Aircraft A and B travel in the xy-plane, with A moving on a straight path and B moving on a circular path having a radius of 4000 ft, have velocity vectors of:

!vA = 300 i − 400 j( ) ft / s!vB = −300 i − 600 j( ) ft / s

respectively. With what speed is B traveling as seen by an observer on aircraft A? SOLUTION

!vB/ A = !vB − !vA = −300 i − 600 j( )− 300 i − 400 j( ) = −600i − 200 j ⇒!vB/ A = observed speed = 6002 + 2002 = 200 10 ft / s