Michael R. Lindeburg, PE

Professional Publications, Inc. • Belmont, California

FEPRACTICEPROBLEMS

CIVIL

for the Civil Fundamentals of Engineering Exam

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28 Kinetics of RotationalMotion

PRACTICE PROBLEMS

1. A 1530 kg car is towing a 300 kg trailer. The coeffi-cient of friction between all tires and the road is 0.80.The car and trailer are traveling at 100 km/h around abanked curve of radius 200 m. What is most nearly thenecessary banking angle such that tire friction will NOTbe necessary to prevent skidding?

(A) 8.0�

(B) 21�

(C) 36�

(D) 78�

2. Why does a spinning ice skater’s angular velocityincrease as she brings her arms in toward her body?

(A) Her mass moment of inertia is reduced.

(B) Her angular momentum is constant.

(C) Her radius of gyration is reduced.

(D) all of the above

3. A 1 m long uniform rod has a mass of 10 kg. It ispinned at one end to a frictionless pivot. What is mostnearly the mass moment of inertia of the rod takenabout the pivot point?

(A) 0.83 kg�m2

(B) 2.5 kg�m2

(C) 3.3 kg�m2

(D) 10 kg�m2

4. In the linkage mechanism shown, link AB rotateswith an instantaneous counterclockwise angular velocityof 10 rad/s.

34 C

BA

D

5 m

�AB � 10 rad/s (counterclockwise)

5 m

5 m

What is most nearly the instantaneous angular velocityof link BC when link AB is horizontal and link CD isvertical?

(A) 2.3 rad/s (clockwise)

(B) 3.3 rad/s (counterclockwise)

(C) 5.5 rad/s (clockwise)

(D) 13 rad/s (clockwise)

5. Two 2 kg blocks are linked as shown.

A

B

30°

3 m

3 m/s

Assuming that the surfaces are frictionless, what is mostnearly the velocity of block B if block A is moving at aspeed of 3 m/s?

(A) 0 m/s

(B) 1.3 m/s

(C) 1.7 m/s

(D) 5.2 m/s

6. A car travels on a perfectly horizontal, unbankedcircular track of radius r. The coefficient of frictionbetween the tires and the track is 0.3. If the car’s veloc-ity is 10 m/s, what is most nearly the smallest radius thecar can travel without skidding?

(A) 10 m

(B) 34 m

(C) 50 m

(D) 68 m

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Dynamics

7. A uniform rod (AB) of length L and weight W ispinned at point C. The rod starts from rest and accel-erates with an angular acceleration of 12g/7L.

L

L4

A

C

B

�

What is the instantaneous reaction at point C at themoment rotation begins?

(A)W4

(B)W3

(C)4W7

(D)7W12

8. A wheel with a 0.75 m radius has a mass of 200 kg.The wheel is pinned at its center and has a radius ofgyration of 0.25 m. A rope is wrapped around the wheeland supports a hanging 100 kg block. When the wheel isreleased, the rope begins to unwind. What is mostnearly the angular acceleration of the wheel as the blockdescends?

(A) 5.9 rad/s2

(B) 6.5 rad/s2

(C) 11 rad/s2

(D) 14 rad/s2

9. A car travels around an unbanked 50 m radius curvewithout skidding. The coefficient of friction between thetires and road is 0.3. What is most nearly the car’smaximum velocity?

(A) 14 km/h

(B) 25 km/h

(C) 44 km/h

(D) 54 km/h

10. A uniform rod (AB) of length L and weight W ispinned at point C and restrained by cable OA. The cableis suddenly cut. The rod starts to rotate about point C,with point A moving down and point B moving up.

L

L4

A

O

C

B

The instantaneous linear acceleration of point B is

(A)3g

16

(B)g

4

(C)3g

7

(D)3g

4

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28-2 F E C I V I L P R A C T I C E P R O B L E M S

Dynamics

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1. The necessary superelevation angle without relyingon friction is

�¼ arctanv2

gr

¼ arctan100

kmh

� �1000

mkm

� �� �2

9:81ms2

� �ð200 mÞ 3600

sh

� �2

¼ 21:47� ð21�Þ

The answer is (B).

2. As the skater brings her arms in, her radius of gyra-tion and mass moment of inertia decrease. However, inthe absence of friction, her angular momentum, H, isconstant.

! ¼ HI

Since angular velocity, !, is inversely proportional to themass moment of inertia, the angular velocity increaseswhen the mass moment of inertia decreases.

The answer is (D).

3. The mass moment of inertia of the rod taken aboutone end is

I rod ¼ ML2

3¼ ð10 kgÞð1 mÞ2

3

¼ 3:33 kg�m2 ð3:3 kg�m2Þ

The answer is (C).

4. Find the instantaneous center of rotation. The abso-lute velocity directions at points B and C are known.The instantaneous center is located by drawing perpen-diculars to these velocities, as shown. The angular veloc-ity of any point on rigid body link BC is the same at thisinstant.

4 m

3 m

O

IC

C

B

3

4ωBC

vB = 50 m/s

vC

The velocity of point B is

vB ¼ AB!AB ¼ ð5 mÞ 10rads

� �¼ 50 m=s

The angular velocity of link BC is

!BC ¼ vBOB

¼50

ms

4 m

¼ 12:5 rad=s ð13 rad=sÞ ½clockwise�

The answer is (D).

5. The instantaneous center of rotation for the sliderrod assembly can be found by extending perpendicularsfrom the velocity vectors, as shown. Both blocks can beassumed to rotate about point C with angularvelocity !.

A

BC

30∘

3 sin 30∘

3 cos 30∘

3 mvA

vB

The velocity of block B is

!¼ vACA

¼ vBBC

vB ¼ vABC

CA¼

3ms

� �ð3 mÞcos 30�

ð3 mÞsin 30�¼ 5:2 m=s

The answer is (D).

6. If there is no skidding, the frictional force, Ff, willequal the centrifugal force, Fc. From the equations forcentrifugal force and frictional force, the smallest possi-ble radius is

Fc ¼ mv2

r

Ff ¼ �N ¼ �mg

mv2

r¼ �mg

r ¼ v2

�g¼

10ms

� �2

ð0:3Þ 9:81ms2

� � ¼ 34 m

The answer is (B).

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K I N E T I C S O F R O T A T I O N A L M O T I O N 28-3

Dynamics

7. The mass moment of inertia of the rod about itscenter of gravity is

ICG ¼ ML2

12¼ W

g

� �L2

12

� �

Take moments about the center of gravity of the rod.All moments due to gravitational forces will cancel. Theonly unbalanced force acting on the rod will be thevertical reaction force, RC, at point C.

åMCG ¼ RCL4

� �¼ ICG�CG

RCL4

� �¼ W

g

� �L2

12

� �� �12g

7L

� �

RC ¼ 4W7

The angular velocity is zero, so the center of the massdoes not have a component of acceleration in the hori-zontal direction. There is no horizontal force componentat point C.

The answer is (C).

8. From the equation for the radius of gyration, themass moment of inertia of the wheel is

r ¼ffiffiffiffiffiffiffiffiffiffiffiffiffi

Imwheel

r

I ¼ mwheelr2

The unbalanced moment on the wheel is

M ¼ FR ¼ ðmg �maÞR ¼ mRðg � aÞ¼ mblockRðg � R�Þ

The acceleration is given by

M ¼ I�

mblockRðg � R�Þ ¼ mwheelr2�

Combine the equations and solve.

�¼ mblockRg

mwheelr2 þmblockR2

¼ð100 kgÞð0:75 mÞ 9:81

ms2

� �

ð200 kgÞð0:25 mÞ2 þ ð100 kgÞð0:75 mÞ2¼ 10:7 rad=s2 ð11 rad=s2Þ

The answer is (C).

9. If the car does not skid, the frictional force and thecentrifugal force must be equal. From the equations forcentrifugal force and frictional force, the car’s maximumvelocity is

Fc ¼ mv2

r

Ff ¼ �N ¼ �mg

mv2

r¼ �mg

v¼ ffiffiffiffiffiffiffiffi�gr

p

¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ð0:3Þ9:81

ms2

� �60

smin

� �260

minh

� �2

1000mkm

0B@

1CA

� 50 m

1000mkm

0@

1A

vuuuuuuuuuuut

¼ 43:67 km=h ð44 km=hÞ

The answer is (C).

10. Point C is L/4 from the center of gravity of the rod.The mass moment of inertia about point C is

IC ¼ ICG þMd2 ¼ ML2

12þM

L4

� �2

¼ 748

� �ML2

The sum of moments on the rod is

åMC ¼åFr ¼ 3W4

� � 3L42

0B@

1CA� W

4

� � L42

0B@

1CA

¼ WL4

¼ MgL

4

The angular acceleration is

�¼åMC

IC¼

MgL

4748

� �ML2

¼ 12g

7L

The tangential acceleration of point B is

at;B ¼ r� ¼ L4

� � 12g

7L

� �¼ 3g

7

The answer is (C).

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28-4 F E C I V I L P R A C T I C E P R O B L E M S

Dynamics