AP Physics Rotational Motion Introduction

AP Physics Rotational Motion

Introduction: Which moves with greater speed on a merry-go-round - a horse near the center or one near the outside? Your

answer probably depends on whether you are considering the translational or rotational motion of the horses. Have you ever

linked arms with friends at a skating rink while making a turn? If you have, you probably noticed that the person on the

inside moved very little while the person on the outside had to run to keep up. The outside person traveled a greater distance

per period of time and therefore had the greater translational speed. During the same period of time all skaters rotated

through the same angle per period of time and had the same rotational speed.

In our previous study of motion we discussed translational motion - that is the motion of bodies moving as a whole

without regard to rotation. In this unit we will extend our ideas of motion to include the rotation of a rigid body about a fixed

axis. If the axis is inside the body we tend to say the body rotates about its axis. If the axis is outside the body, we say the

body revolves about an axis. An example of this would be the earth which daily rotates about its axis and yearly revolves

around the sun.

An object rotating about an axis tends to remain rotating about the same axis unless acted upon by a net external

influence. This property of a body to resist changes in its rotational state is called rotational inertia. The rotational inertia of

a body depends on the amount of mass the body possesses and on the distribution of that mass with respect to the axis of

rotation. The greater the distance of the bulk of the mass from the axis of rotation - the greater the rotational inertia.

A long pendulum has a greater rotational inertia than a short one. The period of a pendulum is directly proportional

to the square root of the length of the pendulum. It takes more time to change the rotational inertia of a long pendulum as it

swings back and forth. People and animals with long legs tend to walk with slower strides than those with short legs for the

same reason. Have you ever tried running with your legs straight?

Performance Objectives: Upon completion of the readings and activities of the unit and when asked to respond either orally

or on a written test, you will be able to:

• state relationships between linear and angular variables.

• recognize that the rotational kinematics formulas are analogous to the translational ones. Use these formulas to

solve problems involving rotating bodies.

• define rotational inertia or moment of inertia. Calculate the rotational inertia for a point mass, a system of point

masses, and rigid bodies.

• use the parallel axis theorem to find the moment of inertia about an axis other than the center of mass.

• calculate the kinetic energy of a rotating body.

• define torque. Calculate the net torque acting on a body.

• state Newton’s second law for rotation. Recognize that the rotational dynamics formulas are analogous to the

translational ones. Use these formulas to solve problems involving rotating bodies.

• use the work-kinetic energy theorem for rotation to solve problems.

Textbook Reference: Tipler: Chapter 9

Glencoe Physics: Chapter 8

"To every thing -- turn, turn, turn

there is a season -- turn, turn, turn

and a time for every purpose under heaven."

-- The Byrds (with a little help from Ecclesiastes)

Recall: From the definition of a radian (arc length/radius) θ = s/r, where s is the arc length, r is the radius and θ is the angle

measure in radians. The following quantities are called the bridges between linear and angular measurements: s = rθ

v = rω aT = rα aR = v2/r = rω2

Definitions and Conversions:

1. What angle in radians is subtended by an arc 3.0 m in

length, on the circumference of a circle whose radius is2.0

m? 1.5 rad

2. What angle in radians is subtended by an arc of length

78.54 cm on the circumference of a circle of diameter

100.0 cm? What is the angle in degrees?

1.57 rad 90˚

3. The angle between two radii of a circle of radius 2.00

m is 0.60 rad. What length of arc is intercepted on the

circumference of the circle by the two radii? 1.2 m

4. What is the angular velocity in radians per second of a

flywheel spinning at the rate of 7230 revolutions per

minute? 757 rad/sec

5. If a wheel spins with an angular velocity of 625 rad/s,

what is its frequency in revolutions per minute?

5968 rpm

6. Compute the angular velocity in rad/s, of the

crankshaft of an automobile engine that is rotating at 4800

rev/min. 503 rad/sec

Rotational Kinematics: Rotational motion is described

with kinematic formulas just like the translational motion

formulas. To get the rotational kinematic formulas,

substitute the rotational variables.

7. A flywheel accelerates uniformly from rest to an

angular velocity of 94 radians per second in 6.0 seconds.

What is the angular acceleration of the flywheel in radians

per second squared? 16 rad/s2

8. a) Calculate the angular acceleration in radians per

second squared of a wheel that starts from rest and attains

an angular velocity of 545 revolutions per minute in 1.00

minutes. b) What is the angular displacement in radians

of the wheel during the first 0.500 minutes? c) During

the second 0.500 minutes?

0.95 rad/s2 428 rad. 1283 rad

9. A fly wheel requires 3.0 seconds to rotate through 234

rad. Its angular velocity at the end of this time is 108

rad/s. Find a) the angular velocity at the beginning of the

3 second interval; b) the constant angular acceleration.

48 rad/s 20.0 rad/s2

10. A playground merry-go-round is pushed by a child.

The angle the merry-go-round turns through varies with

time according to θ(t) = 2t + 0.05t3, where θ is in radians

and t is in seconds. a) Calculate the angular velocity of the

merry-go-round as a function of time. ω = 2 + 0.15t2

b) What is the initial value of the angular velocity?

2 rad/s

c) Calculate the instantaneous velocity at t = 5.0 sec.

d) Calculate the average angular velocity for the time

interval t = 0 to t = 5 seconds. 5.75 rad/s 3.25 rad/s

11. A bicycle wheel of radius 0.33 m turns with angular

acceleration α = 1.2 – 0.4t, where α is in rad/s2 and t is in

seconds. It is at rest at t = 0.

a) Calculate the angular velocity and angular

displacement as functions of time.

b) Calculate the maximum positive angular velocity and

maximum positive angular displacement of the wheel.

ω = 1.2t - 0.2t2 θ = 0.6t2 -0.067t3 1.8 rad/s 7.2 rad

12. A roller in a printing press turns through an angle θ

given by θ(t) = 2.50t2 – 0.400t3.

a) Calculate the angular velocity of the roller as a function

of time. ω(t) = 5t – 1.2t2

b) Calculate the angular acceleration of the roller as a

function of time. α(t) = 5 – 2.4t

c) What is the maximum positive angular velocity and at

what value of t does it occur?

5.21 rad/s 2.08 sec

13. A wheel rotates with a constant angular velocity of 10

rad/s. a) Compute the radial acceleration of a point 0.5 m

from the axis using the relation, a⊥ = rω2. 50 m/s2

b) Find the tangential velocity of the point, and compute

its radial acceleration from the relation,

ac = v2/r. 5m/s 50 m/s2

Rotational Inertia is the resistance of a rotating body to

changes in its angular velocity. According to Newton's

First Law a body tends to resist a change in its motion.

The amount of inertia a body possesses is directly related

to the mass. For rotational motion, an analogous

situation exists. However, rotational inertia depends on

the mass and on the distribution of the mass about the

axis of rotation. This quantity that relates mass and

position of the mass relative to the axis of rotation

is called the moment of inertia and has units of kg-m2.

The symbol for moment of inertia is I. For a point mass m

a distance r from the axis of rotation, the moment of

inertia will be I = mr2. For bodies made up of several

small masses just add all the moments of inertia together.

For bodies which are not composed of discrete point

masses but are continuous distributions of matter, the

methods of calculus must be used to find the moment of

inertia.

14. Small blocks, each of mass 2.0 kg, are clamped at the

ends and at the center of a light rod 1.2 m long. Compute

the moment of inertia of the system about an axis passing

through a point one-third of the length from one end of

the rod if the moment of inertia of the light rod can be

neglected. 1.68 kg-m2

15. Four small spheres, each of mass 0.200 kg, are

arranged in a square 0.400 m on a side and connected by

light rods of negligible mass. Find the moment of inertia

of the system about an axis

a) perpendicular to the plane of the square through the

center. 0.0640 kg-m2

b) bisecting two opposite sides of the square.

0.0320 kg-m2

16. What is the rotational inertia of a solid ball 0.50 m in

radius that weighs 80.0 N if it is rotated about a diameter?

0.816 kg-m2

17. What is the rotational inertia of a thick ring that is

rotating about an axis perpendicular to the plane of the

ring passing through its center? The ring has a mass of

1.20 kg and a diameter of 45.0 cm. The hole in the ring is

15.0 cm wide. 0.0340 kg-m2

18. Find the moment of inertia about each of the

following axes for a rod that is 4.00 cm in diameter and

2.00 m long and has a mass of 8.00 kg. a) An axis

perpendicular to the rod and passing through its center.

b) An axis perpendicular to the rod and passing through

one end. c) A longitudinal axis passing through the center

of the rod.

2.67 kg-m2 10.67 kg-m2 0.0016 kg-m2

Parallel Axis Theorem: The moment of inertia of any

object about an axis through its center of mass is the

minimum moment of inertia for an axis in that direction of

space. The moment of inertia about any axis parallel to

that axis through the center of mass is given by:

Iparallel axis = Icom + Md2

…where Iparallel axis is the moment of inertia about an new

axis (parallel to the original axis of rotation), Icom is the

moment of inertia about the center of mass, M is the mass

of the object and d is the distance between the original

axis of rotation about the center of mass and the new

proposed axis of rotation.

19. Use the parallel axis theorem to calculate the moment

of inertia of a uniform thin rod of mass M and length l for

an axis perpendicular to the rod at one end. Ml2/3

20. Use the parallel axis theorem to calculate the moment

of inertia of a square sheet of metal of side length a and

mass M for an axis perpendicular to the sheet and passing

through one corner. 2Ma2/3

21. The four objects shown in the figure below have

equal masses m.

Object A is a solid cylinder of radius R. Object B is a

hollow, thin cylinder of radius R. Object C is a solid

square whose length of side = 2R. Object D is the same

size as C, but hollow (i.e., made up of four thin sticks).

The objects have axes of rotation perpendicular to the

page and through the center of gravity of each object.

a) Which object has the smallest moment of inertia?

b) Which object has the largest moment of inertia?

Kinetic Energy of Rotation: Because a rotating rigid

body consists of particles in motion, it has kinetic energy.

This kinetic energy is computed using the moment of

inertia of the body and the angular velocity. KE = ½Iω2

22. The rotor of an electric motor has a rotational inertia

of 45 kg-m2. What is its kinetic energy if it turns at 1500

revolutions per minute? 555 kJ

23. A grinding wheel in the shape of a solid disk is 0.200

m in diameter and has a mass of 3.00 kg. The wheel is

rotating at 3600 rev/min about an axis through its center.

a) What is its kinetic energy? 1066 J

b) How far would it have to drop in free fall to acquire the

same kinetic energy? 36.3 m

24. The flywheel of a gasoline engine is required to give

up 300.0 J of kinetic energy while its angular speed

decreases from 660 rev/min to 540 rev/min. What

moment of inertia is required for the wheel? 0.380 kg-m2

25. A phonograph turntable has a kinetic energy of

0.0700 J when turning at 78 rpm. What is the moment of

inertia of the turntable about the rotation axis?

26. Energy is to be stored in a large flywheel in the shape

of a disk with radius of 1.20 m and a mass of 80.0 kg. To

prevent structural failure of the flywheel, the maximum

allowed radial acceleration of a point on its rim is 5000

m/s2. What is the maximum kinetic energy that can be

stored in the flywheel? 1.20 x 105 J

27. Two blocks, one of mass 4.0 kg and the other of mass

2.0 kg are connected

by a light rope that passes

over a pulley as shown in the

figure to the right. The pulley

has radius 0.20 m and moment

of inertia 0.32 kg-m2. The

rope does not slip on the

pulley rim. The larger mass

is 5.0 m above the floor and

released from rest. Use

energy methods to calculate the velocity of the 4-kg block

just before it strikes the floor. 3.74 m/s

Torque: To change the translational inertia of a body

you have to apply a net external force. To change the

rotational inertia of a body you have to apply a torque

(rhymes with fork). If you studied torque in previous

science courses it was probably defined as the product of

the force and the length of the torque arm. The torque

arm (sometimes called lever arm) is the perpendicular

distance between the line of action of the force and the

axis of rotation.

In order to solve problems involving torque, you

need to understand how torque is calculated and then be

able to calculate the net torque acting on a body.

28. Calculate the torque (magnitude and direction) about

point 0 due to the force F in each of the situations

sketched in the figure. In each case the object to which

the force is applied has length 4.00 m, and the force

F =20.0 N.

a) 80.0 m-N ccw b) 69.3 m-N ccw

c) 40.0 m-N ccw d) 34.6 m-N cw e) 0 f) 0

29. Calculate the resultant torque about point O for the

two forces applied in the figure below. 28 m-N cw

30. Calculate the net torque (magnitude and direction) on

the beam shown in the figure below about

a) an axis through O, perpendicular to the figure.

29.5 m-N ccw

b) an axis through C, perpendicular to the figure.

35.6 m-N ccw

31. Find the net torque on

the wheel in the figure

about the axle through O if

a = 10 cm and b = 25 cm.

3.55 m-N cw

Rotational Dynamics: In studying translational

dynamics we made use of Newton's Second Law, which

related the acceleration of a body and the forces applied

to the body. An analogous relationship exists between

angular acceleration and a quantity we call a torque.

Qualitatively speaking, torque is the tendency of a force

to cause a rotation of the body on which it acts.

Mathematically speaking, torque is defined as the cross

product of the moment arm and the applied force. The

moment arm is the perpendicular distance between the

force applied and the axis of rotation. The unit for torque

is a meter-newton. The symbol for torque is the lower-

case Greek letter tau, τ.

32. A net force of 10.0 N is applied tangentially to the

rim of a wheel having a 0.25 m radius. If the rotational

inertia of the wheel is 0.500 kg m2, what is its angular

acceleration? 5 rad/s2

33. A solid ball is rotated by applying a force of 4.7 N

tangentially to it. The ball has a radius of 14 cm and a

mass of 4.0 kg. What is the angular acceleration of the

ball? 21 rad/s2

34. A fly wheel in the shape of a thin ring has a mass of

30.0 kg and a diameter of 0.96 m. A torque of 13 m-N is

applied tangentially to the wheel. How long will it take

for the flywheel to attain an angular velocity of 10.0

rad/s? 5.3 sec

35. A cord is wrapped around the rim of a flywheel

0.5 m in radius, and a steady pull of 50.0 N is exerted on

the cord. The wheel is mounted with frictionless bearings

on a horizontal shaft through its center. The moment of

inertia of the wheel is 4.0 kg-m2. Compute the angular

acceleration of the wheel. 6.25 rad/s2

36. A grindstone in the shape of a solid disk with a

diameter of 1.0 m and a mass of 50.0 kg, is rotating at 900

rev/min. A tool is pressed against the rim with a normal

force of 200.0 N, and the grindstone comes to rest in 10.0

s. Find the coefficient of friction between the tool and the

grindstone. Neglect friction in the bearings. 0.589

37. A 5.0 kg block rests on

a frictionless horizontal

surface. A cord attached to

the block passes over a pulley

whose diameter is 0.2 m, to

a hanging block also of mass

5.0 kg. The system is

released from rest, and the

blocks are observed to move 4.0 m in 2.0 seconds. a)

What is the tension in each part of the cord?

10 N 39N

b) What is the moment of inertia of the pulley?

0.145 kg-m2

38. Two blocks, one of mass 4.0 kg

and the other of mass 2.0 kg are

connected by a light rope that passes

over a pulley as shown in the figure

to the right. The pulley has radius

0.10 m and moment of inertia

0.20 kg-m2. Find the linear

accelerations of Blocks A and B,

the angular acceleration of wheel C, and the tension in

each side of the cord

a) if the surface of the wheel is frictionless;

aA = aB = 3.27 m/s2; aC = 0; TA = TB = 26.1 N

b) if there is no slipping between the cord and the surface

of the wheel.

aA = aB = 0.754 m/s2; aC = 7.54 rad/s2;

TA = 36.2 N; TB = 21.1 N

Equilibrium of Rigid Body:

Recall that we said the first condition for

equilibrium existed when the sum of the forces acting on

the body was zero. Now we introduce the second

condition for equilibrium which exists when the sum of

the torques of all the forces acting on the body, with

respect to any specified axis is zero. This means that the

body is not accelerating and it is not rotating. If it were

rotating then it would experience a centripetal

acceleration.

39. A 200.0 N weight is hung on the end of a horizontal

pole 2.0 m long. What is the torque around the other end

of the pole caused by this weight? Around the center of

the pole? 400 mN 200 mN

40. Two men carry a 1500 N load by hanging it from a

horizontal pole that rests on one shoulder of each man. If

the men are 3.00 m apart and the load is 1.00 m from one

of them, how much load does each man support? The

weight of the pole is 500 N. 1250 N 750 N

Conceptual Questions:

1. Does a record player needle ride faster or slower over the

groove at the beginning or the end of the record? If fidelity

increases with translational speed, what part of the record

produces the highest fidelity?

2. Suppose the first and last selections on a phonograph record

are 3-minutes cuts. Which, if either, of these cuts is wider on

the record? (That is, which contains more grooves along a radial

direction?)

3. Which moves faster on a merry-go-round, a horse near the

center or one near the outside.

4. If you use large diameter tires on your car, how will your

speedometer reading differ?

5. Why are the front wheels located so far out in front on the

racing vehicle?

6. Which will roll down a hill faster, a cylinder or a sphere of

equal radii? A hollow cylinder or a solid cylinder of equal radii?

Explain.

7. Why do buses and heavy trucks have large steering wheels?

8. Which is easier for turning stubborn screws, a screwdriver

with a thick handle or one with a long handle? Explain.

9. Why is the middle seating most comfortable in a bus

traveling on a bumpy road?

10. Explain why a long pole is more beneficial to a tightrope

walker if it droops.

11. Why do you bend forward when carrying a heavy load on

your back?

12. Why is it easier to carry the same amount of water in two

buckets, one in each hand, then in a single bucket?

13. Using the ideas of torque and center of gravity, explain why

a ball rolls down a hill.

14. Why is it dangerous to roll open the top drawers of a fully

loaded file cabinet that is not secured to the floor?

15. Why is less effort required in doing sit-ups when your arms

are extended in front of you? Why is it more difficult when

your arms are placed in back of your head?

16. For a rotating wheel, how do the directions of the linear

velocity vector and the angular velocity vector compare at the

same instant of time?

Answers to conceptual questions:

1. The phonograph needle rides faster at the beginning of the

record. Since fidelity is enhanced with translational speed, then

fidelity would be best at the beginning of a record.

2. Both three minute selections would have the same width

because they would make the same number of revolutions

during a three minute time period.

3. The horse on the outer rail has a greater translational

(tangential) speed, while both have the same rotational speed.

4. The circumference of a large diameter tire is greater,

meaning it will move a greater distance per revolution, which

results in a greater speed than that shown on the speedometer.

5. The long distance to the front wheels increases the rotational

inertia of the vehicle without appreciably adding to its weight.

As the back wheels are driven clockwise, the rest of the car

tends to rotate counter-clockwise. This would lift the front

wheels off the ground.

6. A sphere will roll faster because it has less rotational inertia

than a cylinder. A solid cylinder will roll faster than a hollow

cylinder for the same reason.

7. The large radius of a large steering wheel allows the driver to

exert more torque for a given force.

8. More torque can be exerted by the screw driver having a

thick handle.

9. A rocking bus rocks about its center of gravity which is

around the center of the bus. It works something like a see-saw

- the farther from the center, the more you go up and down.

10. The long drooping pole lowers the center of gravity of the

pole and the tightrope walker. The pole contributes to his

rotational inertia.

11. You bend forward to shift the center of gravity of you and

the back pack. If you did not shift the center of gravity over the

support, you would topple over.

12. There is no need to adjust your center of gravity if the water

is distributed between the two buckets.

13. When a ball is on an incline its center of gravity is not

above the point of support. The weight acts some distance from

the point of support and produces a torque about the point of

support.

14. The center of gravity could be adjusted so that it is no

longer above the support.

15. When your arms are extended in front of you while doing

sit-ups, not only are they not lifted as far, they are closer to the

axis of rotation and give you less rotational inertia. When

behind your head they are lifted farther and their farther distance

from the axis of rotation increases your rotational inertia.

Introduction: Rolling: In the last unit we studied the rotation of a rigid body about a fixed axis. We will now extend our

study to include cases where the axis of rotation moves, that is, where one body experiences both rotational and translational

motion at the same time. Common examples would be a yo-yo unwinding down a string or a ball rolling across the floor.

Newton’s Second Law of Rotation is still valid if the axis of rotation passes through the center of mass of the body and does

not change its direction.

Another useful concept to employ when looking at a body rotating about a moving axis is the total kinetic energy of

the body. The combined kinetic energy for a body of mass m moving with a center-of-mass velocity v and rotating with

angular velocity ω about an axis through the center of mass, the total kinetic energy of the body is:

...where Icom is the moment of inertia about the axis through the center of mass.

Torque: In the previous unit we defined torque qualitatively as the tendency of a force to cause the rotation of a

rigid body on which the force acts. All the particles moved in circular paths around a fixed axis. Mathematically we defined

torque as the cross product of the moment or lever arm and the applied force, where the moment arm is the perpendicular

distance between where the force is applied and the axis of rotation. In this unit we will expand the definition of torque to

apply to a particle that moves along any path relative to a fixed point. We will explore the vector nature of torque and

consider the cross product of the position vector and the force vector.

Angular Momentum: Every rotational quantity that we have seen in the previous unit is the analog of some

quantity of motion of a particle. The analog of momentum of a particle is angular momentum, a vector quantity denoted by

L. Its relation to p, the linear momentum, is exactly the same as the relation of torque to force. The rate of change of

angular momentum of a particle equals the torque of the net force acting on it. The direction of the angular momentum is

given by the right-hand rule for angular velocity. The angular momentum for a system of particles is the sum of all the

angular momenta of the particles. The angular momentum for a solid body is given by L = Iω.

Angular momentum also forms the basis for a very important conservation principle, the principle of conservation of

angular momentum. Like the conservation laws of energy and of linear momentum, this principle appears to be a universal

conservation law, valid at all scales from atomic and nuclear systems to the motions of galaxies. The total angular

momentum of an isolated system is constant.

Performance Objectives: Upon completion of the readings and activities of the unit and when asked to respond either

orally or on a written test, you will be able to:

• describe the rolling motion of an object as the rotational motion about an axis through the center of mass of the

object while the center of mass moves in linear motion.

• know that the linear speed of a point on the rim of a rolling object varies from zero at the point of contact to twice

the speed of the center of mass.

• explain that static friction provides the torque to rotate the moving object. Calculate the torque using the coefficient

of static friction, the normal force and the torque arm.

• solve problems involving forces acting on a rolling object.

• show that the kinetic energy for a rolling object is the sum of the linear and angular kinetic energies of the object.

Solve problems using conservation of energy.

• know and explain that for a body rolling down an incline the speed at the bottom depends upon a dimension less

quantity β, the coefficient in the rotational inertia equation. not I, m, or R alone.

• define angular momentum and show that it is analogous to the definition for linear momentum.

• understand and explain that a particle has angular momentum relative to a point at the origin if a particle is moving

with constant angular velocity in a circular path around the origin or if the particle is moving in a straight line with

constant linear speed relative to the origin.

• determine the angular momentum for a rigid body about a fixed axis and for a system of particles.

• state the law of conservation of angular momentum. Use the law to solve problems for a rigid body and for a system

of particles.

Textbook Reference: Tipler: Chapter 9, Section 6 and Chapter 10

Glencoe Physics: Chapter 8

Problem Solving Strategies for Rotation About a Moving Axis: Recall that when Newton’s Second Law was used to

solve problems, an equation of motion (Fnet = Fapplied – Fopposing or τnet = τapplied – τopposing) was needed for each moving body.

Now one body is moving with two motions so two equations will be needed, one for translational motion and one for

rotational motion. Energy methods can also be used to solve problems involving the rolling motion of a rigid body. If the

body rolls without slipping, mechanical energy is conserved and Ui + Ki = Uf + Kf. Although there is no rotational potential

energy, the potential energy of the center of mass of the rolling body is considered. Remember that a rolling body has both

translational and rotational kinetic energy.

41. A string is wrapped several times around the rim of a

small hoop. The hoop has radius 0.08 m and mass 1.2 kg.

If the free end of the string is held in place and the hoop is

released from rest, calculate

a) the tension in the string while the hoop is descending;

5.88 N

b) the time it takes the hoop to descend 0.5 m; 0.452 s

c) the angular velocity of the rotating hoop after it has

descended 0.5 m. 27.7 rad/s

42. A string is wrapped several times around a solid

cylinder of mass 6.00 kg and radius 0.150 m. The end of

the string is held stationary while the cylinder is released

from rest. Find the downward linear acceleration of the

cylinder and the tension in the string.

6.53 m/s2 19.6 N

43. A string is wound around a

uniform solid disk of radius R and

mass M. The disk is released from

rest with the string vertical and its top

end tied to a fixed support as shown

in the figure. As the disk descends,

find: (a) the tension in the string;

(b) the acceleration of the center of

mass, and (c) the velocity of the center of mass.

T = mg/3 a = 2g/3 v = (4gh/3)½.

(Did you notice that the linear acceleration downward for

the last three problems was 2g/3? This is true for all

solid disks or solid cylinders that ever were and ever will

be.)

The Yo-Yo:

44. A yo-yo consists of two disks of total mass M and

radius R connected by a massless shaft of radius r. What

is the liner acceleration of the yo-yo when released?

45. A yo-yo has a rotational inertia of 950 g-cm2 and a

mass of 120 g. Its axle radius is 3.2 mm, and its string

is120 cm long. The yo-yo rolls from rest down to the end

of the string.

a) What is the magnitude of its linear acceleration?

0.13 m/s2

b) How long does it take to reach the end of the string?

4.4 s

As it reaches the end of the string, what are its

c) linear speed, 0.55 m/s

d) translational kinetic energy, 0.018 J

e) rotational kinetic energy, 1.4 J

f) angular speed? 170 rad/s

Rolling Motion of a Rigid Body: When a body rolls, it experiences rotational motion about a moving axis. If a cylinder

rolls in a straight path, the center of mass moves

in a straight line while the rest of the body

experiences rotational motion about the axis

through the center of mass. It is convenient to

look at this motion as a combination of rotation

about the center of mass and the translation of the

center of mass.

Rolling motion is only possible

if a frictional force is present between the object

and the surface. The frictional force (static) is

necessary to produce a net torque about the

center of mass.

Despite the presence of friction, there is no loss of mechanical energy since the contact point is at rest relative to the surface

at any instant. On the other hand, if the rigid body were to slide, mechanical energy would be lost as motion

progressed. Mechanical energy is conserved if a body rolls without slipping. Rolling without slipping means that:

vcom = Rω and acom = Rα

If an object is set into motion on a surface without these equalities, sliding (kinetic) friction will act on the object as it slips

until rolling without slipping is established.

Another way to look at rolling without slipping is from the point of view of a

stationary observer watching a bug on the rim of a rolling bicycle wheel. The stationary

observer sees the bug rotate about a stationary axis at point P. The angular speed ω of the bug

about point P is the same as the angular speed of the wheel about the axis at the center of mass.

The instantaneous linear velocity vectors are in a direction perpendicular to the radius drawn

from the bug to the contact point P. At any instant, the point P is at rest relative to the surface

since sliding does not occur. The bug on the rim moves in a complex path, called a cycloid.

The linear speed of the bug varies from zero at the point of contact at the bottom of the wheel

to 2vcom at the top of the wheel.

The total kinetic energy of a rolling body is the sum of the rotational kinetic energy about the center of mass, and the

translational kinetic energy of the center of mass.

Translational Motion + Rotational Motion = Rolling Motion

Rolling Without Slipping Problems

46. A solid cylinder of mass 10 kg rolls without slipping

on a rough surface. At the instant its center of mass has a

speed of 10 m/s, determine

a) the translational kinetic energy of its center of mass,

b) the rotational kinetic energy about its center of mass,

c) its total kinetic energy.

d) What fraction of the kinetic energy is associated with

the motion of translation and what fraction with the

motion of rotation about the axis through the center of

mass? 500 J 250 J 750 J 67 % 33 %

47. A solid sphere has a radius of 0.2 m and a mass of

150 kg. How much work is required to get the sphere

rolling with an angular speed of 50 rad/s on a horizontal

surface (Assume the sphere starts from rest and rolls

without slipping.) 10,500 J

48. The center of mass of a uniform cylinder of mass M

and radius R is moving with speed v on a horizontal

surface. The cylinder rolls without slipping. Find the

total kinetic energy of the cylinder with respect to the

reference frame fixed to the horizontal surface.

0.750 Mv2

49. A bowling ball has a mass of 8.00 kg. If it rolls down

the alley, without slipping, at 7.00 m/s, calculate

a) the linear kinetic energy; 196 J

b) the rotational kinetic energy; 78.4 J

c) the total kinetic energy. 274.4 J

50. A metal ring, mass 2.50 kg, rolls along a horizontal

surface with a constant velocity of 5.25 m/s. What is the

total kinetic energy of the moving ring? 69 J

51. A hoop of radius 3.0 m has a mass of 140 kg. It rolls

without slipping along a horizontal floor so that its center

of mass has a speed of 0.15 m/s. How much work must

be done on the hoop to stop it? -3.15 J

Things begin to get tricky when we look at

objects rolling down inclines. You can save yourself a lot

of wasted effort if you spend some time thinking about

what happens when you release a basketball down a

ramp. If the ramp were frictionless, the ball would just

slip along and not rotate. We know that doesn't happen,

so the ramp must apply a frictional force to the ball (and

vice-versa).

52. The figure below shows a hollow cylinder rolling

down a ramp inclined at an angle θ.

a) Label the forces acting on the ball.

b) Which friction force (static or kinetic) is acting on the

cylinder as it rolls down the incline?

c) Write a translational and a rotational equation of

motion for the rolling cylinder.

d) Solve for the linear acceleration down the ramp.

53. A solid cylinder of mass 4.00 kg rolls, without

slipping down a 30˚ slope. Find (a) the acceleration,

(b) the friction force, and (c) the minimum coefficient of

friction needed to prevent slipping. 3.26 m/s2 6.5 N 0.19

54. A uniform solid disk and a uniform hoop are

placed side by side at the top of a rough incline of height

h. a) If they are released from rest and roll without

slipping, determine their velocities when they reach the

bottom. vd = (4gh/3)½ vh = (gh)½

55. A 2.0 kg disk starts at the

top of a ramp 5.0 m high and

rolls down to the bottom.

How fast is it moving at the

bottom of the ramp on the level

plane? 8.08 m/s

Rolling With Slipping

When an object initially slides before it rolls, the nonslip

condition vcom = rω does not hold. Kinetic friction due to

the slipping reduces the linear speed vcom and increases

the angular speed ω until the nonslip condition

vcom = rω is reached. When the not slip condition is

reached the object will continue to roll without slipping.

You may have observed a bowling ball released

without angular velocity. It initially skids along the alley

until it reaches the rolling without slipping condition and

then it rolls the rest of the way down the alley.

56. A bowler throws a bowling ball of radius R = 11 cm

along a lane. The ball slides on the lane, with initial

speed vcom, 0 = 8.5 m/s and initial angular speed ω0 = 0.

The coefficient of kinetic friction between the ball and the

lane is 0.21. The kinetic frictional force acting on the ball

causes a linear acceleration of the ball while producing a

torque that causes an angular acceleration of the ball.

a) When speed vcom has decreased enough and angular

speed ω had increased enough, the ball stops sliding and

then rolls smoothly. What then is vcom in terms of ω?

b) During the sliding, what is the ball’s linear

acceleration? – 2.1 m/s2

c) During the sliding, what is the ball’s angular

acceleration? 47 rad/s2

d) How long does the ball slide? 1.2 sec

e) How far does the ball slide? 8.6 m

f) What is the speed of the ball when smooth rolling

begins? 6.1 m/s

Torque Revisited: Up to this point we calculated the

torque for a rigid body that rotated about a fixed axis. All

the points in that body were moving in a circle about the

axis with a moment or torque arm equal to the radius of

the circle. The same definition of torque can be applied to

an individual particle that moves along any path relative

to a fixed point (rather than a fixed axis). The position of

the particle is given by the position vector r. The fixed

point could be the origin or some other coordinate.

57. What is the magnitude and direction of the torque

about the origin on a plum located at coordinates

(– 2.0 m,0, 4.0 m) due to a force F whose only component

is a) Fx = 6.0 N b) Fx = – 6.0 N c) Fz = 6.0 N

d) Fz = – 6.0 N

24j m-N -24j m-N 12j m-N -12j m-N

58. What is the magnitude and direction of the torque

about the origin on a particle located at coordinates

(0. –4.0 m, 3.0 m) due to

a) F1 = (2.0 N, 0, 0) 10 m-N 53̊ wrt y-axis

b) F2 = (0, 2.0 N, 4.0 N) 22 m-N in negative x direction

If the position of the particle is given relative to

the origin and you are asked to calculate the torque about

a different coordinate point, you have to calculate a new

position vector relative to that point: r’ = ro – p, where r’

is the new position vector, ro is the position relative to the

origin, and the torque is to be calculated about p.

Angular Momentum and Angular Impulse:

The angular momentum of a rigid body rotating

about a fixed axis is defined as L = Iω.

For a single particle the angular momentum

relative to any point would be:

L = r x p or L = m(r x v)

...where m is the mass of the particle, r is the position

vector from the point to the particle and v is the

translational velocity.

The product of the torque and the time interval

during which it acts is called the angular impulse, Jθ. The

angular impulse acting on the body causes a change in

the angular momentum of the body about the same axis.

For a torque that varies with time, the angular impulse is

defined as:

59. Calculate the angular momentum of a uniform sphere

of radius 0.20 m and mass 4.0 kg if it is rotating about an

axis along a diameter at 6.0 rad/s.

0.384 kg-m2/sec

60. A solid wooden door 1.0 m wide and 2.0 m high is

hinged along one side and has a total mass of 50.0 kg.

Initially open and at rest, the door is struck at its center

with a hammer. During the blow, an average force of

2000.0 N acts for 0.01 seconds. Find the angular velocity

of the door after the impact. 0.60 rad/s

61. A man of mass 70.0 kg is standing on the rim of a

large disk that is rotating at 0.500 rev/s about an axis

through its center. The disk has mass 120.0 kg and radius

4.00 m. Calculate the total angular momentum of the

man-plus-disk system. 6530 kg-m2/s

62. A rock of mass 2.0 kg has

a constant velocity of 12m/s.

When it is at point P as shown

in the figure, what is its

angular momentum relative to

point 0? (Remember constant

velocity means no change in

speed or direction.) 116 kg-m2/s

63. A 4.0 kg particle moves in an xy-plane. At the

instant when the particle's position and velocity are

r = (2.0i + 4.0j) m and v = – 4.0j m/s, the force on the

particle is F = – 3.0i N. At this instant, determine

a) the particle's angular momentum about the origin,

b) the particle's angular momentum about the point (0, 4)

c) the torque acting on the particle about the origin, and

d) the torque acting on the particle about the point (0, 4)

– 32 kg-m2/s k + 32 kg-m2/s k 12 m-N k 0

Conservation of Angular Momentum: Conservation of

angular momentum states that when the net external

torque on a system is zero, the angular momentum of the

system remains constant. This principle of conservation

of angular momentum ranks with the principles of

conservation of linear momentum and conservation of

energy as one of the most fundamental of physical laws.

64. On an old-fashioned rotating piano stool, a man sits

holding a pair of dumbbells at a distance of 0.60 m from

the axis of rotation of the stool. He is given an angular

velocity of 5.0 rad/s, after which he pulls the dumbbells in

until they are only 0.2 m distant from the axis. The man's

moment of inertia about the axis of rotation is 5.0 kg-m2

and may be considered constant. Each dumbbell has a

mass of 5.0 kg and may be considered a point mass.

Friction is negligible.

a) What is the initial angular momentum of the system?

43.0 kg-m2/s

b) What is the angular velocity of the system after the

dumbbells are pulled in toward the axis? 7.96 rad/s

c) Compute the kinetic energy of the system before and

after the dumbbells are pulled in. Account for the

difference, if any. 108 J 171 J

65. A puck on a frictionless

air-hockey table has a mass of

0.05kg and is attached to a cord

passing through a hole in the

table surface, as in the figure.

The puck is originally revolving

at a distance of 0.20 m from the

hole, with an angular velocity

of 3.0 rad/s. The cord is then pulled from below,

shortening the radius of the circle in which the puck

revolves to 0.10 m. The puck may be considered a point

mass.

a) What is the new angular velocity? 12 rad/s

b) Find the change in kinetic energy of the puck. 0.027 J

c) How much work was done by the person who pulled

the cord?

66. A large wooden turntable of radius 2.0 m and total

mass 120 kg is rotating about a vertical axis through its

center, with an angular velocity of 3.0 rad/s. From a very

small height a sandbag of mass 100 kg is dropped

vertically onto the turntable, at a point near the outer

edge.

a) Find the angular velocity of the turntable after the

sandbag is dropped. 1.12 rad/s

b) Compute the kinetic energy of the system before and

after the sandbag is dropped.

before: 1080 J after: 405 J

c) Why are these kinetic energies not equal?

67. The outstretched arms of a figure skater preparing for

a spin can be considered a slender rod pivoting about an

axis through its center. When her arms are brought in and

wrapped around her body to execute the spin, they can be

considered a thin-walled hollow cylinder. If her original

angular velocity is 6.28 rad/s, what is her final angular

velocity? Her arms have a combined mass of 8.0 kg.

When outstretched they span 1.8 m; when wrapped, they

form a cylinder of radius 25 cm. The moment of inertia of

the remainder of her body is constant and equal to 3.0 kg-

m2. 9.26 rad/s

68. Two 2.00 kg balls are

attached to the ends of a thin

rod of negligible mass, 50.0 cm

long. The rod is free to rotate

in a vertical plane without

friction about a horizontal axis

through its center. With the rod initially horizontal, as

shown is the figure below, a 50.0 g wad of wet putty

drops onto one of the balls, hitting it with a speed of 3.00

m/s and then sticking to it.

a) What is the angular speed of the system just after the

putty wad hits? 0.148 rad/s

b) What is the ratio of the kinetic energy of the entire

system after the collision to that of the putty wad just

before? 0.0123

c) Through what angle will the system rotate until it

momentarily stops? 181 ̊

Object

Rotational

Inertia

Icom

Percentage of Energy in

Translation Rotation

Hoop MR2 50% 50%

Disk 1/2 MR2 67% 33%

Sphere 2/5 MR2 71% 29%

General

βMR2 Where β is the

coefficient in

the rotational inertia equation