6 raymond a serway, john w jewett physics for scientists and engineers with modern physics 13

Courtesy Tourism Malaysia 10.1 Angular Position, Velocity, and Acceleration 10.2 Rotational Kinematics: The Rigid Object Under Constant Angular Acceleration 10.3 Angular and Translationa

higher than if his back were straight As a model, consider

the jumper as a thin, uniform rod of length L When the

rod is straight, its center of mass is at its center Now bend

the rod in a circular arc so that it subtends an angle of

90.0° at the center of the arc as shown in Figure P9.40b.

In this configuration, how far outside the rod is the

cen-ter of mass?

Section 9.6 Motion of a System of Particles

41. A 2.00-kg particle has a velocity m/s, and

a 3.00-kg particle has a velocity m/s.

Find (a) the velocity of the center of mass and (b) the

total momentum of the system.

42. The vector position of a 3.50-g particle moving in the xy

plane varies in time according to

At the same time, the vector position of a 5.50-g particle

varies as , where t is in s and r is in

cm At t 2.50 s, determine (a) the vector position of the

center of mass, (b) the linear momentum of the system,

(c) the velocity of the center of mass, (d) the acceleration

of the center of mass, and (e) the net force exerted on

the two-particle system.

43. Romeo (77.0 kg) entertains Juliet (55.0 kg) by playing his

guitar from the rear of their boat at rest in still water,

2.70 m away from Juliet, who is in the front of the boat.

After the serenade, Juliet carefully moves to the rear of

the boat (away from shore) to plant a kiss on Romeo’s

cheek How far does the 80.0-kg boat move toward the

shore it is facing?

44. A ball of mass 0.200 kg has a velocity of 1.50 m/s; a ball

of mass 0.300 kg has a velocity of 0.400 m/s They

meet in a head-on elastic collision (a) Find their

veloci-ties after the collision (b) Find the velocity of their

cen-ter of mass before and afcen-ter the collision.

Section 9.7 Deformable Systems

45.  For a technology project, a student has built a vehicle,

of total mass 6.00 kg, that moves itself As shown in Figure

P9.45, it runs on two light caterpillar tracks that pass

around four light wheels A reel is attached to one of the

axles, and a cord originally wound on the reel passes over

a pulley attached to the vehicle to support an elevated

load After the vehicle is released from rest, the load

descends slowly, unwinding the cord to turn the axle and

make the vehicle move forward Friction is negligible in

the pulley and axle bearings The caterpillar tread does

not slip on the wheels or the floor The reel has a conical

shape so that the load descends at a constant low speed

iˆ iˆ

while the vehicle moves horizontally across the floor with constant acceleration, reaching final velocity 3.00 m/s (a) Does the floor impart impulse to the vehicle? If so, how much? (b) Does the floor do work on the vehicle? If

so, how much? (c) Does it make sense to say that the final momentum of the vehicle came from the floor? If not, from where? (d) Does it make sense to say that the final kinetic energy of the vehicle came from the floor? If not, from where? (e) Can we say that one particular force causes the forward acceleration of the vehicle? What does cause it?

46.  A 60.0-kg person bends his knees and then jumps straight up After his feet leave the floor his motion is unaffected by air resistance and his center of mass rises by

a maximum of 15.0 cm Model the floor as completely solid and motionless (a) Does the floor impart impulse

to the person? (b) Does the floor do work on the person? (c) With what momentum does the person leave the floor? (d) Does it make sense to say that this momentum came from the floor? Explain (e) With what kinetic energy does the person leave the floor? (f) Does it make sense to say that this energy came from the floor? Explain.

47.  A particle is suspended from a post on top of a cart by

a light string of length L as shown in Figure P9.47a The

cart and particle are initially moving to the right at

con-stant speed v i, with the string vertical The cart suddenly comes to rest when it runs into and sticks to a bumper as shown in Figure P9.47b The suspended particle swings through an angle u (a) Show that the original speed of

the cart can be computed from v i 

(b) Find the initial speed implied by L 1.20 m and u  35.0° (c) Is the bumper still exerting a horizontal force

on the cart when the hanging particle is at its maximum angle from the vertical? At what moment in the observ- able motion does the bumper stop exerting a horizontal force on the cart?

Figure P9.47

48. On a horizontal air track, a glider of mass m carries a

-shaped post The post supports a small dense sphere, also

of mass m, hanging just above the top of the glider on a cord of length L The glider and sphere are initially at

rest with the cord vertical (Fig P9.47a shows a cart and a sphere similarly connected.) A constant horizontal force

of magnitude F is applied to the glider, moving it through displacement x1; then the force is removed During the time interval when the force is applied, the sphere moves

through a displacement with horizontal component x2 (a) Find the horizontal component of the velocity of the center of mass of the glider-sphere system when the force

is removed (b) After the force is removed, the glider

con-tinues to move on the track and the sphere swings back

and forth, both without friction Find an expression for

the largest angle the cord makes with the vertical.

49.  Sand from a stationary hopper falls onto a moving

con-veyor belt at the rate of 5.00 kg/s as shown in Figure

P9.49 The conveyor belt is supported by frictionless

rollers It moves at a constant speed of 0.750 m/s under

the action of a constant horizontal external force

sup-plied by the motor that drives the belt Find (a) the sand’s

rate of change of momentum in the horizontal direction,

(b) the force of friction exerted by the belt on the sand,

(c) the external force , (d) the work done by in

1 s, and (e) the kinetic energy acquired by the falling

sand each second due to the change in its horizontal

motion (f) Why are the answers to (d) and (e) different?

is required? (b) If a different fuel and engine design could give an exhaust speed of 5 000 m/s, what amount

of fuel and oxidizer would be required for the same task? This exhaust speed is 2.50 times higher than that in part (a) Explain why the required fuel mass is 2.50 times smaller, or larger than that, or still smaller.

Additional Problems

54. Two gliders are set in motion on an air track A spring of

force constant k is attached to the back end of the second glider The first glider, of mass m1, has velocity , and the

second glider, of mass m2, moves more slowly, with velocity

, as shown in Figure P9.54 When m1collides with the

spring attached to m2 and compresses the spring to its

maximum compression xmax, the velocity of the gliders is

In terms of , , m1, m2, and k, find (a) the velocity

at maximum compression, (b) the maximum

compres-sion xmax, and (c) the velocity of each glider after m1has lost contact with the spring.

Section 9.8 Rocket Propulsion

50. Model rocket engines are sized by thrust, thrust duration,

and total impulse, among other characteristics A size C5

model rocket engine has an average thrust of 5.26 N, a

fuel mass of 12.7 g, and an initial mass of 25.5 g The

duration of its burn is 1.90 s (a) What is the average

exhaust speed of the engine? (b) This engine is placed in

a rocket body of mass 53.5 g What is the final velocity of

the rocket if it is fired in outer space? Assume the fuel

burns at a constant rate.

51. The first stage of a Saturn V space vehicle consumed fuel

and oxidizer at the rate of 1.50 10 4 kg/s, with an exhaust

speed of 2.60 10 3 m/s (a) Calculate the thrust produced

by this engine (b) Find the acceleration the vehicle had

just as it lifted off the launch pad on the Earth, taking the

vehicle’s initial mass as 3.00 10 6 kg Note: You must

include the gravitational force to solve part (b).

52. Rocket science A rocket has total mass M i 360 kg,

includ-ing 330 kg of fuel and oxidizer In interstellar space, it

starts from rest at the position x 0, turns on its engine

at time t  0, and puts out exhaust with relative speed v e

1 500 m/s at the constant rate k 2.50 kg/s The fuel will

last for an actual burn time of 330 kg/(2.5 kg/s)  132 s,

but define a “projected depletion time” as T p  M i /k

360 kg/(2.5 kg/s)  144 s (which would be the burn time

if the rocket could use its payload and fuel tanks, and

even the walls of the combustion chamber as fuel) (a)

Show that during the burn the velocity of the rocket as a

function of time is given by

(b) Make a graph of the velocity of the rocket as a

func-tion of time for times running from 0 to 132 s (c) Show

that the acceleration of the rocket is

Figure P9.54

55. An 80.0-kg astronaut is taking a space walk to work on the engines of his ship, which is drifting through space with a constant velocity The astronaut, wishing to get a better view of the Universe, pushes against the ship and much later finds himself 30.0 m behind the ship Without a thruster or tether, the only way to return to the ship is to throw his 0.500-kg wrench directly away from the ship If

he throws the wrench with a speed of 20.0 m/s relative to the ship, after what time interval does the astronaut reach the ship?

56.  An aging Hollywood actor (mass 80.0 kg) has been cloned, but the genetic replica is far from perfect The

clone has a different mass m, his stage presence is poor,

and he uses foul language The clone, serving as the

actor’s stunt double, stands on the brink of a cliff 36.0 m

high, next to a sturdy tree The actor stands on top of a

Humvee, 1.80 m above the level ground, holding a taut

rope tied to a tree branch directly above the clone When

the director calls “action,” the actor starts from rest and

swings down on the rope without friction The actor is

momentarily hidden from the camera at the bottom of

the arc, where he undergoes an elastic head-on collision

with the clone, sending him over the cliff Cursing vilely,

the clone falls freely into the ocean below The actor is

prosecuted for making an obscene clone fall, and you are

called as an expert witness at the sensational trial.

(a) Find the horizontal component R of the clone’s

dis-placement as it depends on m Evaluate R (b) for m 

79.0 kg and (c) for m  81.0 kg (d) What value of m

gives a range of 30.0 m? (e) What is the maximum

possi-ble value for R, and (f) to what value of m does it

corre-spond? What are (g) the minimum values of R and

(h) the corresponding value of m? (i) For the actor–clone–

Earth system, is mechanical energy conserved throughout

the action sequence? Is this principle sufficient to solve

the problem? Explain (j) For the same system, is

momen-tum conserved? Explain how this principle is used.

(k) What If? Show that R does not depend on the value of

the gravitational acceleration Is this result remarkable?

State how one might make sense of it.

57. A bullet of mass m is fired into a block of mass M initially

at rest at the edge of a frictionless table of height h (Fig.

P9.57) The bullet remains in the block, and after impact

the block lands a distance d from the bottom of the table.

Determine the initial speed of the bullet.

wedge after the block reaches the horizontal surface?

(b) What is the height h of the wedge?

59.  A 0.500-kg sphere moving with a velocity given by

strikes another sphere of mass 1.50 kg that is moving with an initial velocity

the 0.500-kg sphere after the collision is given by

Find the final velocity of the 1.50-kg sphere and identify the kind of collision (elas- tic, inelastic, or perfectly inelastic) (b) Now assume the velocity of the 0.500-kg sphere after the collision is

Find the final velocity

of the 1.50-kg sphere and identify the kind of collision.

(c) What If? Take the velocity of the 0.500-kg sphere after

value of a and the velocity of the 1.50-kg sphere after an

elastic collision.

60. A 75.0-kg firefighter slides down a pole while a constant friction force of 300 N retards her motion A horizontal 20.0-kg platform is supported by a spring at the bottom of the pole to cushion the fall The firefighter starts from rest 4.00 m above the platform, and the spring constant is

4 000 N/m Find (a) the firefighter’s speed immediately before she collides with the platform and (b) the maxi- mum distance the spring is compressed Assume the fric- tion force acts during the entire motion.

61. George of the Jungle, with mass m, swings on a light

vine hanging from a stationary tree branch A second vine

of equal length hangs from the same point, and a gorilla

of larger mass M swings in the opposite direction on it.

Both vines are horizontal when the primates start from rest at the same moment George and the gorilla meet at the lowest point of their swings Each is afraid that one vine will break, so they grab each other and hang on They swing upward together, reaching a point where the vines make an angle of 35.0° with the vertical (a) Find

the value of the ratio m/M (b) What If? Try the following

experiment at home Tie a small magnet and a steel screw

to opposite ends of a string Hold the center of the string fixed to represent the tree branch, and reproduce a model of the motions of George and the gorilla What changes in your analysis will make it apply to this situa-

tion? What If? Next assume the magnet is strong so that it

noticeably attracts the screw over a distance of a few timeters Then the screw will be moving faster immedi- ately before it sticks to the magnet Does this extra mag- net strength make a difference?

cen-62.  A student performs a ballistic pendulum experiment using an apparatus similar to that shown in Figure 9.9b.

She obtains the following average data: h  8.68 cm, m1 

68.8 g, and m2 263 g The symbols refer to the

quanti-ties in Figure 9.9a (a) Determine the initial speed v 1Aof the projectile (b) The second part of her experiment is

to obtain v 1A by firing the same projectile horizontally (with the pendulum removed from the path) and measur-

ing its final horizontal position x and distance of fall y

(Fig P9.62) Show that the initial speed of the projectile

is related to x and y by the equation

58. A small block of mass m1 0.500 kg is released from rest

at the top of a curve-shaped, frictionless wedge of mass

m2 3.00 kg, which sits on a frictionless horizontal

sur-face as shown in Figure P9.58a When the block leaves the

wedge, its velocity is measured to be 4.00 m/s to the right

as shown in the figure (a) What is the velocity of the

What numerical value does she obtain for v 1A based on

her measured values of x  257 cm and y  85.3 cm?

What factors might account for the difference in this

value compared with that obtained in part (a)?

63.  Lazarus Carnot, an artillery general, managed the

mili-tary draft for Napoleon Carnot used a ballistic pendulum

to measure the firing speeds of cannonballs In the

sym-bols defined in Example 9.6, he proved that the ratio

of the kinetic energy immediately after the collision to

the kinetic energy immediately before is m1/(m1m2 ).

(a) Carry out the proof yourself (b) If the cannonball has

mass 9.60 kg and the block (a tree trunk) has mass 214 kg,

what fraction of the original energy remains mechanical

after the collision? (c) What is the ratio of the

momen-tum immediately after the collision to the momenmomen-tum

immediately before? (d) A student believes that such a

large loss of mechanical energy must be accompanied by

at least a small loss of momentum How would you

con-vince this student of the truth? General Carnot’s son Sadi

was the second most important engineer in the history of

ideas; we will study his work in Chapter 22.

64.  Pursued by ferocious wolves, you are in a sleigh with no

horses, gliding without friction across an ice-covered lake.

You take an action described by these equations:

(a) Complete the statement of the problem, giving the

data and identifying the unknowns (b) Find the values of

v 1f and v 2f (c) Find the work you do.

65 Review problem. A light spring of force constant

3.85 N/m is compressed by 8.00 cm and held between a

0.250-kg block on the left and a 0.500-kg block on the

right Both blocks are at rest on a horizontal surface The

blocks are released simultaneously so that the spring

tends to push them apart Find the maximum velocity

each block attains if the coefficient of kinetic friction

between each block and the surface is (a) 0, (b) 0.100,

and (c) 0.462 Assume the coefficient of static friction is

greater than the coefficient of kinetic friction in every

case.

66. Consider as a system the Sun with the Earth in a circular

orbit around it Find the magnitude of the change in the

velocity of the Sun relative to the center of mass of the

system over a 6-month period Ignore the influence of

other celestial objects You may obtain the necessary

astro-nomical data from the endpapers of the book.

2 = intermediate; 3 = challenging;  = SSM/SG;  = ThomsonNOW;  = symbolic reasoning;  = qualitative reasoning

Then, a total force N acts on the object for 5.00 s (a) Calculate the object’s final velocity, using the impulse–momentum theorem (b) Calculate its accel- eration from (c) Calculate its accelera- tion from (d) Find the object’s vector dis- placement from (e) Find the work done

on the object from (f) Find the final kinetic energy from (g) Find the final kinetic energy from (h) State the result of comparing the answers to parts b and c, and the answers to parts f and g.

69. A chain of length L and total mass M is released from rest

with its lower end just touching the top of a table as shown in Figure P9.69a Find the force exerted by the table on the chain after the chain has fallen through a

distance x as shown in Figure P9.69b (Assume each link

comes to rest the instant it reaches the table.)

L  x

x L

Figure P9.69

Answers to Quick Quizzes

9.1(d) Two identical objects (m1 m2 ) traveling at the same

speed (v1 v2 ) have the same kinetic energies and the

same magnitudes of momentum It also is possible,

how-ever, for particular combinations of masses and velocities

to satisfy K1  K2 but not p1  p2 For example, a 1-kg

object moving at 2 m/s has the same kinetic energy as a

4-kg object moving at 1 m/s, but the two clearly do not

have the same momenta Because we have no information

about masses and speeds, we cannot choose among (a),

(b), or (c).

9.2(b), (c), (a) The slower the ball, the easier it is to catch If

the momentum of the medicine ball is the same as the

momentum of the baseball, the speed of the medicine

ball must be 1/10 the speed of the baseball because the

medicine ball has 10 times the mass If the kinetic

ener-gies are the same, the speed of the medicine ball must be

the speed of the baseball because of the squared

speed term in the equation for K The medicine ball is

hardest to catch when it has the same speed as the

base-ball.

9.3 (i),(c), (e) Object 2 has a greater acceleration because of

its smaller mass Therefore, it travels the distance d in a

shorter time interval Even though the force applied to

objects 1 and 2 is the same, the change in momentum is

less for object 2 because t is smaller The work W  Fd

done on both objects is the same because both F and d

are the same in the two cases Therefore, K1  K2

(ii),(b), (d) The same impulse is applied to both objects,

so they experience the same change in momentum.

Object 2 has a larger acceleration due to its smaller mass.

Therefore, the distance that object 2 covers in the time

interval is larger than that for object 1 As a result, more

work is done on object 2 and K2 K1

9.4 (a) All three are the same Because the passenger is

brought from the car’s initial speed to a full stop, the

change in momentum (equal to the impulse) is the same

regardless of what stops the passenger (b) Dashboard,

seat belt, air bag The dashboard stops the passenger very

quickly in a front-end collision, resulting in a very large

force The seat belt takes somewhat more time, so the

force is smaller Used along with the seat belt, the air bag

can extend the passenger’s stopping time further, notably

for his head, which would otherwise snap forward.

1 > 110

9.5(a) If all the initial kinetic energy is transformed or ferred away from the system, nothing is moving after the collision Consequently, the final momentum of the sys- tem is necessarily zero and the initial momentum of the system must therefore be zero Although (b) and (d)

trans-together would satisfy the conditions, neither one alone

does.

9.6 (b) Because momentum of the two-ball system is served, Because the table-tennis ball bounces back from the much more massive bowling ball with approximately the same speed, As a con- sequence, Kinetic energy can be expressed as

con-K  p2/2m Because of the much larger mass of the

bowl-ing ball, its kinetic energy is much smaller than that of the table-tennis ball.

9.7(b) The piece with the handle will have less mass than the piece made up of the end of the bat To see why, take the origin of coordinates as the center of mass before the bat was cut Replace each cut piece by a small sphere located

at each piece’s center of mass The sphere representing the handle piece is farther from the origin, but the prod- uct of less mass and greater distance balances the product

of greater mass and less distance for the end piece as shown.

the other way (ii), (b) Once they stop running, the

momentum of the system is the same as it was before they started running; you cannot change the momentum of an isolated system by means of internal forces In case you are thinking that the passengers could run to the stern

repeatedly to take advantage of the speed increase while

they are running, remember that they will slow the ship down every time they return to the bow!

The Malaysian pastime of gasing involves the spinning of tops that can

have masses up to 5 kg Professional spinners can spin their tops so that

they might rotate for more than an hour before stopping We will study the

rotational motion of objects such as these tops in this chapter (Courtesy

Tourism Malaysia)

10.1 Angular Position,

Velocity, and Acceleration

10.2 Rotational Kinematics:

The Rigid Object Under Constant Angular Acceleration

10.3 Angular and

Translational Quantities

10.4 Rotational Kinetic

Energy

When an extended object such as a wheel rotates about its axis, the motion cannot

be analyzed by modeling the object as a particle because at any given time

differ-ent parts of the object have differdiffer-ent linear velocities and linear accelerations We

can, however, analyze the motion of an extended object by modeling it as a

collec-tion of particles, each of which has its own linear velocity and linear acceleracollec-tion

In dealing with a rotating object, analysis is greatly simplified by assuming the

object is rigid A rigid object is one that is nondeformable; that is, the relative

loca-tions of all particles of which the object is composed remain constant All real

objects are deformable to some extent; our rigid-object model, however, is useful

in many situations in which deformation is negligible

and Acceleration

Figure 10.1 illustrates an overhead view of a rotating compact disc, or CD The

disc rotates about a fixed axis perpendicular to the plane of the figure and passing

through the center of the disc at O A small element of the disc modeled as a

par-ticle at P is at a fixed distance r from the origin and rotates about it in a circle of

radius r (In fact, every particle on the disc undergoes circular motion about O.) It

is convenient to represent the position of P with its polar coordinates (r, u), where

Rotation of a Rigid Object About a Fixed Axis

r s

u

Figure 10.1 A compact disc rotating

about a fixed axis through O

per-pendicular to the plane of the figure (a) To define angular position for the disc, a fixed reference line is chosen.

A particle at P is located at a distance

r from the rotation axis at O (b) As the disc rotates, a particle at P moves through an arc length s on a circular path of radius r.

r is the distance from the origin to P and u is measured counterclockwise from some

reference line fixed in space as shown in Figure 10.1a In this representation, the

angle u changes in time while r remains constant As the particle moves along the

circle from the reference line, which is at angle u  0, it moves through an arc of

length s as in Figure 10.1b The arc length s is related to the angle u through the

relationship

(10.1a) (10.1b)

Because u is the ratio of an arc length and the radius of the circle, it is a pure

number Usually, however, we give u the artificial unit radian (rad), where one

radian is the angle subtended by an arc length equal to the radius of the arc.

Because the circumference of a circle is 2pr, it follows from Equation 10.1b that 360° corresponds to an angle of (2pr/r) rad  2p rad Hence, 1 rad  360°/2p 57.3° To convert an angle in degrees to an angle in radians, we use that p rad 180°, so

For example, 60° equals p/3 rad and 45° equals p/4 rad

Because the disc in Figure 10.1 is a rigid object, as the particle moves through

an angle u from the reference line, every other particle on the object rotates

through the same angle u Therefore, we can associate the angle U with the entire

angular position of a rigid object in its rotational motion We choose a reference

line on the object, such as a line connecting O and a chosen particle on the

object The angular position of the rigid object is the angle u between this

refer-ence line on the object and the fixed referrefer-ence line in space, which is often

cho-sen as the x axis Such identification is similar to the way we define the position of

an object in translational motion as the distance x between the object and the erence position, which is the origin, x  0

ref-As the particle in question on our rigid object travels from position  to tion  in a time interval t as in Figure 10.2, the reference line fixed to the object

posi-sweeps out an angle u  uf ui This quantity u is defined as the angular

The rate at which this angular displacement occurs can vary If the rigid objectspins rapidly, this displacement can occur in a short time interval If it rotatesslowly, this displacement occurs in a longer time interval These different rotation

rates can be quantified by defining the average angular speed vavg (Greek letteromega) as the ratio of the angular displacement of a rigid object to the time inter-val t during which the displacement occurs:

(10.2)

In analogy to linear speed, the instantaneous angular speed v is defined as the

limit of the average angular speed as t approaches zero:

(10.3)

Angular speed has units of radians per second (rad/s), which can be written as s1because radians are not dimensional We take v to be positive when u is increasing(counterclockwise motion in Figure 10.2) and negative when u is decreasing(clockwise motion in Figure 10.2)

Remember the Radian

In rotational equations, you must

use angles expressed in radians.

Don’t fall into the trap of using

angles measured in degrees in

Figure 10.2 A particle on a rotating

rigid object moves from  to 

along the arc of a circle In the time

interval t  t f  t i, the radial line of

length r moves through an angular

displacement u  uf ui.

Average angular speed 

Instantaneous angular speed 

Quick Quiz 10.1 A rigid object rotates in a counterclockwise sense around a

fixed axis Each of the following pairs of quantities represents an initial angular

position and a final angular position of the rigid object (i) Which of the sets can

only occur if the rigid object rotates through more than 180°? (a) 3 rad, 6 rad

(b) 1 rad, 1 rad (c) 1 rad, 5 rad (ii) Suppose the change in angular position for

each of these pairs of values occurs in 1 s Which choice represents the lowest

aver-age angular speed?

If the instantaneous angular speed of an object changes from vi to vf in the

time interval t, the object has an angular acceleration The average angular

the change in the angular speed to the time interval t during which the change

in the angular speed occurs:

(10.4)

In analogy to linear acceleration, the instantaneous angular acceleration is

defined as the limit of the average angular acceleration as t approaches zero:

(10.5)

Angular acceleration has units of radians per second squared (rad/s2), or

sim-ply s2 Notice that a is positive when a rigid object rotating counterclockwise is

speeding up or when a rigid object rotating clockwise is slowing down during

some time interval

When a rigid object is rotating about a fixed axis, every particle on the object

rotates through the same angle in a given time interval and has the same angular

charac-terize the rotational motion of the entire rigid object as well as individual particles

in the object

Angular position (u), angular speed (v), and angular acceleration (a) are

anal-ogous to translational position (x), translational speed (v), and translational

accel-eration (a) The variables u, v, and a differ dimensionally from the variables x, v,

and a only by a factor having the unit of length (See Section 10.3.)

We have not specified any direction for angular speed and angular acceleration

Strictly speaking, v and a are the magnitudes of the angular velocity and the

angu-lar acceleration vectors1 and , respectively, and they should always be positive

Because we are considering rotation about a fixed axis, however, we can use

non-vector notation and indicate the non-vectors’ directions by assigning a positive or

nega-tive sign to v and a as discussed earlier with regard to Equations 10.3 and 10.5

For rotation about a fixed axis, the only direction that uniquely specifies the

rota-tional motion is the direction along the axis of rotation Therefore, the directions

of and are along this axis If a particle rotates in the xy plane as in Figure 10.2,

the direction of for the particle is out of the plane of the diagram when the

rotation is counterclockwise and into the plane of the diagram when the rotation

is clockwise To illustrate this convention, it is convenient to use the right-hand rule

demonstrated in Figure 10.3 When the four fingers of the right hand are wrapped

in the direction of rotation, the extended right thumb points in the direction of

The direction of follows from its definition It is in the same

direc-tion as if the angular speed is increasing in time, and it is antiparallel to if the

angular speed is decreasing in time

Specify Your Axis

In solving rotation problems, you must specify an axis of rotation This new feature does not exist in our study of translational motion The choice is arbitrary, but once you make it, you must maintain that choice consistently throughout the problem In some problems, the physical situation suggests a natural axis, such as the center of

an automobile wheel In other problems, there may not be an obvious choice, and you must exer- cise judgment.

1 Although we do not verify it here, the instantaneous angular velocity and instantaneous angular

accel-eration are vector quantities, but the corresponding average values are not because angular

displace-ments do not add as vector quantities for finite rotations.

v

v

Figure 10.3 The right-hand rule for

determining the direction of the angular velocity vector.

10.2 Rotational Kinematics: The Rigid Object

Under Constant Angular Acceleration

When a rigid object rotates about a fixed axis, it often undergoes a constant lar acceleration Therefore, we generate a new analysis model for rotational

angu-motion called the rigid object under constant angular acceleration This model is

the rotational analog to the particle under constant acceleration model Wedevelop kinematic relationships for this model in this section Writing Equation

10.5 in the form dv  a dt and integrating from t i  0 to t f  t gives

tions for translational motion by making the substitutions x S u, v S v, and a S a.

Table 10.1 compares the kinematic equations for rotational and translationalmotion

object in Quick Quiz 10.1 If the object starts from rest at the initial angular tion, moves counterclockwise with constant angular acceleration, and arrives at thefinal angular position with the same angular speed in all three cases, for whichchoice is the angular acceleration the highest?

Just Like Translation?

Equations 10.6 to 10.9 and Table

10.1 suggest that rotational

matics is just like translational

kine-matics That is almost true, with

two key differences (1) In

rota-tional kinematics, you must specify

a rotation axis (per Pitfall

Pre-vention 10.2) (2) In rotational

motion, the object keeps returning

to its original orientation;

there-fore, you may be asked for the

number of revolutions made by a

rigid object This concept has no

meaning in translational motion.

TABLE 10.1

Kinematic Equations for Rotational and Translational Motion Under Constant Acceleration

Rotational Motion About a Fixed Axis Translational Motion

10.3 Angular and Translational Quantities

In this section, we derive some useful relationships between the angular speed and

acceleration of a rotating rigid object and the translational speed and acceleration

of a point in the object To do so, we must keep in mind that when a rigid object

Section 10.3 Angular and Translational Quantities 273

angular displacement of the object:

¢u uf ui vi t1

2at2

E X A M P L E 1 0 1

A wheel rotates with a constant angular acceleration of 3.50 rad/s2

(A) If the angular speed of the wheel is 2.00 rad/s at t i  0, through what angular displacement does the wheelrotate in 2.00 s?

SOLUTION

a constant rate You start your stopwatch when the disc is rotating at 2.00 rad/s This mental image is a model for themotion of the wheel in this example

angu-lar acceleration model

Rotating Wheel

(B)Through how many revolutions has the wheel turned during this time interval?

SOLUTION

Substitute the known values to find the angular

displace-ment at t 2.00 s:  11.0 rad  111.0 rad2 157.3°>rad2  630°

¢u 12.00 rad>s2 12.00 s2 1

213.50 rad>s22 12.00 s22

(C)What is the angular speed of the wheel at t 2.00 s?

SOLUTION

Multiply the angular displacement found in part (A) by

a conversion factor to find the number of revolutions:

¢u 630°a1 rev

360°b  1.75 rev

the particle is 2.00 m/s at t i 0, through what displacement does the particle move in 2.00 s? What is the velocity of

the particle at t 2.00 s?

mathematical solution follows exactly the same form For the displacement,

and for the velocity,

There is no translational analog to part (B) because translational motion under constant acceleration is not repetitive

rotates about a fixed axis as in Active Figure 10.4, every particle of the object

moves in a circle whose center is on the axis of rotation.

Because point P in Active Figure 10.4 moves in a circle, the translational velocity vector is always tangent to the circular path and hence is called tangential velocity The magnitude of the tangential velocity of the point P is by definition the tangen- tial speed v  ds/dt, where s is the distance traveled by this point measured along the circular path Recalling that s  ru (Eq 10.1a) and noting that r is constant,

per-lar speed Therefore, although every point on the rigid object has the same

angu-lar speed, not every point has the same tangential speed because r is not the same

for all points on the object Equation 10.10 shows that the tangential speed of apoint on the rotating object increases as one moves outward from the center ofrotation, as we would intuitively expect For example, the outer end of a swinginggolf club moves much faster than the handle

We can relate the angular acceleration of the rotating rigid object to the

tan-gential acceleration of the point P by taking the time derivative of v:

(10.11)

That is, the tangential component of the translational acceleration of a point on arotating rigid object equals the point’s perpendicular distance from the axis ofrotation multiplied by the angular acceleration

In Section 4.4, we found that a point moving in a circular path undergoes a

radial acceleration a rdirected toward the center of rotation and whose magnitude

is that of the centripetal acceleration v2/r (Fig 10.5) Because v  r v for a point

P on a rotating object, we can express the centripetal acceleration at that point in

terms of angular speed as

(10.12)

The total acceleration vector at the point is , where the magnitude

of is the centripetal acceleration a c Because is a vector having a radial and a

tangential component, the magnitude of at the point P on the rotating rigid

object is

(10.13)

a horse at the outer rim of the circular platform, twice as far from the center of

the circular platform as Brian, who rides on an inner horse (i) When the

merry-go-round is rotating at a constant angular speed, what is Alex’s angular speed?(a) twice Brian’s (b) the same as Brian’s (c) half of Brian’s (d) impossible todetermine (ii)When the merry-go-round is rotating at a constant angular speed,describe Alex’s tangential speed from the same list of choices

v

u

ACTIVE FIGURE 10.4

As a rigid object rotates about the

fixed axis through O, the point P has a

tangential velocity that is always

tan-gent to the circular path of radius r.

Sign in at www.thomsonedu.comand

go to ThomsonNOW to move point P

and observe the tangential velocity as

the object rotates.

v

S

Relation between tangential 

and angular acceleration

x y

O

a r

a t P

a

Figure 10.5 As a rigid object rotates

about a fixed axis through O, the

point P experiences a tangential

com-ponent of translational acceleration

a tand a radial component of

transla-tional acceleration a r The total

accel-eration of this point is aS

 aS

 aS

.

¢u 11.8  105 rad2 a 1 rev

2p radb  2.8  104 rev

Section 10.3 Angular and Translational Quantities 275

E X A M P L E 1 0 2 CD Player

On a compact disc (Fig 10.6), audio information is stored digitally in a series of

pits and flat areas on the surface of the disc The alternations between pits and

flat areas on the surface represent binary ones and zeroes to be read by the CD

player and converted back to sound waves The pits and flat areas are detected by

a system consisting of a laser and lenses The length of a string of ones and zeroes

representing one piece of information is the same everywhere on the disc,

whether the information is near the center of the disc or near its outer edge So

that this length of ones and zeroes always passes by the laser–lens system in the

same time interval, the tangential speed of the disc surface at the location of the

lens must be constant According to Equation 10.10, the angular speed must

therefore vary as the laser–lens system moves radially along the disc In a typical

CD player, the constant speed of the surface at the point of the laser–lens system

is 1.3 m/s

(A)Find the angular speed of the disc in revolutions per minute when information is being read from the innermost

first track (r  23 mm) and the outermost final track (r  58 mm).

SOLUTION

in a time interval of about 3 s Now trace your finger around the circle marked “58 mm” in the same time interval.Notice how much faster your finger is moving relative to the page around the larger circle If your finger representsthe laser reading the disc, it is moving over the surface of the disc much faster for the outer circle than for the innercircle

identify analysis models

Use Equation 10.10 to find the angular speed that

gives the required tangential speed at the position of

the inner track:

Do the same for the outer track: vf v

r f 1.3 m>s5.8 102 m 22 rad>s  2.1  102 rev>min

The CD player adjusts the angular speed v of the disc within this range so that information moves past the objectivelens at a constant rate

(B) The maximum playing time of a standard music disc is 74 min and 33 s How many revolutions does the discmake during that time?

SOLUTION

constant We can then use the rigid object under constant angular acceleration model

is (74 min)(60 s/min)  33 s  4 473 s We are looking for the angular displacement u during this time interval.Use Equation 10.9 to find the angular displacement

10.4 Rotational Kinetic Energy

In Chapter 7, we defined the kinetic energy of an object as the energy associatedwith its motion through space An object rotating about a fixed axis remains sta-tionary in space, so there is no kinetic energy associated with translational motion.The individual particles making up the rotating object, however, are movingthrough space; they follow circular paths Consequently, there is kinetic energyassociated with rotational motion

Let us consider an object as a collection of particles and assume it rotates about

a fixed z axis with an angular speed v Figure 10.7 shows the rotating object and identifies one particle on the object located at a distance r ifrom the rotation axis

If the mass of the ith particle is m i and its tangential speed is v i, its kinetic energy is

To proceed further, recall that although every particle in the rigid object has the

same angular speed v, the individual tangential speeds depend on the distance r i from the axis of rotation according to Equation 10.10 The total kinetic energy of the

rotating rigid object is the sum of the kinetic energies of the individual particles:

We can write this expression in the form

(10.14)

where we have factored v2 from the sum because it is common to every particle

We simplify this expression by defining the quantity in parentheses as the moment

of inertia I:

(10.15)

From the definition of moment of inertia,2 we see that it has dimensions of ML2

(kg·m2in SI units) With this notation, Equation 10.14 becomes

(10.16)

Although we commonly refer to the quantity as rotational kinetic energy, it is

not a new form of energy It is ordinary kinetic energy because it is derived from a

10.6 gives the value of the constant angular acceleration Another approach is to use Equation 10.4 to find the age angular acceleration In this case, we are not assuming that the angular acceleration is constant The answer isthe same from both equations; only the interpretation of the result is different

acceleration:

avf vi

t  22 rad>s  57 rad>s

4 473 s  7.8  103 rad>s2

required for the angular speed to change from the initial value to the final value In reality, the angular acceleration

of the disc is not constant Problem 20 allows you to explore the actual time behavior of the angular acceleration

Figure 10.7 A rigid object rotating

about the z axis with angular speed v.

The kinetic energy of the particle of

mass m iis The total kinetic

energy of the object is called its

rota-tional kinetic energy.

I y a

i

sum over individual kinetic energies of the particles contained in the rigid object

The mathematical form of the kinetic energy given by Equation 10.16 is

conven-ient when we are dealing with rotational motion, provided we know how to

calcu-late I.

It is important to recognize the analogy between kinetic energy associated

with translational motion and rotational kinetic energy The quantities I and

v in rotational motion are analogous to m and v in translational motion,

respec-tively (In fact, I takes the place of m and v takes the place of v every time we

com-pare a translational motion equation with its rotational counterpart.) Moment of

inertia is a measure of the resistance of an object to changes in its rotational

motion, just as mass is a measure of the tendency of an object to resist changes in

its translational motion

No Single Moment of Inertia

There is one major difference between mass and moment of iner- tia Mass is an inherent property of

an object The moment of inertia

of an object depends on your choice of rotation axis Therefore, there is no single value of the moment of inertia for an object.

There is a minimum value of the

moment of inertia, which is that calculated about an axis passing through the center of mass of the object.

E X A M P L E 1 0 3

Four tiny spheres are fastened to the ends of two rods of

negligible mass lying in the xy plane (Fig 10.8) We shall

assume the radii of the spheres are small compared with

the dimensions of the rods

(A)If the system rotates about the y axis (Fig 10.8a) with

an angular speed v, find the moment of inertia and the

rotational kinetic energy of the system about this axis

SOLUTION

that helps conceptualize the system of spheres and how it

spins

because it is a straightforward application of the

defini-tions discussed in this section

Four Rotating Objects

a b

b

(b)

y m

Figure 10.8 (Example 10.3) Four spheres form an unusual baton.

(a) The baton is rotated about the y axis (b) The baton is rotated about the z axis.

Apply Equation 10.15 to the system:

That the two spheres of mass m do not enter into this result makes sense because they have no motion about the axis

of rotation; hence, they have no rotational kinetic energy By similar logic, we expect the moment of inertia about

the x axis to be I x  2mb2with a rotational kinetic energy about that axis of K R  mb2v2

(B) Suppose the system rotates in the xy plane about an axis (the z axis) through O (Fig 10.8b) Calculate the

moment of inertia and rotational kinetic energy about this axis

to include all four spheres and distances because all four spheres are rotating in the xy plane Based on the

work–kinetic energy theorem, that the rotational kinetic energy in part (A) is smaller than that in part (B) indicates

it would require less work to set the system into rotation about the y axis than about the z axis.

Evaluate the rotational kinetic energy using Equation

10.16:

2I zv21

212Ma2 2mb22v2 1Ma2 mb22v2

10.5 Calculation of Moments of Inertia

We can evaluate the moment of inertia of an extended rigid object by imaginingthe object to be divided into many small elements, each of which has mass m i Weuse the definition and take the limit of this sum as m iS0 In thislimit, the sum becomes an integral over the volume of the object:

L

Thin spherical shell

I  13 ML2