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Ch8. Rotational KinematicsRotational Motion

and Angular Displacement

Angular displacement When a rigid body rotates

about a fixed axis, the angular displacements is

the angle swept out by a line passing

through any point on the body and intersecting

the axis of rotation perpendicularly. By

convention, the angular displacement is positive

if it is counterclockwise and negative if it is

clockwise. SI Unit of Angular Displacement

radian (rad)

Angular displacement is expressed in one of three

units

- Degree (1 full turn 3600

degree) - Revolution (rev) RPM
- Radian (rad) SI unit

(in radians)

For 1 full rotation,

Example 1. Adjacent Synchronous Satellites

Synchronous satellites are put into an orbit

whose radius is r 4.23107m. The orbit is in

the plane of the equator, and two adjacent

satellites have an angular separation of

. Find the arc length s.

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Conceptual example 2. A Total Eclipse of the Sun

The diameter of the sun is about 400 times

greater than that of the moon. By coincidence,

the sun is also about 400 times farther from the

earth than is the moon. For an observer on earth,

compare the angle subtended by the moon to the

angle subtended by the sun, and explain why this

result leads to a total solar eclipse.

Since the angle subtended by the moon is nearly

equal to the angle subtended by the sun, the moon

blocks most of the suns light from reaching the

observers eyes.

Since they are very far apart.

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total eclipse

Since the angle subtended by the moon is nearly

equal to the angle subtended by the sun, the moon

blocks most of the suns light from reaching the

observers eyes.

Check Your Understanding 1

Three objects are visible in the night sky. They

have the following diameters (in multiples of d

and subtend the following angles (in multiples of

q 0) at the eye of the observer. Object A has a

diameter of 4d and subtends an angle of 2q 0.

Object B has a diameter of 3d and subtends an

angle of q 0/2. Object C has a diameter of d/2

and subtends an angle of q 0/8. Rank them in

descending order (greatest first) according to

their distance from the observer.

B, C, A

CONCEPTS AT A GLANCE To define angular velocity,

we use two concepts previously encountered. The

angular velocity is obtained by combining the

angular displacement and the time during which

the displacement occurs. Angular velocity is

defined in a manner analogous to that used for

linear velocity. Taking advantage of this analogy

between the two types of velocities will help us

understand rotational motion.

DEFINITION OF AVERAGE ANGULAR VELOCITY

SI Unit of Angular Velocity radian per second

(rad/s)

Example 3. Gymnast on a High Bar

A gymnast on a high bar swings through two

revolutions in a time of 1.90s. Find the average

angular velocity (in rad/s) of the gymnast.

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Instantaneous angular velocity w is the angular

velocity that exists at any given

instant. The magnitude of the instantaneous

angular velocity, without reference to whether it

is a positive or negative quantity, is called the

instantaneous angular speed. If a rotating object

has a constant angular velocity, the

instantaneous value and the average value are the

same.

In linear motion, a changing velocity means that

an acceleration is occurring. Such is also the

case in rotational motion a changing angular

velocity means that an angular acceleration is

occurring. CONCEPTS AT A GLANCE The idea of

angular acceleration describes how rapidly or

slowly the angular velocity changes during a

given time interval.

DEFINITION OF AVERAGE ANGULAR ACCELERATION

SI Unit of Average Angular Acceleration radian

per second squared (rad/s2)

The instantaneous angular acceleration a is the

angular acceleration at a given instant.

Example 4. A Jet Revving Its Engines

A jet awaiting clearance for takeoff is

momentarily stopped on the runway. As seen from

the front of one engine, the fan blades are

rotating with an angular velocity of 110 rad/s,

where the negative sign indicates a clockwise

rotation .

As the plane takes off, the angular velocity of

the blades reaches 330 rad/s in a time of 14 s.

Find the average angular velocity, assuming that

the orientation of the rotating object is given

by..

The Equations of Rotational Kinematics

In example 4, assume that the orientation of the

rotating object is given by q 0 0 rad at time

t0 0 s. Then, the angular displacement becomes

Dq q q 0 q , and the time interval

becomes Dt t t0 t.

The Equations of Kinematics for Rational and

Linear Motion

Rotational Motion (a constant)

Linear Motion (a constant)

Symbols Used in Rotational and Linear Kinematics

Rotational Motion Quantity LinearMotion

q Displacement x

w0 Initial velocity v0

w Final velocity v

a Acceleration a

t Time t

Example 5. Blending with a Blender

The blades of an electric blender are whirling

with an angular velocity of 375 rad/s while the

puree button is pushed in. When the blend

button is pressed, the blades accelerate and

reach a greater angular velocity after the blades

have rotated through an angular displacement of

44.0 rad (seven revolutions). The angular

acceleration has a constant value of 1740

rad/s2. Find the final angular velocity of the

blades.

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Check Your Understanding 2

The blades of a ceiling fan start from rest and,

after two revolutions, have an angular speed of

0.50 rev/s. The angular acceleration of the

blades is constant. What is the angular speed

after eight revolutions?

What can be found next?

after eight revolution,

1.0 rev/s

Angular Variables and Tangential Variables

For every individual skater, the vector is drawn

tangent to the appropriate circle and, therefore,

is called the tangential velocity vT. The

magnitude of the tangential velocity is referred

to as the tangential speed.

If time is measured relative to t0 0 s, the

definition of linear acceleration is given by

Equation 2.4 as aT (vT vT0)/t, where vT and

vT0 are the final and initial tangential speeds,

respectively.

Example 6. A Helicopter Blade

A helicopter blade has an angular speed of w

6.50 rev/s and an angular acceleration of a

1.30 rev/s2. For points 1 and 2 on the blade,

find the magnitudes of (a) the tangential speeds

and (b) the tangential accelerations.

(a)

Centripetal Acceleration and Tangential

Acceleration

(centripetal acceleration)

The centripetal acceleration can be expressed in

terms of the angular speed w by using vT rw

While the tangential speed is changing, the

motion is called nonuniform circular motion.

Since the direction and the magnitude of the

tangential velocity are both changing, the

airplane experiences two acceleration components

simultaneously.

aT

aC

Check Your Understanding 3

The blade of a lawn mower is rotating at an

angular speed of 17 rev/s. The tangential speed

of the outer edge of the blade is 32 m/s. What is

the radius of the blade?

0.30 m

Example 7. A Discus Thrower

Discus throwers often warm up by standing with

both feet flat on the ground and throwing the

discus with a twisting motion of their bodies. A

top view of such a warm-up throw. Starting from

rest, the thrower accelerates the discus to a

final angular velocity of 15.0 rad/s in a time

of 0.270 s before releasing it. During the

acceleration, the discus moves on a circular arc

of radius 0.810 m.

Find (a) the magnitude a of the total

acceleration of the discus just before it is

released and (b) the angle f that the total

acceleration makes with the radius at this moment.

(a)

(b)

Check Your Understanding 4

A rotating object starts from rest and has a

constant angular acceleration. Three seconds

later the centripetal acceleration of a point on

the object has a magnitude of 2.0 m/s2. What is

the magnitude of the centripetal acceleration of

this point six seconds after the motion begins?

after six second,

at six second

8.0 m/s2

Rolling Motion

Linear speed

Tangential speed, vT

Linear acceleration

Tangential acceleration, aT

Example 8. An Accelerating Car

An automobile starts from rest and for 20.0 s has

a constant linear acceleration of 0.800 m/s2 to

the right. During this period, the tires do not

slip. The radius of the tires is 0.330 m. At the

end of the 20.0-s interval, what is the angle

through which each wheel has rotated?

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The Vector Nature of Angular Variables

Right-Hand Rule Grasp the axis of rotation with

your right hand, so that your fingers circle the

axis in the same sense as the rotation. Your

extended thumb points along the axis in the

direction of the angular velocity vector.

Angular acceleration arises when the angular

velocity changes, and the acceleration vector

also points along the axis of rotation. The

acceleration vector has the same direction as the

change in the angular velocity.

Concepts Calculations Example 9. Riding a

Mountain Bike

A rider on a mountain bike is traveling to the

left. Each wheel has an angular velocity of 21.7

rad/s, where, as usual, the plus sign indicates

that the wheel is rotating in the

counterclockwise direction.

- To pass another cyclist, the rider pumps harder,

and the angular velocity of the wheels increases

from 21.7 to 28.5 rad/s in a time of 3.50 s. - After passing the cyclist, the rider begins to

coast, and the angular velocity of the wheels

decreases from 28.5 to 15.3 rad/s in a time of

10.7 s. In both instances, determine the

magnitude and direction of the angular

acceleration (assumed constant) of the wheels.

(a)

The angular acceleration is positive

(counterclockwise).

(b)

The angular acceleration is negative (clockwise).

Concepts Calculations Example 10. A Circular

Roadway and the Acceleration of Your Car

Suppose you are driving a car in a

counterclockwise direction on a circular road

whose radius is r 390 m (see Figure 8.20). You

look at the speedometer and it reads a steady 32

m/s (about 72 mi/h). (a) What is the angular

speed of the car? (b) Determine the acceleration

(magnitude and direction) of the car. (c) To

avoid a rear-end collision with a vehicle ahead,

you apply the brakes and reduce your angular

speed to 4.9 102 rad/s in a time of 4.0 s.

What is the tangential acceleration (magnitude

and direction) of the car?

(a)

(b)

(c)

Problem 5

REASONING AND SOLUTION Using Equation 8.4 and

the appropriate conversion factors, the average

angular acceleration of the CD in rad/s2 is

The magnitude of the average angular acceleration

is 6.4 10-3 rad/s2

Problem 7

REASONING AND SOLUTION Equation 8.4 gives the

desired result. Assuming t0 0 s, the final

angular velocity is

Problem 13

REASONING AND SOLUTION The baton will make four

revolutions in a time t given by

Half of this time is required for the baton to

reach its highest point. The magnitude of the

initial vertical velocity of the baton is then

With this initial velocity the baton can reach a

height of

Problem 14

REASONING AND SOLUTION

The figure above shows the relevant angles and

dimensions for either one of the celestial bodies

under consideration.

a. Using the figure above

-3

-3

b. Since the sun subtends a slightly larger

angle than the moon, as measured by a person

standing on the earth, the sun cannot be

completely blocked by the moon. Therefore,

.

c. The relevant geometry is shown below.

The apparent circular area of the sun as measured

by a person standing on the earth is given by

, where Rsun is the radius of

the sun. The apparent circular area of the sun

that is blocked by the moon is

, where Rb is shown in the figure above.

Also from the figure above, it follows that

Rsun (1/2) ssun and Rb (1/2) sb

Therefore, the fraction of the apparent circular

area of the sun that is blocked by the moon is

The moon blocks out 95.1 percent of the apparent

circular area of the sun.

Problem 17

REASONING AND SOLUTION Since the angular speed

of the fan decreases, the sign of the angular

acceleration must be opposite to the sign for the

angular velocity. Taking the angular velocity to

be positive, the angular acceleration, therefore,

must be a negative quantity. Using Equation 8.4

we obtain

Problem 21

REASONING Equation 8.8

from the equations of rotational kinematics

can be employed to find the final angular

velocity ?. The initial angular velocity is

?0 0 rad/s since the top is initially at rest,

and the angular acceleration is given as

? 12 rad/s2. The angle ? (in radians) through

which the pulley rotates is not given, but it can

be obtained from Equation 8.1 (? s/r ), where

the arc length s is the 64-cm length of the

string and r is the 2.0-cm radius of the top.

SOLUTION Solving Equation 8.8 for the final

angular velocity gives

We choose the positive root, because the angular

acceleration is given as positive and the top is

at rest initially. Substituting ? s/r from

Equation 8.1 gives

Problem 29

REASONING AND SOLUTION Equation 8.9 gives the

desired result

-3

Problem 39

REASONING Since the car is traveling with a

constant speed, its tangential acceleration must

be zero. The radial or centripetal acceleration

of the car can be found from Equation 5.2. Since

the tangential acceleration is zero, the total

acceleration of the car is equal to its radial

acceleration.

SOLUTION

a. Using Equation 5.2, we find that the cars

radial acceleration, and therefore its total

acceleration, is

b The direction of the cars total acceleration

is the same as the direction of its radial

acceleration. That is, the direction is

Problem 42

REASONING The drawing shows a top view of the

race car as it travels around the circular turn.

Its acceleration a has two perpendicular

components a centripetal acceleration ac that

arises because the car is moving on a circular

path and a tangential acceleration aT due to the

fact that the car has an angular acceleration and

its angular velocity is increasing.

We can determine the magnitude of the centripetal

acceleration from Equation 8.11 as ac rw2,

since both r and w are given in the statement of

the problem. As the drawing shows, we can use

trigonometry to determine the magnitude a of the

total acceleration, since the angle (35.0?)

between a and ac is given.

SOLUTION Since the vectors ac and a are one side

and the hypotenuse of a right triangle, we have

that

The magnitude of the centripetal acceleration is

given by Equation 8.11 as ac rw2, so the

magnitude of the total acceleration is

Problem 46

REASONING AND SOLUTION

a. If the wheel does not slip, a point on the rim

rotates about the axle with a speed vT v

15.0 m/s

For a point on the rim w vT/r (15.0

m/s)/(0.330 m)

b. vT rw (0.175 m)(45.5 rad/s)

Problem 50

REASONING The angle through which the tire

rotates is equal to its average angular velocity

multiplied by the elapsed time t, q

t . According to Equation 8.6, this angle is

related to the initial and final angular

velocities of the tire by

The tire is assumed to roll at a constant angular

velocity, so that w0 w and q wt. Since the

tire is rolling, its angular speed is related to

its linear speed v by Equation 8.12, v rw,

where r is the radius of the tire. The angle of

rotation then becomes

The time t that it takes for the tire to travel a

distance x is equal to t x/v, according to

Equation 2.1. Thus, the angle that the tire

rotates through is

SOLUTION Since 1 rev 2p rad, the angle (in

revolutions) is

Problem 51

REASONING Assuming that the belt does not slip

on the platter or the shaft pulley, the

tangential speed of points on the platter and

shaft pulley must be equal therefore,

SOLUTION Solving the above expression for

gives

Problem 60

REASONING AND SOLUTION

a. The tangential acceleration of the train is

given by Equation 8.10 as

The centripetal acceleration of the train is

given by Equation 8.11 as

The magnitude of the total acceleration is found

from the Pythagorean theorem to be

b. The total acceleration vector makes an angle

relative to the radial acceleration of

Problem 67

REASONING AND SOLUTION By inspection, the

distance traveled by the "axle" or the center of

the moving quarter is

where r is the radius of the quarter. The

distance d traveled by the "axle" of the moving

quarter must be equal to the circular arc length

s along the outer edge of the quarter. This arc

length is , where is the

angle through which the quarter rotates. Thus,

so that . This is

equivalent to