Ch8. Rotational Kinematics Rotational Motion and Angular Displacement

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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)
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Angular displacement is expressed in one of three
units

1. Degree (1 full turn 3600
   degree)
2. Revolution (rev) RPM
3. Radian (rad) SI unit

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(in radians)
For 1 full rotation,
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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
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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.
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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.
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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.
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B, C, A
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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.
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DEFINITION OF AVERAGE ANGULAR VELOCITY
SI Unit of Angular Velocity radian per second
(rad/s)
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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.
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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.
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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.
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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..
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The Equations of Rotational Kinematics
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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.
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The Equations of Kinematics for Rational and
Linear Motion
Rotational Motion (a constant) 
Linear Motion (a constant) 
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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
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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?
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after eight revolution,
1.0 rev/s
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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.
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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.
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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.
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(a)
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Centripetal Acceleration and Tangential
Acceleration
(centripetal acceleration)
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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
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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
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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.
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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)
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(b)
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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?
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after six second,
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at six second
8.0 m/s2
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Rolling Motion
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Linear speed
Tangential speed, vT
Linear acceleration
Tangential acceleration, aT
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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.
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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.
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1. 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.
2. 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).
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(b)
The angular acceleration is negative (clockwise).
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Concepts Calculations Example 10.  A Circular
Roadway and the Acceleration of Your Car
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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)
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(b)
(c)
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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 
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Problem 7
REASONING AND SOLUTION Equation 8.4 gives the
desired result. Assuming t0 0 s, the final
angular velocity is
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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
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With this initial velocity the baton can reach a
height of
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Problem 14
REASONING AND SOLUTION
The figure above shows the relevant angles and
dimensions for either one of the celestial bodies
under consideration.
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a. Using the figure above
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-3
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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,
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c. The relevant geometry is shown below.
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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.
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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
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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
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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
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Problem 29
REASONING AND SOLUTION Equation 8.9 gives the
desired result
-3
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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
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b The direction of the cars total acceleration
is the same as the direction of its radial
acceleration. That is, the direction is
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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.
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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
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The magnitude of the centripetal acceleration is
given by Equation 8.11 as ac rw2, so the
magnitude of the total acceleration is
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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)
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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
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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
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Problem 51
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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
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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
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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
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Problem 67
REASONING AND SOLUTION By inspection, the
distance traveled by the "axle" or the center of
the moving quarter is
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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