Here are the key steps to solve this problem: 1) Find the linear acceleration (a = 0.800 m/s2) 2) Find the time of acceleration (t = 20.0 s) 3) Use the equation for linear acceleration (a = rα) to find the angular acceleration: a / r = α 0.800 m/s2 / 0.330 m = α α = 2.42 rad/s2 4) Use the equation for angular velocity (ω = ω0 + αt) to find the final angular velocity: ω = 0 + 2.42 rad/s2 * 20.0 s ω = 48.4 rad/s

1. Ch8. Rotational Kinematics
Rotational Motion and Angular
Displacement
Angular displacement: When a rigid body
rotates about a fixed axis, the angular
displacements is the angle ∆θ 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)
1

2. 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 2

3. arc length s
θ (in radians) = =
Radius r
2πr
For 1 full rotation, θ= = 2π rad
r
∴ 360 deg ree = 2π rad
°
360°
1 rad = = 57.3°
2π 3

4. Example 1.
Adjacent Synchronous Satellites
Synchronous satellites are put
into an orbit whose radius is r
= 4.23*107m. The orbit is in the
plane of the equator, and two
adjacent satellites have an
angular separation of θ = 2.0° .
Find the arc length s.
4

5. 2π radians
2.0° = (2.0 deg rees)( ) = 0.0349radians
360 deg rees
s = rθ = (4.23 × 10 m)(0.0349rad )
7
= 1.48 × 10 m (920 miles)
6
5

6. Conceptual example 2.
A Total Eclipse of the Sun
6

7. 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 sun’s light from reaching the
observer’s eyes.
7

8. smoon
θ moon = smoon ≈ diametermoon
rmoon
Since they are very far apart.
ssun
θ sun = ssun ≈ diametersun
rsun
ssun ≈ 400 smoon rsun ≈ 400r moon
8

9. diametermoon
θ moon =
rmoon
diametersun 400diametermoon
θ sun = =
rsun 400rmoon
diametermoon smoon
= ≈
rmoon rmoon
9

10. θ sun ≈ θ moon 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 sun’s light from reaching the
observer’s eyes.
10

11. 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 θ 0) at the eye of the
observer. Object A has a diameter of 4d and subtends an
angle of 2θ 0. Object B has a diameter of 3d and subtends
an angle of θ 0/2. Object C has a diameter of d/2 and
subtends an angle of θ 0/8. Rank them in descending
order (greatest first) according to their distance from the
observer.
11

12. s d
rθ = s r= ≈
θ θ
4d 2d 3d 6d
rA = = rB = =
2θ 0 θ 0 θ0 θ0
2
d
rC = 2 = 8d = 4 d rB > rC > rA
θ 0 2θ 0 θ 0
8 B, C, A
12

13. 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.
13

14. DEFINITION OF AVERAGE ANGULAR VELOCITY
SI Unit of Angular Velocity: radian per second (rad/s)
14

15. 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.
15

16.  2π radians 
∆θ = −2.00revolutions 
 1 revolution 
 
= −12.6radians
16

17. Instantaneous angular velocity ω is the angular velocity
that exists at any given instant.
t t t
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.
17

18. 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.
18

19. DEFINITION OF AVERAGE ANGULAR ACCELERATION
SI Unit of Average Angular Acceleration: radian per second
squared (rad/s2)
The instantaneous angular acceleration α is the angular
acceleration at a given instant.
19

20. 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…..
20

21. The Equations of Rotational
Kinematics
21

22. In example 4, assume that the orientation of the rotating
object is given by θ 0 = 0 rad at time t0 = 0 s. Then, the
angular displacement becomes ∆ θ = θ – θ 0 = θ , and
the time interval becomes ∆t = t – t0 = t.
22

23. The Equations of Kinematics for Rational
and Linear Motion
Rotational Motion Linear Motion
(α = constant) (a = constant)
ω = ω 0 + αt v = v0 + at
1 1
θ = (ω0 + ω )t x = (v0 + v)t
2 2
1 2 1 2
θ = ω 0 t + αt x = v0t + at
2 2
ω = ω0 + 2αθ
2 2
v = v0 + 2ax
2 2
23

24. Symbols Used in Rotational and Linear
Kinematics
Rotational Quantity LinearMotion
Motion
θ Displacement x
ω0 Initial velocity v0
ω Final velocity v
α Acceleration a
t Time t
1 2
θ = ω0 t + αt ω = ω + 2αθ
2 2
0
2 24

25. 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. 25

26. θ α ω ω0 t
+44.0 rad +1740 rad/s2 ? +375 rad/s
ω = + ω + 2αθ
2
0
= (375rad / s ) + 2(1740rad / s )(44.0rad )
2 2
= +542rad / s 26

27. 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?
ω0 = 0 θ = 2revolution = 2× 2π radian
ω f = 0.5rev / s = 0.5 × 2π rad / s = π rad / s
What can be found next?
α
27

28. ω = ω + 2αθ
2
f
2
0
π = 0 + 2α × 2 × 2π
2
π × π 3.14
α= = rad / s 2
8π 8
after eight revolution, θ = 8 × revolution = 8 × 2π
π
ω = ω + 2αθ
2
f
2
0 0 + 2 × × 8 × 2π = (2π ) 2
2
8
ω f = 2π rad / s = 1rev / s 1.0 rev/s
28

29. 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.
29

30. vT = rω (ω in rad / s )
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.
vT − vT 0 (rω ) − (rω0 )  ω − ω0 
aT = = = r 
t t  t 
aT = rα (α in rad / s ) 2
30

31. Example 6.
A Helicopter Blade
A helicopter blade has an angular
speed of ω = 6.50 rev/s and an
angular acceleration of α = 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.
31

32. (a)
(b)
32

33. Centripetal Acceleration and
Tangential Acceleration
2
v
ac = T (centripetal
acceleration)
r 33

34. The centripetal acceleration can be expressed in terms of the
angular speed ω by using vT = rω
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
aT = rα tan Φ =
aC
aT
a = a +a 2
c
2
T aC
Φ
34

35. 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?
ω = 17 rev / s = 17 × 2π rad / s
v = rω vT = 32m / s
32 = r × 17 × 2π
0.30 m
32
r= = 0.3
17 × 2π
35

36. 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.
36

37. Find (a) the magnitude a of the total acceleration of the
discus just before it is released and (b) the angle φ that the
total acceleration makes with the radius at this moment.
(a)
37

38. (b)
α α
t
38

39. 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?
ω0 = 0 α = cast
ac = 2.0m / s 2
t = 3.0
vT = rω 39

40. 2
v
rω = 2.0m / s
2 2
ac = = rω 2
r
ω = ω 0 + αt
r ( 3α ) = 2.0
2
ω( t =3.0 ) = 3α
after six second,
ω = ω0 + 6 × α = 6α
(=0)
40

41. at six second
rω = r ( 6α )
2 2
( t = 6. 0 )
= r ( 2 × 3α )
2
= 4r ( 3α )
2
[
= 4 r ( 3α )
2
]
8.0 m/s 2 = 4 × 2.0 = 8.0m / s 2
41

42. Rolling Motion
42

43. v = rω (ω in rad / s )
Linear speed Tangential speed, vT
a = rα (α in rad / s ) 2
Linear Tangential acceleration, aT
acceleration
43

44. 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?
44

45. θ α w w0 t
? –2.42 rad/s2 0 rad/s 20.0 s
1 2
θ = ω 0 t + αt
2
1
= (0rad / s )(20.0 s ) + (−2.42rad / s )(20.0 s )
2 2
2
= −484rad
45

46. 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. 46

47. 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.
47

48. (a) 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.
(b) 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).
48

49. (b)
The angular acceleration is negative (clockwise).
49

50. Concepts & Calculations
Example 10.
A Circular Roadway and the
Acceleration of Your Car
50

51. 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 × 10–2
rad/s in a time of 4.0 s. What is the tangential acceleration
(magnitude and direction) of the car?
(a)
51

52. (b)
(c)
52

53. 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
∆ω 210rev / min − 480rev / min 2πrad 1 min 2
α= =( )( )( )
∆t 74 min 1rev 60 s
−3
= −6.4 ×10 rad / s 2
The magnitude of the average angular acceleration
is 6.4 × 10-3 rad/s2
53

54. Problem 7
REASONING AND SOLUTION Equation 8.4 gives the
desired result. Assuming t0 = 0 s, the final angular
velocity is
2
ω = ω 0 + α t = 0 rad/s + (328 rad/s )(1.50 s) = 492 rad/s
54

55. Problem 13
REASONING AND SOLUTION The baton will make four
revolutions in a time t given by θ
t=
ω
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
 θ 
v0 = g ( )
1t
2
= g 
 2ω 
55

56. With this initial velocity the baton can reach a height of
h=
2
v0
=
gθ 2
=
( 9.80 m/s ) ( 8π rad )
2 2
= 6.05 m
2g 2 2
8ω  rev   2π rad  
8 1.80  
 s   1 rev  
56

57. Problem 14
REASONING AND SOLUTION
r
s θ
person on earth
celestial body
The figure above shows the relevant angles and
dimensions for either one of the celestial bodies
under consideration.
57

58. a. Using the figure above
-3
-3
58

59. 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, a "total"eclipse of the sun is not really total
.
c. The relevant geometry is shown below.
r sun
r moon
R sun R
b
s sun s
b θ sun moon θ moon
person on earth
59

60. The apparent circular area of the sun as measured by a person
2
standing on the earth is given by: Asun = π Rsun, where Rsun is the
radius of the sun. The apparent circular area of the sun that is
2
blocked by the moon is Ablocked = π Rb , 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
2 2 2 2
Ablocked π Rb π ( sb / 2)  sb   θmoon rsun 
= 2
= 2
=
s  =
  θ r 

Asun π Rsun π ( ssun / 2)  sun   sun sun 
2 2
 θmoon   9.04 × 10−3 rad 
=
θ  =
  = 0.951
 9.27 × 10−3 rad 
 sun   
The moon blocks out 95.1 percent of the apparent circular area
60
of the sun.

61. 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
ω0 = ω − αt = 83.8rad / s − (−42.0rad / s )(1.75s ) 2
= 157.3rad / s
61

62. Problem 21
( )
REASONING Equation 8.8 ω 2 = ω 0 + 2αθ from the equations
2
of rotational kinematics can be employed to find the final angular
velocity ω . The initial angular velocity is ω 0 = 0 rad/s since the
top is initially at rest, and the angular acceleration is given as
α = 12 rad/s2. The angle θ (in radians) through which the pulley
rotates is not given, but it can be obtained from Equation 8.1
(θ = 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
ω = ± ω 0 + 2αθ
2
62

63. We choose the positive root, because the angular
acceleration is given as positive and the top is at rest
initially. Substituting θ = s/r from Equation 8.1 gives
s  64 cm 
( 0 rad/s ) + 2 ( 12 rad/s ) 
2
ω = + ω + 2α   = +
2
0
2
 = 28 rad/s
r  2.0 cm 
63

64. Problem 29
REASONING AND SOLUTION Equation 8.9 gives
the desired result
-3
64

65. 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 car’s radial
acceleration, and therefore its total acceleration, is
2
vT(75.0 m/s) 2
a = aR = = = 9.00 m/s 2
r 625 m
65

66. b The direction of the car’s total acceleration is the same as the
direction of its radial acceleration. That is, the direction is
radially inward
66

67. 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.
67

68. We can determine the magnitude of the centripetal
acceleration from Equation 8.11 as ac = rω 2, since both r
and ω 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
ac
a=
cos 35.0°
68

69. The magnitude of the centripetal acceleration is given by
Equation 8.11 as ac = rω 2, so the magnitude of the total
acceleration is
a=
ac rω 2
(23.5m)(0.571rad / s ) 2
= =
cos 35.0° cos 35.0° cos 35.0°
= 9.35m / s 2
69

70. 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 ω = vT/r = (15.0 m/s)/(0.330 m)
= 45.5 rad/s
b. vT = rω = (0.175 m)(45.5 rad/s) = 7.96 m/s
70

71. Problem 50
REASONING The angle through which the tire rotates is
equal to its average angular velocity ω multiplied by the
elapsed time t, θ = ω t . According to Equation 8.6, this
angle is related to the initial and final angular velocities of the
tire by
θ =ωt = 1
2(ω 0 +ω)t
The tire is assumed to roll at a constant angular velocity, so
that ω 0 = ω and θ = ω t. Since the tire is rolling, its angular
speed is related to its linear speed v by Equation 8.12, v = rω ,
where r is the radius of the tire. The angle of rotation then
becomes
  v
θ =ωt = t
r
71

72. 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
v v x x
θ = t =  =
r  r v  r
SOLUTION Since 1 rev = 2π rad, the angle (in revolutions)
is
3
x 96 000 × 10 m
θ= = = 3.1 × 108 rad
r 0.31 m
( 8
θ = 3.1 × 10 rad  )
 1 rev  7
 = 4.9 × 10 rev
 2π rad 
72

73. Problem 51
73

74. 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,
rsω s = rpω p
SOLUTION Solving the above expression for ωs gives
rpω p (3.49 rad/s)(0.102 m)
ωs = = −2
= 28.0 rad/s
rs 1.27 × 10 m
74

75. Problem 60
REASONING AND SOLUTION
a. The tangential acceleration of the train is given by Equation
8.10 as
2 −3 2 2
aT = r α = (2.00 ×10 m)(1.50 ×10 rad/s ) = 0.300 m/s
The centripetal acceleration of the train is given by Equation
8.11 as
2 2 2 2
ac = r ω = (2.00 × 10 m)(0.0500 rad/s) = 0.500 m/s
75

76. The magnitude of the total acceleration is found from the
Pythagorean theorem to be
a = aT + ac = 0.583 m/s 2
2 2
b. The total acceleration vector makes an angle relative to
the radial acceleration of


−1  aT

 = −1  0.300 m/s 2 

θ= tan  
tan  2
= 31.0°
a   0.500 m/s 
 c   
76

77. Problem 67
REASONING AND
SOLUTION By inspection,
the distance traveled by the
"axle" or the center of the
moving quarter is
d = 2 π (2 r ) = 4π r
77

78. 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 s = rθ , where θ is the angle through which the quarter
rotates. Thus,
4 π r = rθ
so that θ = 4π rad . This is equivalent to
 1 rev 
(4π rad)  = 2 revolutions
 2π rad 
78