PHY300

1. Giambattista Physics
Chapter 5
©McGraw-Hill Education. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribution permitted without the prior written consent of McGraw-Hill Education.

2. ©McGraw-Hill Education
Chapter 5: Circular Motion
• 5.1 Description of Uniform Circular Motion
• 5.2 Radial Acceleration
• 5.3 Unbanked and Banked Curves
• 5.4 Circular Orbits of Satellites and Planets
• 5.5 Nonuniform Circular Motion
• 5.6 Angular Acceleration
• 5.7 Apparent Weight and Artificial Gravity

3. ©McGraw-Hill Education
5.1 Description of Uniform Circular Motion
Rotation of a Rigid Body
To describe circular motion, we could use the familiar definitions
of displacement, velocity, and acceleration.
But much of the circular motion around us occurs in the rotation
of a rigid object.
A rigid object (or body) is one for which the distance between
any two points of the body remains the same when the body is
translated or rotated.

4. ©McGraw-Hill Education
Example: DVD
When a DVD spins inside a DVD
player, different points on the DVD
have different velocities and
accelerations.
The velocity and acceleration of a
given point keep changing
direction as the DVD spins.
It is much simpler to say “the DVD spins at 210 rpm” than to say
“a point 6.0 cm from the rotation axis of the DVD is moving at
1.3 m/s.”

5. ©McGraw-Hill Education
Angular Displacement and Angular Velocity
To simplify the description of circular motion, we concentrate
on angles instead of distances.
Instead of displacement, we
speak of angular displacement
Δθ, the angle through which the
DVD turns.
Counterclockwise rotation is
positive;
Clockwise is negative.

6. ©McGraw-Hill Education
Definitions: Angular Displacement and Velocity
Definition of angular displacement:
f i
  
  
Definition of average angular velocity:
av




t
Definition of instantaneous angular velocity:
0
lim


 

 

t t

7. ©McGraw-Hill Education
Radian Measure
In many situations the most
convenient angle measure is
the radian.
(in radians)
 
s
r
2
2 rad

 
  
s r
r r
2 rad
360 
 

8. ©McGraw-Hill Education
Radian Measure Continued
(in radians)
 
s
r
In equations that relate linear variables to angular variables,
think of r as the number of meters of arc length per radian of
angle subtended.
That is, think of r as having units of meters per radian. Doing so,
the radians cancel out in these equations.
2.0 rad
 
 r
s
m
1.2
rad
 2.4 m


9. ©McGraw-Hill Education
Example 5.1
Earth is rotating about its axis. What is its angular speed in
rad/s? (The question asks for angular speed, so we do not have
to worry about the direction of rotation.)
Strategy
The Earth’s angular velocity is
constant, or nearly so.
Therefore, we can calculate the
average angular velocity for any
convenient time interval and, in
turn, the Earth’s instantaneous
angular speed |ω|.

10. ©McGraw-Hill Education
Example 5.1 Continued
1 day 24 h 24 h 3600 s / h 86400 s
   
2 rad 5
7.3 10 rad/s
86400 s

 
  

11. ©McGraw-Hill Education
Relation Between Linear and Angular Speed
For a point moving in a circular path of radius r, the linear
distance traveled along the circular path during an angular
displacement of Δθ (in radians) is the arc length s, where
f i
| | | (angles in radians)
|
  
   
s r r
av
| |
( in radians)



  
 
s r
v
t t
Taking the limit as Δt approaches zero,
| | ( in radians)
 

v r

12. ©McGraw-Hill Education
Example: Linear and Angular Speed
| | ( in radians per unit time)
 

v r
A person standing at the
equator is moving much
faster than another person
standing at the Arctic
Circle, but their angular
speeds are the same.

13. ©McGraw-Hill Education
Period and Frequency
When the speed of a point moving in a circle is constant, its
motion is called uniform circular motion.
The time for the point to travel completely around the circle is
called the period of the motion, T.
The frequency of the motion, which is the number of revolutions
per unit time, is defined as
1

f
T

14. ©McGraw-Hill Education
Period and Frequency Continued
The speed is the total distance traveled divided by the time
taken:
2
2


 
r
v rf
T
Then, for uniform circular motion
2 ( in radians per unit time)
  
 
v
f
r
where, in SI units, angular velocity  is measured in rad/s and
frequency f is measured in hertz (Hz).

15. ©McGraw-Hill Education
Example 5.2
A centrifuge is spinning at 5400 rpm.
(a) Find the period (in s) and frequency (in Hz) of the motion.
(b) If the radius of the centrifuge is 14 cm, how fast (in m/s) is an
object at the outer edge moving?

16. ©McGraw-Hill Education
Example 5.2 Strategy
Remember that rpm means revolutions per minute.
5400 rpm is the frequency, but in a unit other than Hz.
After a unit conversion, the other quantities can be found using
the relations already discussed.

17. ©McGraw-Hill Education
Example 5.2 Solution
(a) rev 1 min
5400 90 rev/s
min 60 s
  
f
1
90 Hz 90 s
 
f
1/ 0.011 s
 
T f
(b) rad rev
2 2 90 180 rad/s
rev s
   
 
 
f
1
180 s 0.14 m 79 m/s
  
   
v r

18. ©McGraw-Hill Education
Rolling Without Slipping
Rotation and Translation Combined
When an object is rolling, it is both rotating and translating.
What is the relationship
between the angular speed
of the wheel and the linear
speed of the axle?

19. ©McGraw-Hill Education
Rolling Without Slipping Continued
axle
2

r
v
T
2
 
T
axle ( in radians per unit time)
 

v r

20. ©McGraw-Hill Education
Example 5.3
Kevin is riding his motorcycle at a speed of 13.0 m/s.
If the diameter of the rear tire is 65.0 cm, what is the angular
speed of the rear wheel? Assume that it rolls without slipping.
Strategy
The given diameter of the tire enables us to find the
circumference and, thus, the distance traveled in one revolution
of the wheel.
From the speed of the motorcycle we can find how many
revolutions the tire must make per second.

21. ©McGraw-Hill Education
Example 5.3 Continued
Solution distance 2
time



r
T
v
/
2

T r v
2
 
 
2
 


 
t T
2
 
2
13.0 m/s rad
40.0
(0.650 m) / 2 s
/
  
v
r
r v

22. ©McGraw-Hill Education
5.2 Radial Acceleration
For a particle undergoing uniform circular motion, the direction
of the velocity continually changes, so the particle has a nonzero
acceleration.

23. ©McGraw-Hill Education
Direction of Radial Acceleration
The acceleration has the same direction as the change
in velocity (in the limit t  0):
0
lim
t t
 



v
a
The acceleration of an object undergoing
uniform circular motion is directed radially
inward – along a radius. The radial
acceleration is the component of the
acceleration directed toward the center.
(A synonym for radial acceleration is
centripetal acceleration. Centripetal means
“toward the center”.)

24. ©McGraw-Hill Education
Magnitude of Radial Acceleration
arc length
radius of circle angle subtended
v v t
 

 
   
Δv
( in radians per unit time)
r
a v
t
 
  

Δv
a
v r


2
2
or ( in radians per unit time)
r r
a
v
a r
r
 
 

25. ©McGraw-Hill Education
Example 5.4
Centrifuges are used to establish the maximum acceleration a
pilot can withstand without “blacking out”.
If the pilot undergoes a radial acceleration of 4.00 g (as
measured at her head) and the radial distance from her head to
the axis of rotation is 12.5 m, what is the period of rotation of
the centrifuge?
Strategy The radial acceleration can be found from the radius of
the circular path and either the linear or the angular speed. The
period is the time for one complete revolution; in one evolution
the distance traveled is the circumference of the circle.

26. ©McGraw-Hill Education
Example 5.4 Solution
2
2 2
r r
v
v a r
r
r r
v T
T v
a
 
  
  
2
2 12.5m
2 2 3.55 s
4.00 9.80m/s
r
r
r r
T
a
a r

 
    


27. ©McGraw-Hill Education
Problem-Solving Strategy for an Object in
Uniform Circular Motion
1. Begin as for any Newton’s second law problem: identify all
the forces acting on the object and draw an FBD.
2. Choose perpendicular axes as the point of interest so that
one is radial and the other is tangent to the circular path.
3. Find the radial component of each force.
4. Apply Newton’s second law as follows:
r r
F ma


2
2
r
v
r
r
a 
 

28. ©McGraw-Hill Education
Example 5.5
An athlete whirls a 4.00-kg hammer six or seven times around
and then releases it. Although the purpose of whirling it around
several times is to increase the hammer’s speed, assume that
just before the hammer is released, it moves at constant speed
along a circular arc of radius 1.7 m.
At the instant she releases the hammer, it is 1.0 m above the
ground and its velocity is directed 40 above the horizontal. The
hammer lands a horizontal distance of 74.0 m away.
What force does the athlete apply to the grip just before she
releases it? Ignore air resistance.

29. ©McGraw-Hill Education
Example 5.5 Strategy
By analyzing the projectile motion of the hammer, we can find
the speed of the hammer just after its release.
Just before release, the forces acting on the hammer are the
tension in the cable and gravity.

30. ©McGraw-Hill Education
Example 5.5 Solution
During its projectile motion, the initial velocity has magnitude vi
(to be determined) and direction θ = 40 above the horizontal.
2
i i
1
( cos ) and ( sin ) ( )
2
x v t y v t g t
 
       
i
y v
 
i
sin
x
v


2
i
1
2 cos
cos
x
g
v 

 

  
 
2 2 2 2
i i
2 2
i
sin
2 cos 2 cos
cos
2 cos
x
v y v
v

 



 

i
2 cos
x
g
v 

2
 
 
 
 

31. ©McGraw-Hill Education
Example 5.5 Solution Continued 1
2 2 2 2
i i
2 2
i
sin
2 cos 2 cos
cos
2 cos
x
v y v
v

 



 

i
2 cos
x
g
v 

2
 
 
 
 
2 2 2
i (2 cos 2 cos sin ) ( )
v y x g x
  
     
2
i 2
2 2
2
( )
2 cos sin 2 cos
9.80 m/s (74.0 m)
26.9 m/s
2(74.0 m)cos40 sin 40 2( 1.0 m)cos 40
v
g x
x y
  
  


  

 
 

32. ©McGraw-Hill Education
Example 5.5 Solution Continued 2
2
r r
mv
F T ma
r
  

2
4.00 kg (26.9 m/s)
1700 N
1.7m
T



The athlete must apply a force of magnitude 1700 N—almost
400 lb—to the grip.

33. ©McGraw-Hill Education
Example 5.6
Suppose you whirl a stone in a horizontal circle at a slow speed
so that the weight of the stone is not negligible compared with
the tension in the cord. Then the cord cannot be horizontal—
the tension must have a vertical component to cancel the
weight and leave a horizontal net force.
If the cord has length L , the stone has mass m , and the cord
makes an angle  with the vertical direction, what is the
constant angular speed of the stone?

34. ©McGraw-Hill Education
Example 5.6 Strategy
The net force must point toward the center of the circle, since
the stone is moving in uniform circular motion.
With the stone in the position
depicted in the figure, the
direction of the net force is along
the +x-axis. This time the tension
in the cord does not pull toward
the center, but the net force does.

35. ©McGraw-Hill Education
Example 5.6 Solution
2
sin
x x
F T ma m r
 
  

sin
r L 

2
2
sin sin
x
F T m L
T m L
  

 
 

cos 0 cos
y
mg ma T mg
T  
    
2
c
( ) os
L mg
m  
cos
g
L




36. ©McGraw-Hill Education
5.3 Unbanked and Banked Curves
Unbanked Curves
When you drive an automobile in a
circular path along an unbanked
roadway, friction acting on the tires due
to the pavement acts to keep the
automobile moving in a curved path.
This frictional force acts sideways , toward the center of the
car’s circular path.
The frictional force might also have a tangential component; for
example, if the car is braking.

37. ©McGraw-Hill Education
Assumptions about Friction
For now, assume that the car’s speed
is constant and that the forward or
backward component of the frictional
force is negligibly small.
As long as the tires roll without slipping, there is no relative
motion between the bottom of the tires and the road, so it is
the force of static friction that acts.

38. ©McGraw-Hill Education
Banked Curves: Application of Radial
Acceleration and Contact Forces
To help prevent cars from
going into a skid or losing
control, the roadway is often
banked (tilted at a slight
angle) around curves.

39. ©McGraw-Hill Education
Example 5.7
A car is going around an unbanked curve at the recommended
speed of 11 m/s.

40. ©McGraw-Hill Education
Example 5.7 Continued 1
a) If the radius of curvature of the path is 25 m, and the
coefficient of static friction between the rubber and the
road is s = 0.70, does the car skid as it goes around the
curve?
b) What happens if the driver ignores the highway speed limit
sign and travels at 18 m/s?
c) What speed is safe for traveling around the curve if the road
surface is wet from a recent rainstorm and the coefficient of
static friction between the wet road and the rubber tires is
s = 0.50?

41. ©McGraw-Hill Education
Example 5.7 Continued 2
d) For a car to safely negotiate the curve in icy conditions at a
speed of 13 m/s, what banking angle would be required?

42. ©McGraw-Hill Education
Example 5.7 Strategy
The force of static friction is the only horizontal force acting on
the car when the curve is not banked.
The maximum force of static friction, which depends on road
conditions, determines the maximum possible radial
acceleration (and hence, speed) of the car.

43. ©McGraw-Hill Education
Example 5.7 Solution
a) 2 2
2
(11 m/s)
4.8 m/s
25 m
r
a
v
r
  
2
r r
r s s
v
F ma m
r
F f N

 
 


2
s g
m
v
m
r


Thus, the radial acceleration cannot exceed sg.

44. ©McGraw-Hill Education
Example 5.7 Solution Continued 1
a) continued.
s
v gr


2
0.70 9.80 m/s 25 m 13 m/s
v
    
Since 11 m/s is less than the maximum safe speed of 13 m/s, the
car safely negotiates the curve.
b) At 18 m/s, the car moves at a speed higher than the
maximum safe speed of 13 m/s. The frictional force cannot
supply the radial acceleration needed for the car to go around
the curve—the car goes into a skid.

45. ©McGraw-Hill Education
Example 5.7 Solution Continued 2
c)
s
v gr


0.50
s
 
2
max 0.50 9.80 m/s 25 m 11 m/s
v
    

46. ©McGraw-Hill Education
Example 5.7 Solution Continued 3
d) Icy conditions mean we can assume the road is
frictionless. The normal force provides the needed
radial force. 2
sin
cos 0
x
y
v
F N m
r
F N mg


 
  


2 2
sin /
tan
cos
N mv r v
N mg rg



  
2 2
1 1
2
(13 m/s)
tan tan 35
25 m 9.80 m/s
v
rg
  
    


47. ©McGraw-Hill Education
5.4 Circular Orbits of Satellites and Planets
A satellite can orbit Earth in
a circular path because of
the long-range gravitational
force on the satellite due to
the Earth.
The magnitude of the
gravitational force on the
satellite is
1 2
2
Gm m
F
r


48. ©McGraw-Hill Education
Circular Orbits Continued
Since gravity is the only force acting on the satellite,
E
2
,
r
mM
F G
r


where r is the distance from the center of the Earth to the
satellite.
Then, from Newton’s second law,
2
r r
v
F ma m
r
 

2
E E
2
mM mv GM
v
r r r
G   

49. ©McGraw-Hill Education
Example 5.8
The Hubble Space Telescope is in a circular orbit 613 km above
Earth’s surface. The average radius of the Earth is 6.37 × 103 km
and the mass of Earth is 5.97 × 1024 kg.
What is the speed of the telescope in its orbit?
Strategy
First, find the orbital radius of the telescope. It is not 613 km.
Then, use Newton’s second law, along with what we know about
radial acceleration.

50. ©McGraw-Hill Education
Example 5.8 Solution
2 3
3
10 km 6.37 10 km
6.98 10 km
6.13
r  
 
 
2
E
2
r
GmM mv
F
r r
 

E
v
GM
r

11 2 2 24
6
6.67 10 N m /kg 5.97 10 kg
6.98 10 m
7550 m/s 27200 km/h
v

   





51. ©McGraw-Hill Education
Kepler’s Laws of Planetary Motion
1. The planets travel in elliptical orbits with the Sun at one
focus of the ellipse.

52. ©McGraw-Hill Education
Kepler’s Laws Continued
2. A line drawn from a planet to the Sun sweeps out equal
areas in equal time intervals.
3. The square of the orbital period is proportional to the cube
of the average distance from the planet to the Sun.

53. ©McGraw-Hill Education
Application of Radial Acceleration: Kepler’s
Third Law for a Circular Orbit
2
Sun
2
r
GmM mv
F
r r
 
 Sun 2
GM r
r T
v




Kepler’s third law: the
square of the period of
a planet is directly
proportional to the
cube of the average
orbital radius.
3
Sun
2
r
T
GM


2
2 3 3
Sun
constant
T
r
r r
GM

  

54. ©McGraw-Hill Education
Geostationary Orbits
Many satellites, such as those used for communications, are
placed in a geostationary (or geosynchronous ) orbit—a circular
orbit in Earth’s equatorial plane whose period is equal to Earth’s
rotational period .
A satellite in geostationary orbit remains directly above a
particular point on the equator.

55. ©McGraw-Hill Education
Example 5.9
A 300.0-kg communications satellite is placed in a geostationary
orbit 35,800 km above a relay station located in Kenya.
What is the speed of the satellite in orbit?
Strategy
The period of the satellite is 1 d or approximately 24 h.
To find the speed of the satellite in orbit we use Newton’s law of
gravity and his second law of motion along with what we know
about radial acceleration.

56. ©McGraw-Hill Education
Example 5.9 Solution
2
E E
2
r
GmM mv GM
F v
r r r
   

E
7 7 7
3.58 10 m 0.638 10 m 4.217 10 m
r h R
   




11 2 2 24
7
6 2 2
3
6.67 10 N m /kg 5.97 10 kg
4.217 10 m
9.443 10 m /s
3.07 10 m/s
v

   

 




57. ©McGraw-Hill Education
Example 5.10
A satellite revolves about Earth with an orbital radius of r1 and
speed v1 .
If an identical satellite were set into circular orbit with the same
speed about a planet of mass three times that of Earth, what
would its orbital radius be?

58. ©McGraw-Hill Education
Example 5.10 Solution
2
E 1
2
1 1
GmM mv
r r
 E
1 2
1
GM
r
v
 
2
2
E 1
2
2
3
Gm M mv
r r

 E
2 2
1
3
G M
r
v

 
2
2 E 1
2 1
2
1 E 1
3 /
3 3
/
r G M v
r r
r GM v
   

59. ©McGraw-Hill Education
5.5 Nonuniform Circular Motion
For circular motion, the direction of the acceleration is not radial
if the speed is changing.
2 2
r t
a
a a
 
2
2
( in radians per unit time)
r
a
v
r
r
 
 

60. ©McGraw-Hill Education
Problem-Solving Strategy for an Object in
Nonuniform Circular Motion
For nonuniform circular motion, use the same strategy as for
uniform circular motion. The only difference is that now the
tangential acceleration component at is nonzero:
t t
F ma


The tangential acceleration component at determines how the
speed of the object changes.

61. ©McGraw-Hill Education
Example 5.11
A roller coaster includes a vertical circular loop of radius 20.0 m.
What is the minimum speed at which the car must move at the
top of the loop so that it doesn’t lose contact with the track?

62. ©McGraw-Hill Education
Example 5.11 Solution
2
top
r r
mv
F N mg ma
r
   

2 2
top top
0
mv v
mg m g
r
N
r
 
   
 
 
 

top
v gr

2
top 9.80 m/s 20.0 m 14.0 m/s
v gr
   

63. ©McGraw-Hill Education
Example 5.12
A pendulum is released from rest at point A.
a) Sketch a qualitative motion diagram from B to D.
b) Sketch an FBD and the acceleration vector for the pendulum
bob at points B and C.

64. ©McGraw-Hill Education
Example 5.12 Strategy
a) The pendulum bob moves along the arc of a circle, but not at
constant speed. The spacing between points on the motion
diagram is larger where the bob is moving faster.
b) Two forces appear on each FBD: gravity and the force due to
the cord. The gravitational force is the same at both points
(magnitude mg, direction down), but the force due to the
cord varies in magnitude and in direction.
2
r
a
v
r


65. ©McGraw-Hill Education
Example 5.12 Solution
a)

66. ©McGraw-Hill Education
Example 5.12 Solution Continued
b)

67. ©McGraw-Hill Education
5.6 Angular Acceleration
Average angular acceleration:
2 1
av
2 1
t t t
  

 
 
 
Instantaneous angular acceleration:
0
lim
t t


 



t
v r 

(in the limit 0)
t
t
a
v
r t
t t

 
   
 
t
a r 


68. ©McGraw-Hill Education
Constant Angular Acceleration
The mathematical relationships between
, , and 
are the same as the mathematical relationships between
x, vx, and ax
that were developed in Chapter 4. Each quantity is the
instantaneous rate of change of the preceding quantity.

69. ©McGraw-Hill Education
Relationships Between , , and  for
Constant Angular Acceleration
Constant Acceleration
Along x-axis
f i
1
f i
2
2
1
i 2
2 2
f i
(
( )
)
2
x x x x
x x
x x
x x x
v v v a t
x v v t
x v t a
a x
v
t
v
    
  


   
  
Constant Angular
Acceleration
f i
1
f i
2
2
1
i 2
2 2
f i
(
( )
2
)
t
t
t t
   
  
  
   
    
  
   



 

70. ©McGraw-Hill Education
Example 5.13
A potter’s wheel rotates from rest to 210 rpm in a time of 0.75 s.
a) What is the angular acceleration of the wheel during this
time, assuming constant angular acceleration?
b) How many revolutions does the wheel make during this time
interval?
c) Find the tangential and radial components of the
acceleration of a point 12 cm from the rotation axis when
the wheel is spinning at 180 rpm.

71. ©McGraw-Hill Education
Example 5.13 Strategy
We know the initial and final frequencies, so we can find the
initial and final angular velocities.
We also know the time it takes for the wheel to get to the final
angular velocity.
That is all we need to find the average angular acceleration that,
for constant angular acceleration, is equal to the instantaneous
angular acceleration.

72. ©McGraw-Hill Education
Example 5.13 Solution
a)
i 0 rad/s
 
f
rev 1 min rad
210 2 7.0 rad/s
min 60 s rev
  
   
2
f i
f i
7.0 rad/s 0 7.0 rad/s
29 rad/s
0.75 s 0 0.75 s
t t
   

 
   
 

73. ©McGraw-Hill Education
Example 5.13 Solution Continued
b)
1 1
f i
2 2
( ) (7.0 rad/s 0)(0.75 s) 8.25 rad
t
   
      
8.25 rad
1.3 rev
2 rad/rev


c) rev 1 min rad
180 2 6.0 rad/s
min 60 s rev
  
   
2 2 2
(6.0 rad/s) 0.12 m 43 m/s
r
a r
 
   
2 2
29 rad/s 0.12 m=3.5 m/s
t
a r

  

74. ©McGraw-Hill Education
5.7 Apparent Weight and Artificial Gravity
Comparing an astronaut’s weight in orbit with his or her weight
on Earth’s surface, we find
E
2 2
2
orbit E
E
2 2
E
surface E
2
E
(6400 km)
( )
0.84
( ) (7000 km)
GM m
W R
R h
GM m
W R h
R

   

So, why does the astronaut seem to be weightless? An astronaut
feels weightless when
,

a g
where g is the local gravitational field.

75. ©McGraw-Hill Education
Example 5.14
Dave wants to practice vertical
circles for a flying show exhibition.
a) What must the minimum radius
of the circle be to ensure that
his acceleration at the bottom
does not exceed 3.0 g? The
speed of the plane is 78 m/s at
the bottom of the circle.
b) What is Dave’s apparent weight at the bottom of the circular
path? Express your answer in terms of his true weight.

76. ©McGraw-Hill Education
Example 5.14 Strategy
For the minimum radius, we use the maximum possible radial
acceleration, since
2
r
a
v
r

For the maximum radial acceleration, the tangential
acceleration must be zero - the magnitude of the acceleration is
2 2
r t
a
a a
 

77. ©McGraw-Hill Education
Example 5.14 Solution
a) 2
r
a
v
r

2 2
2
2
3.0
(78 m/s)
210 m
3.0 9.8 m/s
r
r
v v
a g

 



78. ©McGraw-Hill Education
Example 5.14 Solution Continued
b)
3.0
( ) 4.0
y y
y
y
F N mg ma
g
W N m g a mg
a
  
 
    

His apparent weight is 4.0 times his true weight.