Uniform circular motion is the motion of an object traveling at a constant (uniform) speed on a circular path.
r
Period T is the time required to travel once around the circle, that is, to make one complete revolution.
2
 r v

T
1

The wheel of a car has a radius of r = 0.29m and is being rotated at 830 revolutions per minute (rpm) on a tire-balancing machine. Determine the speed
(in m/s) at which the outer edge of the wheel is moving.
2
 r
The speed v can be obtained directly from ,
T but first the period T is needed. It must be v
 expressed in seconds.
2

830 revolutions in one minute
1
830 revolution s / min

1 .
2

10

3 min/ revolution
T=1.2
*
10 -3 min, which corresponds to 0.072s
v

2
 r
T

2

( 0 .
29 m )
0 .
072 s

25 m / s
3

1. The speed , or the magnitude of the velocity vector, is constant .
2. Direction of the vector is not constant .
3. Change in direction, means acceleration
4. “Centripetal acceleration” , it points toward the center of the circle.
4

Magnitude a c of the centripetal acceleration depends on the speed v of the object and the radius r of the circular path. a c
=v 2 /r
5

 v in velocity divided by the elapsed time
  t
Sector of the circle COP.
 t is very small, the arc length OP is approximately a straight line whose length
 traveled by the object.
6

COP is an isosceles triangle. Both triangles
 v v
 v
 t r
 v /
 t a c
=v 2 /r
The direction is toward the center of the circle.
7

An object on a guideline is in uniform circular motion. The object is symbolized by a dot, and at point O it is release suddenly from its circular path.
If the guideline is cut suddenly, will the object move along
OA or OP ?
8

Newton’s first law of motion guides our reasoning. An object continues in a state of rest or in a state of motion at a constant speed along a straight line unless compelled to changes that state by a net force. When the object is suddenly released from its circular path, there is no longer a net force being applied to the object. In the case of a model airplane, the guideline cannot apply a force, since it is cut. Gravity certainly acts on the plane, but the wings provide a lift force that balances the weight of the plane.
9

In the absence of a net force, then, the plane or any object would continue to move at a constant speed along a straight line in the direction it had at the time of release. This speed and direction are given in Figure 5.4 by the velocity vector v.
As a result, the object would move along the straight line between points O and A, not on the circular arc between points O and P.
10

The bobsled track at the 1994
Olympics in Lillehammer,
Norway, contained turns with radii of 33 m and 24 m, as the figure illustrates. Find the centripetal acceleration at each turn for a speed of 34 m/s, a speed that was achieved in the two-man event. Express the answers as multiples of g=9.8m/s 2 .
11

From a c
=v 2 /r it follows that
Radius=33m a c

( 34 m / s )
2
33 m

35 m / s
2 
3 .
6 g
Radius=24m a c

( 34 m / s )
2
24 m

48 m / s
2 
4 .
9 g
12

A car moves at a constant speed, and there are three parts to the motion. It moves along a straight line toward a circular turn, goes around the turn, and then moves away along a straight line. In each of three parts, is the car in equilibrium?
13

An object in equilibrium has no acceleration, according to the definition given in Section 4.11. As the car approaches the turn, both the speed and direction of the motion are constant.
Thus, the velocity vector does not change, and there is no acceleration. The same is true as the car moves away from the turn. For these parts of the motion, then, the car is in equilibrium. As the car goes around the turn, however, the direction of travel changes, so the car has a centripetal acceleration that is characteristic of uniform circular motion.
Because of this acceleration, the car is not in equilibrium during the turn.
In general, an object that is in uniform circular motion can never be in equilibrium.
14

The car in the drawing is moving clockwise around a circular section of road at a constant speed. What are the directions of its velocity and acceleration at (a) position 1 and (b) position 2?
15

(a) The velocity is due south, and the acceleration is due west.
(b) The velocity is due west, and the acceleration is due north.
16

17

Concepts at a glance: Newton’s second law indicates that whenever an object accelerates, there must be a net force to create the acceleration. Thus, in uniform circular motion there must be a net force to produce the centripetal acceleration. As the Concept-at-aglance chart, the second law gives this net force as the product of the object’s mass m and its acceleration v 2 /r. This chart is an expanded version of the chart shown previously in Figure 4.9. The net force causing the centripetal acceleration is called the centripetal force F
C and points in the same direction as the acceleration- that is, toward the center of the circle.
18

F
C
 mv
2 r
“centripetal force” does not denote a new and separate force created by nature.
19

The model airplane has a mass of 0.90 kg and moves at a constant speed on a circle that is parallel to the ground. The path of the airplane and its guideline lie in the same horizontal plane, because the weight of the plane is balanced by the lift generated by its wings. Find the tension T in the guideline(length=17m) for speeds of 19 and 38m/s.
20

Equation 5.3 gives the tension directly:
F
C
=T=mv 2 /r
Speed =19m/s
T

( 0 .
90 kg )( 19 m / s )
2
17 m

19 N
Speed =38m/s
T

( 0 .
90 kg )( 38 m / s )
2
17 m

76 N
21

In a circus, a man hangs upside down from a trapeze, legs bent over the bar and arms downward, holding his partner. Is it harder for the man to hold his partner when the partner hangs straight down and is stationary or when the partner is swinging through the straight-down position?
22

Reasoning and Solution: When the man and his partner are stationary, the man’s arms must support his partner’s weight. When the two are swinging, however, the man’s arms must do an additional job. Then the partner is moving on a circular arc and has a centripetal acceleration. The man’s arms must exert and additional pull so that there will be sufficient centripetal force to produce this acceleration.
Because of the additional pull, it is harder for the man to hold his partner while swinging than while stationary.
23


S
 dry

0 .
9
Compare the maximum speeds at which a car can safely negotiate an unbanked turn (r= 50.0m)

S
--dry = 0.9

S
--icy = 0.1
N
F
S s f max
  s
N
F
S mg
24

f s
MAX  
S
F
N f S
The car does not accelerate ,
F
N
– mg = 0 F
N
= mg.
N mg
  s g
 v
2 r v
  s gr
25

v


( 0 .
900 )( 9 .
80 m / s
2
)( 50 .
0 m )

21 .
0 m / s
 v

( 0 .
100 )( 9 .
80 m / s
2
)( 50 .
0 m )

7 .
00 m / s
As expected, the dry road allows the greater maximum speed.
26

Upward on the wing surfaces with a net lifting force L, the
  lifting force is directed toward the center of the turn.
Greater speeds and/or tighter turns require greater centripetal forces. Banking into a turn also has an application in the construction of high-speed roadways.
27

A car is traveling in uniform circular motion on a section of road whose radius is r. The road is slippery, and the car is just on the verge of sliding.
(a) If the car’s speed was doubled, what would have to be the smallest radius in order that the car does not slide? Express your answer in terms of r.
(b)What would be your answer to part (a) if the car were replaced by one that weighted twice as much?
28

(a) 4r (b) 4r
29

30

A car is going around a friction-free banked curve. The radius of the curve is r.
F
N
F
C

F
N sin
  mv
2 r
F
N vertical direction, this component must balance the weight mg of the car.
F
N cos
  mg
31

F
N
F
N sin
 cos

 mv
2
/ r mg tan
  v
2 rg would slide down a frictionless banked curve: at a speed that is too large, a car would slide off the top.
32

The Daytona 500 is the major event of the
NASCAR (National Association for Stock Car
Auto Racing) season. It is held at the Daytona
International Speedway in Daytona, Florida. The turns in this oval track have a maximum radius(at the top)of r=316m and are banked
 
31
 radius turns were frictionless. At what speed would the cars have to travel around them?
33

From Equation 5.4, it follows that v
 rg tan
 
( 316 m )( 9 .
80 m / s
2
) tan 31
 
43 m / s
(96 mph)
Drivers actually negotiate the turns at speeds up to
195 mph, however, which requires a greater centripetal force than that implied by Equation 5.4 for frictionless turns.
34

F
C

G mM
E r
2
 mv
2 r v

GM
E r
35

If the satellite is to remain in an orbit of radius r, the speed must have precisely this value.
The closer the satellite is to the earth, the smaller is the value for r and the greater the orbital speed must be.
Mass m of the satellite does not appear. For a given orbit, a satellite with a large mass has exactly the same orbital speed as a satellite with a small mass.
36

Determine the speed of the Hubble Space
Telescope orbiting at a height of 598 km above the earth’s surface.
Orbital radius r must be determined relative to the
center of the earth. The radius of the earth is approximately 6.38
*
10 6 m, r=6.98
*
10 6 m
37

The orbital speed is v

GM
E r

( 6 .
67

10

11
N .
m
2
/ kg
2
)( 5 .
98

10
24
6 .
98

10
6 m kg ) v

7 .
56

10
3 m / s ( 16900 mi / h )
38

39

The Hubble Telescope has detected the light being emitted from different regions of galaxy
M87. The black circle identifies the center of the galaxy. From the characteristics of this light, astronomers have determined an orbiting speed of 7.5
*
10 5 m/s for matter located at a distance of
5.7
*
10 17 m from the center. Find the mass M of the object located at the galactic center.
40

Replacing M
E with M v

GM r
M
 v
2 r
G

( 7 .
5

10
5 m / s )
2
( 5 .
7

10
17 m )
6 .
67

10

11
N .
m
2
/ kg
2
=4.8
*
10 39 kg
41

The ratio of this incredibly large mass to the mass of our sun is (4.8
*
10 39 kg)/(2.0
*
10 30 kg)=2.4
*
10 9 .
Matter equivalent to 2.4 billion suns is located at the center of galaxy M87. The volume of space in which this matter is located contains relatively few visible star. There are strong evidences for the existence of a super-massive black hole.
“ black hole ” tremendous mass prevents even light from escaping. The light that forms the image comes not from the black hole itself, but from matter that surrounds it.
42

Period T of a satellite is the time required for one orbital revolution. v

2
 r / T
GM
E r

2
 r
T
T

2
 r
3 / 2
GM
E
43

Period is proportional to the three-halves power of the orbital radius is know as Kepler’s third law. (Johannes Kepler, 1571-1630). Kepler’s third law also holds for elliptical orbits.
“ synchronous satellites ”: orbital period is chosen to be one day, the time it takes for the earth to turn once about its axis. Satellites move around their orbits in a way that is synchronized with the rotation of the earth, appearing in fixed positions in the sky and can serve as “stationary”.
44

The period T of a synchronous satellite is one day.
Find the distance r from the center of the earth and the height H of the satellite above the earth’s surface. The earth itself has a radius of 6.38
*
10 6 m.
T=8.64
*
10 4 s T

2
 r
3 / 2
GM
E
45

r
3 / 2 
T GM
E
2


( 8 .
64

10
4 s ) ( 6 .
67

10

11
N .
m
2
/ kg
2
)( 5 .
98

10
24 kg )
2
 r = 4.23
*
10 7 m
H=4.23
*
10 7 m-0.64
*
10 7 m=3.59
*
10 7 m (22300 mi)
46

Two satellites are placed in orbit, one about Mars and the other about Jupiter, such that the orbital speeds are the same. Mars has the smaller mass. Is the radius of the satellite in orbit about Mars less than, greater than, or equal to the radius of the satellite orbiting Jupiter?
v

GM r
Less than.
M
Mars
---smaller
Since v is the same, M small -----r small.
47

The idea of life on board an orbiting satellite conjures up visions of astronauts floating around in a state of “weightlessness”
48

Objects in uniform circular motion continually accelerate or “fall” toward the center of the circle, in order to remain on the circular path.
The only difference between the satellite and the elevator is that the satellite moves on a circle, so that its “falling” does not bring it closer to the earth. True weight is the gravitational force
(F=GmM
E
/r 2 ) that the earth exerts on an object and is not zero.
49

At what speed must the surface of the space station (r=1700m) move in the figure, so that the astronaut at point P experiences a push on his feet that equals his earth weight?
50

F
C
=mv 2 /r
Earth weight of the astronaut (mass=m) is mg.
F
C
=mg=mv 2 /r v
 rg

( 1700 m )( 9 .
80 m / s
2
)

130 m / s
51

The outer ring (radius=r
0
) simulates gravity on earth, while the inner ring (radius=r
1
) simulates gravity on Mars
A space laboratory is rotating to create artificial gravity. Its period of rotation is chosen so the outer ring (r
0
=2150m) simulates the acceleration due to gravity on earth (9.80 m/s 2 ).
What should be the radius r
1 of the inner ring, so it simulates the acceleration due to gravity on the surface of
Mars (3.72 m/s 2 )?
52

Centripetal acceleration: a c
=v 2 /r, speed v and radius r: v

2
 r / T
T is the period of the motion.
The laboratory is rigid. All points on a rigid object make one revolution in the same time.
Both rings have the same period.
a c
 v
2 r

2
 r
T r
2

4

2 r
T
2
53

9 .
80 m / s
2 
4

2
( 2150 m )
T
2
Outer ring
3 .
72 m / s
2 
4

T
2
2 r
1
Inner ring
Dividing the inner ring expression by the outer ring expression,
3 .
72 m / s
2
9 .
80 m / s
2
 r
1
2150 m r
1
= 816 m
54

The acceleration due to gravity on the moon is onesixth that on earth.
(a) Is the true weight of a person on the moon less than, greater than, or equal to the true weight of the same person on the earth?
(b)Is the apparent weight of a person in orbit about the moon less than, greater than, or equal to the apparent weight of the same person in orbit about the earth?
(a) Less than (b) Equal to
55

Usually, the speed varies in this stunt.
“non-uniform”
56

(1) F
N 1
 mg
 mv
1
2 r
=F
C1
(3)
F
N 3
 mg
 mv
3
2 r
=F
C3
(2) F
N 2
=F
C2
 mv
2
2 r
(4) F
N 4
 mv
4
2 r
=F
C4
The magnitude of the normal force changes, because the speed changes and the weight does not have the same effect at every point.
57

The weight is tangent to the circle at points 2 and 4 and has no component pointing toward the center. If the speed at each of the four places is known, along with the mass and radius, the normal forces can be determined.
They must have at least a minimum speed at the top of the circle to remain on the track. v
3 minimum when F
N3 is zero. v

3 rg is a
Weight mg provides all the centripetal force. The rider experiences an apparent weightlessness.
58

At time t=0 s, automobile A is traveling at a speed of 18 m/s along a straight road and its picking up speed with an acceleration that has a magnitude of 3.5 m/s 2 .
At time t=0 s, automobile A is traveling at a speed of 18 m/s in uniform circular motion as it negotiates a turn. It has a centripetal acceleration whose magnitude is also 3.5 m/s 2 .
Determine the speed of each automobile when t=2.0 s.
59

Which automobile has a constant acceleration?
Both its magnitude and direction must be constant.
A has constant acceleration, a constant magnitude of 3.5 m/s 2 and its direction always points forward along the straight road.
B has an acceleration with a constant magnitude of 3.5 m/s 2 , a centripetal acceleration, which points toward the center of the circle at every instant.
Which automobile do the equations of kinematics apply?
Apply for automobile A.
60

Speed of automobile A at t=2.0 s v=v
0
+at=18 m/s+(3.5 m/s 2 )(2.0 s)=25 m/s
B is in uniform circular motion and goes around the turn. At a time of t=2.0 s, its speed is the same as it was at t=0 s, i.e., v=18 m/s.
61

Ball A is attached to one end of a rigid mass-less rod, while an identical ball B is attached to the center of the rod. Each ball has a mass of m=0.50kg, and the length of each half of the rod is L=0.40m.
This arrangement is held by the empty end and is whirled around in a horizontal circle at a constant rate, so each ball is in uniform circular motion. Ball A travels at a constant speed of v
A
=5.0m/s. Find the tension in each half of the rod.
62

How many tension forces contribute to the centripetal force that acts on ball A?
A single tension force of magnitude T
A acts on ball A, due to the tension in the rod between the two balls. This force alone provides the centripetal force keeping ball A on its circular path of radius 2L
How many tension forces contribute to the centripetal force that acts on ball B?
Two tension forces act on ball B. T
B
-T
A
Is the speed of ball B the same as the speed of ball A?
No, it is not. Because A travels farther than B in the same time. A travels a distance equal to the circumference of its
  one half the speed of ball A , or v
B
=2.5 m/s
63

Ball A
T
A
 mv
A
2
2 L
Centripetal force F
C
Ball B
T
B

T
A
 mv
B
2
L
Centripetal force F
C
T
A
 mv
A
2
2 L

( 0 .
50 kg )( 5 .
0 m / s
2
)

16 N
2 ( 0 .
40 m )
T
B
 mv
B
2
L
 mv
A
2
2 L

( 0 .
50 kg )( 2 .
5 m / s
2
)

0 .
40 m
( 0 .
50 kg )( 5 .
0 m / s
2
)
2 ( 0 .
40 m )
=23N
64

R = 3.6m

= 25
OA = ?
65

REASONING AND SOLUTION Since the speed of the object on and off the circle remains constant at the same value, the object always travels the same distance in equal time intervals, both on and off the circle. Furthermore since the object travels the distance OA in the same time it would have moved from O to P on the circle, we know that the distance OA is equal to the distance along the arc of the circle from O to P.
66

Circumference =
2
 r

2

(3.6 m) = 22.6 m
360 o
1 o
25 o
22.6m
(22.6/360)m
(22.6/360)
*
25m and, from the argument given above, we conclude that the distance OA is 1.6m.
67

REASONING In Example 3, it was shown that the magnitudes of the centripetal acceleration for the two cases are
Radius

33 m a 35 m
C

/ s
2
Radius

24 m a 48
C

m / s
2
According to Newton's second law, the centripetal
F
C
 ma
C
68

SOLUTION a. Therefore, when the sled undergoes the turn of radius 33 m,
F
C
 ma
C

( 350 kg)(35 m / s
2
)

1.2

10
4
N b. Similarly, when the radius of the turn is 24 m,
F
C
 ma
C

( 350 kg)(48 m / s
2
)

1.7

10
4
N
69

REASONING AND SOLUTION The force
P supplied by the man will be largest when the partner is at the lowest point in the swing. The diagram at the right shows the forces acting on the partner in this situation. The centripetal force necessary to keep the partner swinging along the arc of a circle is provided by the resultant of the force supplied by the man and the weight of the partner.
70

From the figure
Therefore
P mg
 mv 2 r
P
 mv 2 r
 mg
Since the weight of the partner, W , is equal to mg , it follows that m = ( W / g ) and
P
 r
2
(6.50 m)
2

(475 N) = 594 N
71