Physics, 6th Edition
Chapter 10. Uniform Circular Motion
Chapter 10. Uniform Circular Motion
Centripetal Acceleration
10-1. A ball is attached to the end of a 1.5 m string and it swings in a circle with a constant speed
of 8 m/s. What is the centripetal acceleration?
v 2 (8 m/s) 2
ac  
R
1.5 m
ac = 42.7 m/s2
10-2. What are the period and frequency of rotation for the ball in Problem 10-1?
v  2 fR;
v
2 R
;
T
f 
2 R 2 (1.5 m)

;
v
8 m/s
T=
1
1

;
T 1.18 s
T = 1.18 s
f = 0.849 rev/s
10-3. A drive pulley 6-cm in diameter is set to rotate at 9 rev/s. What is the centripetal
acceleration of a point on the edge of the pulley? What would be the linear speed of a belt
around the pulley? [ R = (0.06 m/2) = 0.03 m ]
ac  4 2 f 2 R  4 2 (9 rev/s) 2 (0.03 m) ;
ac = 95.9 m/s2
v  2 fR  2 (9 rev/s)(0.03 m) ;
v = 1.70 m/s
10-4. An object revolves in a circle of diameter 3 m at a frequency of 6 rev/s. What is the period
of revolution, the linear speed, and the centripetal acceleration? [ R = (3 m/2) = 1.5 m ]
T
1
1

;
f 6 rev/s
T = 0.167 s ;
v  2 fR  2 (6 rev/s)(1.5 m) ; v = 56.5 m/s
ac 
v 2 (56.5 m/s)2
;

R
(1.5 m)
121
ac = 2130 m/s2
Physics, 6th Edition
Chapter 10. Uniform Circular Motion
10-5. A car moves around a curve 50 m in radius and undergoes a centripetal acceleration of 2
m/s2. What is its constant speed?
v2
a ;
R
v  aR  (2 m/s)2 (50 m) ;
v = 10.0 m/s
10-6. A 1500-kg car moves at a constant speed of 22 m/s along a circular track. The centripetal
acceleration is 6 m/s2. What is the radius of the track and the centripetal force on the car?
a
F
v2
v 2 (22 m/s)2
;
; R 
R
a
6 m/s 2
mv 2 (1500 kg)(22 m/s) 2
;

R
(80.7 m)
R = 80.7 m
Fc = 9000 N
10-7. An airplane dives along a curved path of radius R and velocity v. The centripetal
acceleration is 20 m/s2. If both the velocity and the radius are doubled, what will be the
new acceleration?
v2
a1  ;
R
(2v)2 4v 2
a2 

;
2R
2R
a2 = 2a1 = 2(20 m/s2;
2v 2
a2 
;
R
a = 40 m/s2
Centripetal Force
10-8. A 20-kg child riding a loop-the-loop at the Fair moves at 16 m/s through a track of radius
16 m. What is the resultant force on the child?
F
mv 2 (20 kg)(16 m/s)2

;
R
16 m
122
Fc = 320 N
Physics, 6th Edition
Chapter 10. Uniform Circular Motion
10-9. A 3-kg rock, attached to a 2-m cord, swings in a horizontal circle so that it makes one
revolution in 0.3 s. What is the centripetal force on the rock? Is there an outward force on
the rock?
2
 1 
Fc  4 f mR  4 
 (3 kg)(2 m) ;
 0.3 s 
2
2
2
Fc = 2630 N, No
10-10. An 8-lb object swings in horizontal circle with a speed of 95 ft/s. What is the radius of
the path, if the centripetal force is 2000 lb?
m
R
8 lb
mv 2

0.25
slug;
F

c
32 ft/s 2
R
mv 2 (0.25 slug)(95 ft/s) 2
;

Fc
2000 lb
R = 1.13 ft
*10-11. Two 8-kg masses are attached to the end of a tin rod 400 mm long. The rod is supported
in the middle and whirled in a circle. The rod can support a maximum tension of only 800
N. What is the maximum frequency of revolution? [ R = (400 mm/2) = 200 mm ]
Fc 
mv 2
; v
R
Fc R
(800 N)(0.20 m)

;
m
8 kg
v  2 fR; f 
v
2 R

4.47 m/s
;
2 (0.20 m)
v = 4.47 m/s
f = 3.56 rev/s
*10-12. A 500-g damp shirt rotates against the wall of a washer at 300 rpm. The diameter of the
rotating drum is 70 cm. What is the magnitude and direction of the resultant force on the
shirt?
[ R = (70 cm/2) = 35 cm; f = 300 rpm(60 s/min) = 1800 rev/s ]
Fc  4 2 f 2 mR  4 2 (1800 rev/s) 2 (0.5 kg)(0.35 m) ;
Fc = 2.24 x 107 N, toward the center
123
Physics, 6th Edition
Chapter 10. Uniform Circular Motion
*10-13. A 70-kg runner rounds a track of radius 25 m at a speed of 8.8 m/s. What is the central
force causing the runner to turn and what exerts the force?
mv 2 (70 kg)(8.8 m/s) 2
;
Fc 

R
25 m
Fc = 217 N, friction
*10-14. In Olympic bobsled competition, a team takes a turn of radius 24 ft at a speed of 60 mi/h.
What is the acceleration? How many g’s do passengers experience?
ac 
v 2 (88 ft/s)2
;

R
24 ft
(60 mi/h = 88 ft/s)
ac = 323 ft/s2 or 10.1 g’s
Flat Curves and Banked Curves
10-15. On a rainy day the coefficient of static friction between tires and the roadway is only 0.4.
What is the maximum speed at which a car can negotiate a turn of radius 80 m?
mv 2
  s mg ;
R
vc   s gR  (0.4)(9.8 m/s 2 )(80 m) ;
vc = 17.7 m/s or
63.8 km/h
10-16. A bus negotiates a turn of radius 400 ft while traveling at a speed of 60 mi/h. If slipping
just begins at this speed, what is the coefficient of static friction between the tires and the
road?
(60 mi/h = 88 ft/s)
mv 2
 s mg ;
R
s 
v2
(88 ft/s)2
;

gR (32 ft/s 2 )(400 ft)
s = 0.605
10-17. Find the coefficient of static friction necessary to sustain motion at 20 m/s around a turn
of radius 84 m.
mv 2
 s mg ;
R
s 
v2
(20 m/s)2
;

gR (9.8 m/s 2 )(84 m)
124
s = 0.486
Physics, 6th Edition
Chapter 10. Uniform Circular Motion
*10-18. A 20-kg child sits 3 m from the center of a rotating platform. If s = 0.4, what is the
maximum number of revolutions per minute that can be achieved without slipping?
(Slipping occurs when the centripetal force equals the maximum force of static friction.)
Fc  4 2 f 2 mR  s mg ;
f 
s g
(0.4)(9.8 m/s 2 )
;

4 2 R
4 2 (3 m)
f = 0.182 rev/s (60 s/min);
f = 10.9 rpm
*10-19. A platform rotates freely at 100 rpm. If the coefficient of static friction is 0.5, how far
from the center of the platform can a bolt be placed to without slipping?
f = 100 rev/min (1 min/60 s) = 1.67 rev/s; s = 0.5; R = ?
Fc  4 2 f 2 mR  s mg ;
R
s g
(0.5)(9.8 m/s 2 )
;

4 2 f 2
4 2 (1.67 rev/s)2
R = 21.1 cm
10-20. Find the required banking angle to negotiate the curve of Prob.10-15 without slipping.
v2
(17.7 m/s)2
;
tan  

gR (9.8 m/s2 )(80 m)
 = 21.80
10-21. Find the required banking angle for Problem 10-16 to prevent slipping?
tan  
v2
(88 ft/s)2
;

gR (32 ft/s 2 )(400 ft)
 = 31.20
10-22. The optimum banking angle for a curve of radius 20 m is found to be 280. For what
speed was this angle designed?
tan  
v2
;
gR
v  gR tan   (9.8 m/s 2 )(20 m) tan 280 ; v = 10.3 m/s
125
Physics, 6th Edition
Chapter 10. Uniform Circular Motion
*10-23. A curve in a road 9 m wide has a radius of 96 m. How much higher than the inside edge
should the outside edge be for an automobile to travel at the optimum speed of 40 km/h?
v = 40 km/h = 11.1 m/s;
tan  
v2
(11.1 m/s)2
;

gR (9.8 m/s2 )(96 m)
v2
tan  
gR
h = (9 m) sin ;
 = 7.480; h = (9 m) sin 7.480;
h = 1.17 m
The Conical Pendulum
10-24. A conical pendulum swings in a horizontal circle of radius 30 cm. What angle does the
supporting cord make with the vertical when the liner speed of the mass is 12 m/s?
tan  
v2
(12 m/s)2
;

gR (9.8 m/s 2 )(0.30 m)
 = 88.80
10-25. What is the linear speed of the flyweights in Fig. 10-16 if L = 20 cm and  = 600? What is
the frequency of revolution?
L

L = 20 cm = 0.20 m; R = L sin  = (0.2 m) sin 60 ; R = 0.173 m
0
R
tan  
v2
; v  gR tan   (9.8 m/s 2 )(0.173 m)( tan 600 ) ;
gR
v  2 fR; f 
v
2 R

1.71 m/s
;
2 (0.173 m)
v = 1.71 m/s
f = 1.58 rev/s
10-26. If the length of the L in Fig. 10-16 is 60 cm. What velocity is required to cause the
flyweights to move to an angle of 300 with the vertical?
R = L sin  = (60 cm) sin 300;
tan  
v2
;
gR
R = 30 cm = 0.30 m
v  gR tan   (9.8 m/s 2 )(0.30 m) tan 300 ;
126
v = 1.30 m/s
Physics, 6th Edition
Chapter 10. Uniform Circular Motion
10-27. Each of the flyweights in Fig. 10-16 has a mass of 2 kg. The length L = 40 cm and the
shaft rotates at 80 rpm. What is the tension in each arm? What is the angle ? What is
the height h?
T cos 

2
80 rpm = 1.33 rev/s; T sin = Fc = mv /R
sin  
R
R

;
L 0.4 m
T cos  = mg;
T sin 
mg
 R 
2 2
T
  4 f mR
0.4
m


T = (0.4 m)(4 2)(1.33 rev/s)2(2 kg);

h
R
T = 56.1 N
mg (2 kg)(9.8 m/s 2 )
cos 

;
T
56.1 N
h = L cos  = (0.4 m) cos 69.60 ;
L
h = 0.14 m
 = 69.60
or
h = 14.0 cm
10-28. In Fig. 10-16, assume that L = 6 in., each flyweight is 1.5 lb, and the shaft is rotating at
100 rpm. What is the tension in each arm? What is the angle ? What is the distance h?
100 rpm = 1.67 rev/s; T sin = Fc = mv2/R
T cos 
1.5 lb
m
 0.0469 slug ;
32 ft/s
R
R
sin   
;
L 0.5 ft
T sin 
mg
 R 
2 2
T
  4 f mR
 0.5 ft 
T = (0.5 ft)(4 2)(1.67 rev/s)2(0.0469 slug);
T cos  = mg;

L = 6 in. = 0.50 ft
cos 
T = 2.75 lb
mg (0.0469 slug)(32 ft/s 2 )

;
T
2.57 lb
h = L cos  = (0.5 ft) cos 54.30 ;
127
 = 54.30
h = 0.292 ft
L

R
h
Physics, 6th Edition
Chapter 10. Uniform Circular Motion
10-29. Consider the rotating swings in Fig. 10-17. The length L = 10 m and the distance a = 3
m. What must be the linear velocity of the seat if the rope is to make an angle of 300 with
the vertical?
 = 300 ]
[ L = 10 m; a = 3 m,
b = L sin  = (10 m) sin 300 = 5 m; R = a + b = 7 m;
tan  
v2
;
gR
a

L
b
v  gR tan 
R=a+b
v  (9.8 m/s 2 )(7 m) tan 300
v = 6.73 m/s
10-30. What must be the frequency of revolution for the swing in Problem 10-17 if the angle  is
to be equal to 250?
b = L sin 280 = (10 m) sin 280; b = 4.695 m; R = a + b = 7.695 m
tan  
v2
;
gR
v  2 fR;
v  gR tan  ;
f 
v
2 R

v  (9.8 m/s 2 )(7.695 m) tan 280 ; v = 6.33 m/s
6.33 m/s
;
2 (7.695 m)
f = 0.131 rev/s or 7.86 rpm
Motion in a Vertical Circle
10-31. A rock rests on the bottom of a bucket moving in a vertical circle of radius 70 cm. What
is the least speed the bucket must have as it rounds the top of the circle if it is to remain
in the bucket?
[ Resultant force to center = mv2/R ]
0
mv 2
T  mg 
; Critical speed vc is when T = 0
R
vc  gR  (9.8 m/s 2 )(0.7 m) ;
vc = 6.86 m/s
128
T
mg
R
Physics, 6th Edition
Chapter 10. Uniform Circular Motion
10-32. A 1.2-kg rock is tied to the end of a 90-cm length of string. The rock is then whirled in a
vertical circle at a constant speed. What is the critical velocity at the top of the path if the
v
string is not to become slack?
0
T  mg 
T
mv 2
; Critical speed vc is when T = 0
R
vc  gR  (9.8 m/s 2 )(0.7 m) ;
mg
R
vc = 2.97 m/s
*10-33. Assume that the rock of Problem 10-32 moves in a vertical circle at a constant speed of
8 m/s? What are the tensions in the rope at the top and bottom of the circle.
At Top: T  mg 
mv 2
and
R
T
mv 2
 mg
R
(1.2 kg)(8 m/s) 2
T
 (1.2 kg)(9.8 m/s 2 ) ;
(0.9 m)
At Bottom: T  mg 
mv 2
and
R
T
v
T
T = 73.6 N
mv 2
 mg
R
(1.2 kg)(8 m/s) 2
T
 (1.2 kg)(9.8 m/s 2 ) ;
(0.9 m)
mg
R
T
v
mg
T = 97.1 N
*10-34. A test pilot in Fig. 10-18 goes into a dive at 620 ft/s and pulls out in a curve of radius
2800 ft. If the pilot weighs 160 lb, what acceleration will be experienced at the lowest
point? What is the force exerted by the seat on the pilot?
W = 160 lb; m =160 lb/32 ft/s2 = 5 slugs; v = 620 ft/s;
a
v 2 (620 ft/s)2
;

R
(2800 ft)
N  mg 
mv 2
;
R
a = 137 ft/s2
N 
129
mv 2
 mg
R
Physics, 6th Edition
Chapter 10. Uniform Circular Motion
N 
*10-34. (Cont.)
(5 slugs)(620 ft/s) 2
 160 lb ;
(2800 ft)
N = 846 lb
*10-35. If it is desired that the pilot in Problem 10-34 not experience an acceleration greater
than 7 times gravity (7g), what is the maximum velocity for pulling out of a dive of
radius 1 km?
a
v2
 7 g;
R
v  7 gR  7(9.8 m/s 2 )(1000 m) ;
v = 262 m/s or 943 km/h
Note: The pilot actually “feels” a force that is eight times W:
N – mg = m(7g); N = 8mg
*10-36. A 3-kg ball swings in a vertical circle at the end of an 8-m cord. When it reaches the
top of its path, its velocity is 16 m/s. What is the tension in the cord? What is the critical
speed at the top?
[ R = 8 m; m = 3 kg; v = 16 m/s ]
At Top: T  mg 
T
mv 2
and
R
T
v
T
mv 2
 mg
R
(3 kg)(16 m/s) 2
 (3 kg)(9.8 m/s 2 ) ;
(8 m)
mg
R
T = 66.6 N
When T = 0, vc  gR  (9.8 m/s 2 )(8 m) ;
vc = 8.85 m/s
*10-37. A 36-kg girl rides on the seat of a swing attached to two chains that are each 20 m long.
If she is released from a position 8 m below the top of the swing, what force does the
swing exert on the girl as she passes the lowest point?

8m
cos  
;
20 m
  66.40 ; h = 20 m– 8 m = 12 m
½ mv 2  mgh;
v  2 gh  2(9.8 m/s 2 )(12 m) ;
T
130
v = 15.3 m/s
mg
8m
h
Physics, 6th Edition
Chapter 10. Uniform Circular Motion
At Bottom: T  mg 
*10-37. (Cont.)
T
mv 2
and
R
T
mv 2
 mg
R
(36 kg)(15.34 m/s) 2
 (36 kg)(9.8 m/s 2 ) ;
(20 m)
T = 776 N
Gravitation
10-38. How far apart should a 2-ton weight be from a 3-ton weight if their mutual force of
attraction is equal to 0.0004 lb?
( G = 3.44 x 10-8 lb ft2/slug2 )
m1 
(2 ton)(2000 lb/ton)
(3 ton)(2000 lb/ton)
 125 slugs; m2 
 187.5 slugs
2
32 ft/s
32 ft/s 2
F
Gm1m2
;
R2
R
Gm1m2
(3.44 x 10-8 lb ft 2 /slug 2 )(125 slug)(187.5 slug)

F
0.0004 lb
R = 1.42 ft
10-39. A 4-kg mass is separated from a 2 kg mass by a distance of 8 cm. Compute the
gravitational force of attraction between the two masses.
F
Gm1m2 (6.67 x 10-11 N m 2 /kg 2 )(4 kg)(2 kg)

;
R2
0.08 m
F = 8.34 x 10-8 N
*10-40. A 3-kg mass is located 10 cm away from a 6-kg mass. What is the resultant gravitational
force on a 2-kg mass located at the midpoint of a line joining the first two masses?
F3 
F6 
Gm3m2 (6.67 x 10-11 N m2 /kg 2 )(3 kg)(2 kg)

R2
(0.05 m)2
3 kg
Gm6 m2 (6.67 x 10-11 N m2 /kg 2 )(6 kg)(2 kg)

R2
(0.05 m)2
0.05 m
F3
F3 = -1.6 x 10-7 N, F6 = 3.20 x 10-7 N
FR = -1.6 x 10-7 N + 3.60 x 10-7 N;
131
FR = 1.60 x 10-7 N
2 kg
0.05 m
F6
6 kg
Physics, 6th Edition
Chapter 10. Uniform Circular Motion
*10-41. On a distant planet, the acceleration due to gravity is 5.00 m/s2. and the radius of the
planet is roughly 4560 m. Use the law of gravitation to estimate the mass of this planet.
mg 
Gmmp
R
2
;
gR 2 (5.00 m/s 2 )(4560 m) 2
;
mp 

G
6.67 x 10-11 N m2 / kg 2
mp = 1.56 x 1024 kg
*10-42. The mass of the earth is about 81 times the mass of the moon. If the radius of the earth is
4 times that of the moon, what is the acceleration due to gravity on the moon?
me = 81mm; Re = 4Rm ; Consider test mass m on moon and then on earth:
mg m 
Gmmm
Gm
Gmme
Gm
and g m = 2m ; mg e 
and g e = 2 e
2
2
Rm
Rm
Re
Re
g m mm Re2 mm (4 Rm ) 2
;


ge me Rm2
81mm Rm2
gm
 0.1975 ;
9.8 m/s 2
gm = 1.94 m/s2
*10-43. At 60-kg mass and a 20-kg mass are separated by 10 m. At what point on a line joining
these charges will another mass experience zero resultant force? [ F2 = F6 ]
m
x2
 6
2
(10  x)
m2
Gm2 m ' Gm6 m '
;

(10  x) 2
x2
60 kg
x
F6
m6
x
60 kg


 1.732 ;
(10  x)
m2
20 kg
x = 1.732(10 – x);
m’ 10 - x 20 kg
F2
x = 17.32 – 1.732 x
x = 6.34 m from the 60-kg mass.
Kepler’s Laws and Satellites
10-44. What speed must a satellite have if it is to move in a circular orbit of 800 km above the
surface of the earth? [ The central force Fc must equal the gravitational force Fg.]
Note that : R = Re + h = (6.38 x 106 m) + (0.8 x 106 m) = 7.18 x 106 m
132
Physics, 6th Edition
Chapter 10. Uniform Circular Motion
10-44. (Cont.)
Fc = Fg ;
v
mv 2 Gmme

;
R
R2
v
Gme
;
R
(6.67 x 10-11 N m 2 / kg 2 )(5.98 x 1024 kg)
;
7.18 x 106 m
v = 7450 m/s
10-45. The mass of the Jupiter is 1.90 x 1027 kg and its radius is 7.15 x 107 m. What speed must
a spacecraft have to circle Jupiter at a height of 6.00 x 107 m above the surface of Jupiter?
mv 2 Gmme

;
R = Rj + h = 7.15 x 10 m + 6 x 10 m; R = 1.315 x 10 m ;
R
R2
7
7
8
Gme
(6.67 x 10-11 )(1.9 x 1027 kg)

Rh
1.315 x 108 m
v
v =31,400 m/s
This represents a speed of approximately 69,800 mi/h.
10-46. What is the orbital speed of a satellite that moves in an orbit 1200 km above the earth’s
surface?
Note that : R = Re + h = (6.38 x 106 m) + (1.2 x 106 m) = 7.58 x 106 m
Fc = Fg ;
mv 2 Gmme

;
R
R2
v
Gme
;
R
(6.67 x 10-11 N m 2 / kg 2 )(5.98 x 1024 kg)
v
;
7.58 x 106 m
v = 7254 m/s
10-47. The radius of the moon is 1.74 x 106 m and the acceleration due to its gravity is 1.63
m/s2. Apply the law of universal gravitation to find the mass of the moon.
mg 
Gmmm
;
Rm2
mm 
gRm2 (1.63 m/s 2 )(1.74 x 106 m)2

;
G
6.67 x 10-11 N m2 /kg 2
133
mm = 7.40 x 1022 kg
Physics, 6th Edition
Chapter 10. Uniform Circular Motion
*10-48.A satellite is located at a distance of 900 km above the earth’s surface. What is the period
of the satellites motion? [ R = 6.38 x 106 m + 0.9 x 106 m = 7.28 x 106 m ]
 4 2  3
4 2 (7.28 x 106 )3
T2  
;
R 
(6.67 x 10-11N m2 /kg 2 )(5.98 x 1024kg)
 Gme 
T  3.82 x 107 s2 ;
T = 6180 s
T 2 = 3.82 x 107 s 2
(about an hour and 43 minutes)
*10-49. How far above the earth’s surface must a satellite be located if it is to circle the earth in
a time of 28 h?
T = 28 h (3600 s/h) = 1.01 x 105 s;
T2 = 1.02 x 1010 s2
 4 2  3 3 GmeT 2 (6.67 x 10-11N m2 /kg 2 )(5.98 x 1024kg)(1.02 x 1010s2 )
T2  

R ; R 
2
Gm
4

4 2
e 

R  3 1.03 x 1023m3 ;
R = 4.69 x 107 m; h = R – Re = 4.05 x 107 m
Challenge Problems
10-50. At what frequency should a 6-lb ball be revolved in a radius of 3 ft to produce a
centripetal acceleration of 12 ft/s2? What is the tension in the cord?
ac  4 2 f 2 R;
f2
ac
4 2 R

(12 ft/s2 )
; f  0.1013 s-2 ;
4 2 (3 ft)
 6 lb 
T  mac  
(12 ft/s 2 ) ;
2 
 32 ft/s 
f = 0.318 rev/s
T = 2.25 lb
10-51. What centripetal acceleration is required to move a 2.6 kg mass in a horizontal circle of
radius 300 mm if its linear speed is 15 m/s? What is the centripetal force?
ac 
v 2 (15 m/s)2
;

R (0.300 m)
a = 750 m/s2
Fc = mac = (2.6 kg)(750 m/s2);
134
Fc = 1950 N
Physics, 6th Edition
Chapter 10. Uniform Circular Motion
10-52. What must be the speed of a satellite located 1000 mi above the earth’s surface if it is to
travel in a circular path? [ R = 4000 mi + 1000 mi = 5000 mi ; 5000 mi = 2.64 x 107 ft ]
mv 2 Gmme

;
R
R2
v
v
 0.3048 m 
6
R  2.64 x 107 ft 
  8.047 x 10 m
 1 ft 
Gme
;
R
(6.67 x 10-11 N m 2 / kg 2 )(5.98 x 1024 kg)
;
8.047 x 106 m
v = 7041 m/s
10-53. A 2-kg ball at swings in a vertical circle at the end of a cord 2 m in length. What is the
critical velocity at the top if the orbit is to remain circular?
T
mg
mv 2
; Critical speed vc is when T = 0
T  mg 
R
vc  gR  (9.8 m/s 2 )(2.0 m) ;
R
vc = 4.42 m/s
*10-54. A 4-kg rock swings at a constant speed of 10 m/s in a vertical circle at the end of a 1.4
m cord. What are the tensions in the cord at the top and bottom of the circular path?
At Top: T  mg 
mv 2
and
R
T
mv 2
 mg
R
v
T
(4 kg)(10 m/s)2
T
 (4 kg)(9.8 m/s 2 ) ;
(1.4 m)
At Bottom: T  mg 
mv 2
and
R
T
T = 247 N
mv 2
 mg
R
mg
T
v
mg
T
2
(4 kg)(10 m/s)
 (4 kg)(9.8 m/s 2 ) ;
(1.4 m)
135
T = 325 N
R
Physics, 6th Edition
Chapter 10. Uniform Circular Motion
*10-55. What frequency of revolution is required to raise the flyweights in Fig. 10-16 a vertical
distance of 25 mm above their lowest position. Assume that L = 150 mm.
h = 150 mm – 25 mm = 124 mm; h = 0.124 m
1
f 
2
g
1

h 2
9.8 m/s 2
;
0.125 m
L

h
R
f = 1.41 rev/s = 84.6 rpm
*10-56. The combined mass of a motorcycle and driver is 210 kg. If the driver is to negotiate a
loop-the-loop of radius 6 m, what is the critical speed at the top?
vc  gR  (9.8 m/s 2 )(6.0 m) ;
vc = 7.67 m/s
*10-57. If the speed at the top of the loop in Prob. 10-54 is 12 m/s, what is the normal force at
the top of the loop?
At Top: N  mg 
N 
T
mv 2
and
R
N 
mv 2
 mg
R
(210 kg)(12 m/s) 2
 (210 kg)(9.8 m/s 2 ) ;
(6 m)
mg
R
N = 2980 N
10-58. The speed limit at a certain turn of radius 200 ft is 45 mi/h. What is the optimum
banking angle for this situation. Are roads actually constructed at the optimum angles?
v = 45 mi/h = 66.0 ft/s;
tan  
v2
(66 ft/s)2

;  = 34.20; NO
2
gR (9.8 m/s )(200 ft)
*10-59. For the figure shown in Fig. 10-17, assume that a = 2 m and L = 4 m.
a
0
Find the speed to cause the swing to move out to an angle of 20 ?

L
b = L sin  = (4 m) sin 200 = 1.37 m; R = a + b = 3.37 m;
v2
tan  
;
gR
b
v  gR tan 
v = 3.47 m/s
136
R=a+b
Physics, 6th Edition
Chapter 10. Uniform Circular Motion
Critical Thinking Questions
*10-60. A coin rests on a rotating platform distance of 12 cm from the center of rotation. If the
coefficient of static friction is 0.6, what is the maximum frequency of rotation such that
he coin does not slip? Suppose the frequency is cut in half. Now how far from the
Fc = Fs
center can the coin be placed?
The maximum frequency occurs when Fc = Fs
Fc  4 2 f 2 mR; Fs =s mg ;
f 
f2 
4 2 f 2 mR  s mg ;
s g
(0.6)(9.8 m/s 2 )
;

4 2 R
4 2 (0.12 m)
1.114 rev/s
 0.557 rev/s;
2
(0.6)(9.8 m/s 2 )
;
R2  2
4 (0.557 rev/s)2
f 
s g
4 2 R
f = 1.11 rev/s
4 2 f 22 mR2  s mg ;
R = 0.480 m;
R2 
s g
4 2 f 22
R = 48.0 cm
*10-61. The laboratory apparatus shown in Fig. 10-19 allows a rotating mass to stretch a spring
so that the supporting cord is vertical at a particular frequency of rotation. Assume the
mass of the bob is 400 g and the radius of revolution is 14 cm. With a stop watch the
time for 50 revolutions is found to be 35 s. What is the magnitude and direction of the
force acting on the bob? First find f in (rev/s).
f 
14 cm
50 rev
 1.429 rev/s . Fc  4 2 f 2 mR
35 s
Fc  4 2 (1.429 rev/s) 2 (0.4 kg)(0.14 m) ;
Fc = 4.51 N, directed toward the center.
400 g
The centripetal force is ON the bob.
The outward force ON the spring is NOT the centripetal force.
137
Physics, 6th Edition
Chapter 10. Uniform Circular Motion
*10-62. In Problem 10-2 above, assume that a 100-g mass is added to the 400-g bob? The force
required to stretch the spring should be the same as before, but the rotating mass has
increased. What changes when the experiment is performed again so that the centripetal
force is the same as before? On what does the centripetal force act in this experiment?
Since the centripetal force must be the same, it is
14 cm
necessary that the velocity be reduced so that:
100 g
Fc 
mv 2
; The product mv2 must be the same.
R
m1v12  m2v22 ; v22 
v2 
m1v12 (400 g)v12

m2
(500 g)
(400 g)v12
 0.894 v1 ;
(500 g)
400 g
v1 = 2f1R = 2(1.429 rev/s)(0.14 m) = 1.26 m/s
v2 = (0.894)(1.26 m/s);
v2 = 1.13 m/s
Thus, the object moves slower and the frequency of revolution decreases so that the
centripetal force acting ON the bob does not change: m1v12  m2v22
*10-63. A 10-in. diameter platform turns at 78 rpm. A bug rests on the platform 1 in. from the
outside edge. If the bug weighs 0.02 lb, what force acts on it? What exerts this force?
Where should the bug crawl in order to reduce this force by one-half?
f = 78 rev/min = 1.30 rev/s;
R = 5 in. – 1 in. = 4 in.;
m = W/g = 0.02 lb/32 ft/s2; m = 0.000625 slugs
R = (4/12) ft = 0.333 ft;
Fc  4 2 (1.3 rev/s) 2 (0.000625 slug)(0.333 ft) ;
Fc = 42f2 mR
Fc = 0.0139 lb
The central force ON the bug is exerted BY the platform (Static friction).
Since Fc is proportional to R, halving the radius will also halve the force!
R2 
(4 in.)
 2 in. ; The bug should crawl to a point 2 cm from the center.
2
138
Physics, 6th Edition
Chapter 10. Uniform Circular Motion
*10-64. The diameter of Jupiter is 11 times that of the earth, and its mass is about 320 times that
of earth. What is the acceleration due to gravity near the surface of Jupiter?
mj = 11me = 320(5.98 x 1024 kg); mj = 1.914 x 1027
Rj = 11Re = 11(6.38 x 106 m); Rj = 7.018 x 107 m ;
gj 
Gm j
R 2j

(6.67 x 10-11 N m 2 /kg 2 )(1.914 x 1027 kg)
;
(7.018 x 107 m) 2
mg j 
Gmm j
R 2j
gj = 25.9 m/s2
*10-65. Assume that L = 50 cm and m = 2 kg in Fig. 10-16. How many revolutions per second
are needed to make the angle  = 300? What is the tension in the supporting rod at that
point?
[ h = (50 cm) cos 300 = 0.433 m; m = 2.0 kg;  = 300 ]
L
v2
tan  
; v  gR tan   (9.8 m/s 2 )(0.25 m) tan(300 )
gR
v = 1.19 m/s;
T cos  = mg;
v = 2fR;
T
f 
v
2 R

(1.19 m/s)
;
2 (0.25 m)
mg
(2 kg)(9.8 m/s 2 )

;
cos
cos 300

h
R
f = 0.757 rev/s
T = 22.6 N
*10-66. A 9-kg block rests on the bed of a truck as it turns a curve of radius 86 m. Assume that
k = 0.3 and that s = 0.4. Does the friction force on the block act toward the center of
the turn or away? What is the maximum speed with which the truck can make the turn
without slipping? What if the truck makes the turn at a much greater speed, what would
be the resultant force on the block? [ F = Fc ; m = 9 kg ]
vc  s gR  (0.4)(9.8 m/s2 )(86 m) ;
vc = 18.4 m/s
Fk = kmg = (0.3)(9 kg)(9.8 m/s ); Fk = FR = 26.5 N
2
139
R = 86 m
Fc
Fs