Solucionario Capitulo 10 - Paul E. Tippens

Chapter 10. Uniform Circular Motion Physics, 6th Edition

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?

2 2(8 m/s)1.5 mc

vaR

ac = 42.7 m/s2

10-2. What are the period and frequency of rotation for the ball in Problem 10-1?

2 2 R 2 (1.5 m)2 ; ; T = 8 m/s

Rv fR vT v

; T = 1.18 s

1 11.18 s

fT

; 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 ]

2 2 2 24 4 (9 rev/s) (0.03 m)ca f R ; ac = 95.9 m/s2

2 2 (9 rev/s)(0.03 m)v fR ; 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 ]

121


1 16 rev/s

Tf

; T = 0.167 s ;

2 2 (6 rev/s)(1.5 m)v fR ; v = 56.5 m/s

2 2(56.5 m/s)(1.5 m)c

vaR

; ac = 2130 m/s2

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?

22; (2 m/s) (50 m)va v aR

R

; 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?

2 2 2

2

(22 m/s); 6 m/s

v va RR a

; R = 80.7 m

2 2(1500 kg)(22 m/s)(80.7 m)

mvFR

; Fc = 9000 N

122


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?

2 2 2 2

1 2 2(2 ) 4 2; ; 2 2

v v v va a aR R R R

;

a2 = 2a1 = 2(20 m/s2; 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?

2 2(20 kg)(16 m/s)16 m

mvFR

; Fc = 320 N

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?

22 2 2 14 4 (3 kg)(2 m)

0.3 scF f mR ; 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?

123


2

2

8 lb 0.25 slug; 32 ft/s c

mvm FR

2 2(0.25 slug)(95 ft/s)2000 lbc

mvRF

; 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 ]

2 (800 N)(0.20 m); 8 kg

cc

F RmvF vR m

; v = 4.47 m/s

4.47 m/s2 ; 2 2 (0.20 m)

vv fR fR

; 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 ]

2 2 2 24 4 (1800 rev/s) (0.5 kg)(0.35 m)cF f mR ;

Fc = 2.24 x 107 N, toward the center

*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?

124


2 2(70 kg)(8.8 m/s)25 mc

mvFR

; 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? (60 mi/h = 88 ft/s)

2 2(88 ft/s)24 ftc

vaR

; 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?

22; (0.4)(9.8 m/s )(80 m)s c s

mv mg v gRR

;

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)

2 2 2

2

(88 ft/s); (32 ft/s )(400 ft)s s

mv vmgR gR

; s = 0.605

125


10-17. Find the coefficient of static friction necessary to sustain motion at 20 m/s around a turn

of radius 84 m.

2 2 2

2

(20 m/s); (9.8 m/s )(84 m)s s

mv vmgR gR

; s = 0.486

*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.)

22 2

2 2

(0.4)(9.8 m/s )4 ; 4 4 (3 m)

sc s

gF f mR mg fR

;

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 = ?

22 2

2 2 2 2

(0.5)(9.8 m/s )4 ; 4 4 (1.67 rev/s)

sc s

gF f mR mg Rf

; R = 21.1 cm

10-20. Find the required banking angle to negotiate the curve of Prob.10-15 without slipping.

2 2

2

(17.7 m/s)tan(9.8 m/s )(80 m)

vgR

; = 21.80

126


10-21. Find the required banking angle for Problem 10-16 to prevent slipping?

2 2

2

(88 ft/s)tan(32 ft/s )(400 ft)

vgR

; = 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?

22 0tan ; tan (9.8 m/s )(20 m) tan 28v v gR

gR

; v = 10.3 m/s

*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; h = (9 m) sin ;

2

tan vgR

2 2

2

(11.1 m/s)tan(9.8 m/s )(96 m)

vgR

; = 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?

127


2 2

2

(12 m/s)tan(9.8 m/s )(0.30 m)

vgR

; = 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 = 20 cm = 0.20 m; R = L sin = (0.2 m) sin 600; R = 0.173 m

22 0tan ; tan (9.8 m/s )(0.173 m)( tan 60 )v v gR

gR

; v = 1.71 m/s

1.71 m/s2 ; 2 2 (0.173 m)

vv fR fR

; 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; R = 30 cm = 0.30 m

22 0tan ; tan (9.8 m/s )(0.30 m) tan 30v v gR

gR

; v = 1.30 m/s

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?

80 rpm = 1.33 rev/s; T sin = Fc = mv2/R

128

R

L

h

mg

T cos

R

L

T sin


sin

0.4 mR RL

;

2 240.4 m

RT f mR

T = (0.4 m)(4 2)(1.33 rev/s)2(2 kg); T = 56.1 N

T cos = mg;

2(2 kg)(9.8 m/s )cos56.1 N

mgT

; = 69.60

h = L cos = (0.4 m) cos 69.60 ; h = 0.14 m 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

1.5 lb 0.0469 slug32 ft/s

m ; L = 6 in. = 0.50 ft

sin

0.5 ftR RL

;

2 240.5 ft

RT f mR

T = (0.5 ft)(4 2)(1.67 rev/s)2(0.0469 slug); T = 2.75 lb

T cos = mg;

2(0.0469 slug)(32 ft/s )cos2.57 lb

mgT

; = 54.30

h = L cos = (0.5 ft) cos 54.30 ; h = 0.292 ft

129

h

mg

T cos

R

L

T sin


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? [ L = 10 m; a = 3 m, = 300 ]

b = L sin = (10 m) sin 300 = 5 m; R = a + b = 7 m;

2

tan ; tanv v gRgR

2 0(9.8 m/s )(7 m) tan 30v 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

2

tan ; tanv v gRgR

;

2 0(9.8 m/s )(7.695 m) tan 28v ; v = 6.33 m/s

6.33 m/s2 ; 2 2 (7.695 m)

vv fR fR

; f = 0.131 rev/s or 7.86 rpm

Motion in a Vertical Circle

130

b

a

R = a + b

L


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 ]

2mvT mgR

; Critical speed vc is when T = 0

2(9.8 m/s )(0.7 m)cv gR ; vc = 6.86 m/s

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

string is not to become slack?

2mvT mgR


2(9.8 m/s )(0.7 m)cv gR ; 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:

2mvT mgR

and

2mvT mgR

131

0

0 R

T

mg

v

R

T

mg

v

R

T

mg

vT

mg


22(1.2 kg)(8 m/s) (1.2 kg)(9.8 m/s )

(0.9 m)T

; T = 73.6 N

At Bottom:

2mvT mgR

and

2mvT mgR

22(1.2 kg)(8 m/s) (1.2 kg)(9.8 m/s )

(0.9 m)T

; 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;

2 2(620 ft/s)(2800 ft)

vaR

; a = 137 ft/s2

2 2

; mv mvmg mgR R

N N

*10-34. (Cont.)

2(5 slugs)(620 ft/s) 160 lb(2800 ft)

N; N = 846 lb

132


*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?

227 ; 7 7(9.8 m/s )(1000 m)va g v gR

R

;

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:

2mvT mgR

and

2mvT mgR

22(3 kg)(16 m/s) (3 kg)(9.8 m/s )

(8 m)T

; T = 66.6 N

When T = 0, 2(9.8 m/s )(8 m)cv gR ; 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?

133

v

R

T

mg

8 m

hT

mg


08 mcos ; 66.420 m

; h = 20 m– 8 m = 12 m

2 2½ ; 2 2(9.8 m/s )(12 m)mv mgh v gh ; v = 15.3 m/s

*10-37. (Cont.) At Bottom:

2mvT mgR

and

2mvT mgR

22(36 kg)(15.34 m/s) (36 kg)(9.8 m/s )

(20 m)T

; 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 )

1 22 2

(2 ton)(2000 lb/ton) (3 ton)(2000 lb/ton) 125 slugs; 187.5 slugs32 ft/s 32 ft/s

m m

-8 2 21 2 1 22

(3.44 x 10 lb ft /slug )(125 slug)(187.5 slug); 0.0004 lb

Gm m Gm mF RR F

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.

134


-11 2 21 22

(6.67 x 10 N m /kg )(4 kg)(2 kg)0.08 m

Gm mFR

; 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?

-11 2 23 2

3 2 2

(6.67 x 10 N m /kg )(3 kg)(2 kg)(0.05 m)

Gm mFR

-11 2 26 2

6 2 2

(6.67 x 10 N m /kg )(6 kg)(2 kg)(0.05 m)

Gm mFR

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; FR = 1.60 x 10-7 N

*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.

2 2 2

2 -11 2 2

(5.00 m/s )(4560 m); 6.67 x 10 N m / kg

pp

Gmm gRmg mR G

; 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:

135

3 kg 6 kg

F6F3

0.05 m0.05 m2 kg


m2 2 and g =m m

mm m

Gmm GmmgR R

;

e2 2 and g =e ee

e e

Gmm GmmgR R

2 2

2 2

(4 )81

m m e m m

e e m m m

g m R m Rg m R m R

;

2 0.19759.8 m/s

mg

; 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 ]

62

2 2

''(10 x)

Gm mGm mx

;

26

22(10 x)

mxm

6

2

60 kg 1.732(10 x) 20 kg

mxm

; x = 1.732(10 – x); 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

136

m’60 kg 20 kg

F2F6

10 - xx


10-44. (Cont.) Fc = Fg ;

2

2 ; e eGmm Gmmv vR R R

;

-11 2 2 24

6

(6.67 x 10 N m / kg )(5.98 x 10 kg)7.18 x 10 m

v ; 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?

R = Rj + h = 7.15 x 107 m + 6 x 107 m; R = 1.315 x 108 m ;

2

2 ;eGmmmvR R

-11 27

8

(6.67 x 10 )(1.9 x 10 kg)1.315 x 10 m

eGmvR h

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 ;

2


;

-11 2 2 24

6

(6.67 x 10 N m / kg )(5.98 x 10 kg)7.58 x 10 m

v ; v = 7254 m/s

137


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.

2 2 6 2

2 -11 2 2

(1.63 m/s )(1.74 x 10 m); 6.67 x 10 N m /kg

m mm

m

Gmm gRmg mR G

; mm = 7.40 x 1022 kg

*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 ]

2 2 6 32 3 2 7 2

-11 2 2 24

4 4 (7.28 x 10 ) ; T = 3.82 x 10 s(6.67 x 10 N m /kg )(5.98 x 10 kg)e

T RGm

7 23.82 x 10 sT ; T = 6180 s (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

22 -11 2 2 24 10 22 3 3

2 2

4 (6.67 x 10 N m /kg )(5.98 x 10 kg)(1.02 x 10 s ); R4 4

e

e

Gm TT RGm

3 23 31.03 x 10 mR ; R = 4.69 x 107 m; h = R – Re = 4.05 x 107 m

138


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?

22 2 2 -2

2 2

(12 ft/s )4 ; ; 0.1013 s4 4 (3 ft)

cc

aa f R f fR

; f = 0.318 rev/s

22

6 lb (12 ft/s )32 ft/scT ma

; 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?

2 2(15 m/s)(0.300 m)c

vaR

; a = 750 m/s2

Fc = mac = (2.6 kg)(750 m/s2); Fc = 1950 N

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 ]

2


;

7 60.3048 m2.64 x 10 ft 8.047 x 10 m1 ft

R

-11 2 2 24

6

(6.67 x 10 N m / kg )(5.98 x 10 kg)8.047 x 10 m

v ; v = 7041 m/s

139


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?

2mvT mgR


2(9.8 m/s )(2.0 m)cv gR ; 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:

2mvT mgR

and

2mvT mgR

22(4 kg)(10 m/s) (4 kg)(9.8 m/s )

(1.4 m)T

; T = 247 N

At Bottom:

2mvT mgR

and

2mvT mgR

22(4 kg)(10 m/s) (4 kg)(9.8 m/s )

(1.4 m)T

; T = 325 N

140

R

T

mg

v

R

T

mg

vT

mg


*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

21 1 9.8 m/s2 2 0.125 m

gfh

; 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?

2(9.8 m/s )(6.0 m)cv gR ; 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:

2mvmgR

N and

2mv mgR

N

22(210 kg)(12 m/s) (210 kg)(9.8 m/s )

(6 m) N

; 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?

141

h

R

L

R

T

mg


v = 45 mi/h = 66.0 ft/s;

2 2

2

(66 ft/s)tan ;(9.8 m/s )(200 ft)

vgR

= 34.20; NO

*10-59. For the figure shown in Fig. 10-17, assume that a = 2 m and L = 4 m.

Find the speed to cause the swing to move out to an angle of 200?

b = L sin = (4 m) sin 200 = 1.37 m; R = a + b = 3.37 m;

2

tan ; tanv v gRgR

v = 3.47 m/s

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

center can the coin be placed?

The maximum frequency occurs when Fc = Fs

2 2 2 2s s4 ; = ; 4cF f mR mg f mR mg sF ;

24s gf

R

2

2 2

(0.6)(9.8 m/s )4 4 (0.12 m)

s gfR

; f = 1.11 rev/s

21.114 rev/s 0.557 rev/s;

2f

2 22 2 s 2 2 2

2

4 ; 4

s gf mR mg Rf

142

b

a

R = a + b

L

Fc = Fs


2

2 2 2

(0.6)(9.8 m/s ) 4 (0.557 rev/s)

R

; R = 0.480 m; 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).

50 rev 1.429 rev/s35 s

f .

2 24cF f mR

2 24 (1.429 rev/s) (0.4 kg)(0.14 m)cF ;

Fc = 4.51 N, directed toward the center. The centripetal force is ON the bob.

The outward force ON the spring is NOT the centripetal force.

*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

necessary that the velocity be reduced so that:

2

;cmvFR

The product mv2 must be the same.

143

14 cm

400 g

100 g

14 cm

400 g


2 22 2 2 1 1 1

1 1 2 2 22

(400 g); (500 g)

m v vm v m v vm

v1 = 2f1R = 2(1.429 rev/s)(0.14 m) = 1.26 m/s

21

2 1(400 g) 0.894 (500 g)

vv v ; 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: 2 2

1 1 2 2m v m v

*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; m = W/g = 0.02 lb/32 ft/s2; m = 0.000625 slugs

R = 5 in. – 1 in. = 4 in.; R = (4/12) ft = 0.333 ft; Fc = 42f2 mR

2 24 (1.3 rev/s) (0.000625 slug)(0.333 ft)cF ; 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!

2(4 in.) 2 in.

2R

; The bug should crawl to a point 2 cm from the center.

*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?

144


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 ; 2

jj

j

Gmmmg

R

-11 2 2 27

2 7 2

(6.67 x 10 N m /kg )(1.914 x 10 kg)(7.018 x 10 m)

jj

j

Gmg

R

; 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 ]

22 0tan ; tan (9.8 m/s )(0.25 m) tan(30 )v v gR

gR

v = 1.19 m/s; v = 2fR;

(1.19 m/s)2 2 (0.25 m)

vfR

; f = 0.757 rev/s

T cos = mg;

2

0

(2 kg)(9.8 m/s )cos cos30mgT

; 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

145R = 86 m

h

R

L


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 ]

2(0.4)(9.8 m/s )(86 m)c sv gR ; vc = 18.4 m/s

Fk = kmg = (0.3)(9 kg)(9.8 m/s2); Fk = FR = 26.5 N

146

Fs

Fc

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Solucionario Capitulo 10 - Paul E. Tippens

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