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Unformatted text preview: Homework Solutions Chapter 05 6 Ed. Ch.5: Q: 1, 2, 5; P: 3, 5, 7, 9, 11, 13, 19* 5 Ed. Ch.5: Q: 1, 2, 5, P: 3, x, 8, 7, 10, 12, 5 6 Ed. Ch.5: Q: 18, 20; P: 29, 33*, 35, 43, 47, 53, 57* 5 Ed. Ch.5: Q: x, 16; P: 29, x, 32, 39, x, 44, 52 P3. Calculate the centripetal acceleration of the Earth in its orbit around the Sun, and the net force exerted on the Earth. What exerts this force on the Earth? Assume the distance EarthSun is 1.5 m Solution: m In one year, the Earth travels distance equal to the circumference of a circle with radius m
m m/s m m/s m/s P5. Suppose the space shuttle is in orbit 400 km from the Earth's surface, and circles the Earth about once every 90 minutes. Find the centripetal acceleration of the space shuttle in its orbit. Express your answer in terms of g, the gravitational acceleration near Earth's surface. Solution: 6400 km min
m/s m km km m s period for one revolution m m/s s m/s P7. A ball on the end of a string is revolved at a uniform rate in a vertical circle of radius 72.0 cm, as shown in the diagram. If its speed is 4.00 m/s and its mass is 0.30 kg, calculate the tension in the string when the ball is (a) at the top of its path, and (b) at the bottom of its path. Solution: 1 T mg T mg (a) Top Tension and gravity are in the same direction ! " ! kg (b) Bottom Tension and gravity in the opposite direction. The centripetal force points up (towards the center). " ! kg
m/s m m/s m " m/s N ! m/s N P9. What is the maximum speed with which a 1050 kg car can round a turn of radius 77 m on a flat road if the coefficient of static friction between tires and road is 0.80? Is this results independent of the mass of the car? Solution: If a car needs to turn, a force must provide the centripetal acceleration. In this case, it will be the friction:
# For the car not to skid sideways, the friction must be static: $ # $ ! % ! $ ! $ ! # & m/s m m/s & 2 P11. A device for training astronauts and jet fighter pilots is designed to rotate a trainee in a horizontal circle of radius 12.0 m. If the force felt by the trainee on her back is 7.85 times her own weight, how fast is she rotating? Express your answer in both m/s and rev/s. Solution: The force that provides the centripetal acceleration is the normal force exerted by the wall/mechanism onto the person. It points horizontally and towards the center of the circle. Newton's Second Law: The weight of the person is ! ! The centripetal force is 7.85 times larger than the weight ! ! m/s m m/s The length for one revolution is: ' m For 1 second, the person goes distance m/s s m The number of revolutions is then: m # rev/s ' m Note: The number of revolutions per second is called frequency, and it is related to the time it takes for one revolution, which is called period, T, in the following way: # P13. At what minimum speed must a roller coaster be traveling when upside down at the top of a circle so that the passengers will not fall out? Assume a radius of curvature of 7.4 m Solution
v FN mg Two forces are acting on the roller coaster: Gravity and Normal Force. 3 Newton's Second Law: ! " !( % % The normal force cannot be less than zero, that is, it cannot point upwards. m/s ( ! ! P19. A flat puck (mass M) is rotated in a circle on a frictionless airhockey tabletop, and is held in this orbit by light cord connected to a dangling block (mass m) through a hole as shown in the diagram. Show that the speed of the puck is given by
! ) Solution:
M R T v T m mg Newton's Second Law for the hanging : !" ! Newton's Second Law for the rotating mass: ) ! )
! ) P29. At the surface of a certain planet, the gravitational acceleration g has a magnitude of 12.0 m/s . A 21kg brass ball is transported to this planet. What is (a) the mass of the brass on the Earth and on the planet? (b) the weight of the brass ball on the Earth and on the planet? Solution: (a) mass is the same : 21.0 kg (b) * + kg m/s * ,kg m/s N N " P33. Two objects attract each other gravitationally with a force of 2.5 N when they are 0.25 m apart. Their total mass is 4.0 kg. Find their individual mass. 4 Solution:
. kg N
" kg "
Nm kg kg" m .
" kg " kg " " kg
kg 0 / kg " kg kg kg
kg kg 0 kg kg or or kg kg kg kg " kg " kg kg kg kg P35. What is the distance from the Earth's center to a point outside the Earth where the gravitational acceleration due to the Earth is of its value at the Earth's surface? Solution: . )+ . At the surface: ! ! . )+
m/s m/s + . )+
+ m/s m/s At another distance: !1
! !1 . .
)+ + )+ + + kg km P43. Calculate the speed of a satellite moving in a stable circular orbit about the Earth at a height of 3600 km. Mass of Earth: 5.98 Radius of Earth: Solution: Newton's second Law: . .
)+ 3 kg km )+
" .)+ Nm /kg m 5.98 m kg m/s 5 P47. At what horizontal velocity would a satellite have to be launched from the top of Mt. Everest to be placed in a circular orbit around the Earth? Height of Mt. Everest:8.48 km Radius of Earth: 6400 km Solution: Newton's second Law: . .
)+ )+
" .)+ Nm /kg 5.98 m 8.48 m kg m/s P53. What will a spring scale read for the weight of a 55kg woman in an elevator that moves (a) upward with constant speed of 6.0 m/s (b) downward with constant speed of 6.0 m/s (c) upward with acceleration of 0.33g (d) downward with acceleration 0.33g (e) in free fall Note: This problem is similar to Problem 15 from Chapter 4, and to one of the solved examples from Chapter 4. Answers: (a) (b) (c) (d) (e) N N N N N P57. The asteroid Icarus, though only a few hundred meters across, orbits the Sun like the planets. Its period is 410 d. What is its mean distance from the Sun? Mean distance EarthSun Period of Earth around Sun Solution: Kepler's Third Law: Compare Earth's parameters with those of Icarus: 6 1.5 365 d
8 km d 410 d +" 2 +" 2 R d d 8 km d d 8 km 7 ...
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