Page 123 - Schaum's Outline of Theory and Problems of Applied Physics

108 CIRCULAR MOTION AND GRAVITATION [CHAP. 9

Squaring both sides yields
2
2 2
4π r r g 0
e
=
T 2 r
Multiplying both sides by r gives
2 3
4π r
2
= r g 0
e
T 2
Solving for r, we have
1/3
2 2
r g 0 T
r = e
4π 2
4
Since here T = 1 day = (24 h)(60 min/h)(60 s/min) = 8.64 × 10 s,
6 2 2 4 2 1/3
(6.4 × 10 m) (9.8 m/s )(8.64 × 10 s)
7
r = = 4.23 × 10 m
4π 2
The corresponding altitude above the earth’s surface is
7
7
h = r − r e = (4.23 − 0.64) × 10 m = 3.59 × 10 m = 35,900 km

Multiple-Choice Questions

9.1. A 500-g ball moves in a horizontal circle 40 cm in radius at 4 m/s. Its centripetal acceleration is
(a) 10 m/s 2 (c) 40 m/s 2
(b) 20 m/s 2 (d) 80 m/s 2

9.2. An object in uniform circular motion is being acted upon by a centripetal force F. If the radius of the object’s path is
to be doubled while its velocity remains the same, the new centripetal force must be

(a) F/2 (c) 2F
(b) F (d) 4F

9.3. A 1200-kg car whose velocity is 6 m/s rounds a turn whose radius is 30 m. The centripetal force on the car is
(a) 48 N (c) 240 N
(b) 147 N (d) 1440 N

9.4. If the car of Question 9.3 rounds the same turn at 12 m/s, the required centripetal force is
(a) halved (c) doubled
(b) the same (d) quadrupled

9.5. The maximum centripetal force that friction can provide the car of Question 9.3 on a rainy day is 8000 N. The highest
velocity at which the car can round the turn is
(a) 14 m/s (c) 200 m/s
(b) 77 m/s (d) 40 km/s

9.6. The rain stops, the road dries, and the coefficient of friction between the tires of the car of Question 9.3 and the road
increases to 1.40 times its value in Question 9.5. The highest velocity at which the car can now round the turn is
(a) unchanged (c) 1.40 its former value
(b) 1.18 its former value (d) 1.96 its former value