Page 117 - Schaum's Outline of Theory and Problems of Applied Physics

102 CIRCULAR MOTION AND GRAVITATION [CHAP. 9

Since each revolution takes 2 s, the ball’s velocity is

s 3.77 m
v = = = 1.88 m/s
t 2s
The ball’s centripetal acceleration is therefore

v 2 (1.88 m/s) 2 2
a c = = = 5.9 m/s
r 0.6m

SOLVED PROBLEM 9.2
A 1000-kg car rounds a turn of radius 30 m at a velocity of 9 m/s. (a) How much centripetal force is
required? (b) Where does this force come from?
mv 2 (1000 kg)(9 m/s) 2
(a) F c = = = 2700 N
r 30 m
(b) The centripetal force on a car making a turn on a level road is provided by the road acting via friction on the
car’s tires.

SOLVED PROBLEM 9.3

How much centripetal force is needed to keep a 160-lb skater moving in a circle 20 ft in radius at a
velocity of 10 ft/s?
2
The skater’s mass is m = w/g = 160 lb/32 ft/s = 5 slugs. Hence
mv 2 (5 slugs)(10 ft/s) 2
F c = = = 25 lb
r 20 ft
SOLVED PROBLEM 9.4
The maximum force a road can exert on the tires of a 1500-kg car is 8500 N. What is the maximum
velocity at which the car can round a turn of radius 120 m?
2
Solving the formula F c = mv /r for v gives for the maximum velocity

F c r (8500 N)(120 m)
v = = = 26 m/s
m 1500 m
which is

m 3600 s/h
26 = 94 km/h
s 1000 m/kh
SOLVED PROBLEM 9.5

A car is traveling at 20 mi/h on a level road where the coefficient of static friction between tires and road
is 0.8. Find the minimum turning radius of the car.

The frictional force between the tires and the road is the source of the centripetal force when a car makes a turn
on a level road. The maximum centripetal force that friction can provide here is

F f = µ s N = µ s w = µ s mg