Page 119 - Schaum's Outline of Theory and Problems of Applied Physics

104 CIRCULAR MOTION AND GRAVITATION [CHAP. 9

SOLVED PROBLEM 9.7
(a) Find the proper banking angle for cars moving at 72 km/h to go around a curve 300 m in radius. (b)If
the curve were not banked, what coefficient of friction would be required between the tires and the road?
(a) Here v = (72 km/h)[0.278 (m/s)/(km/h)] = 20 m/s and so, from the solution to Prob. 9.6,
v 2 (20 m/s) 2
tan θ = = 2 = 0.136 θ = 7.7 ◦
gr (9.8 m/s )(300 m)
2
(b) From the solution to Prob. 9.5, r = v /(µ s g). Hence
v 2
µ s = = 0.169
gr

MOTION IN A VERTICAL CIRCLE
When a body moves in a vertical circle at the end of a string, the tension T in the string varies with the body’s
position. The centripetal force F c on the body at any point is the vector sum of T and the component of the
body’s weight w toward the center of the circle. At the top of the circle, as in Fig. 9-2(a), the weight w and the
tension T both act toward the center of the circle, and so
T = F c − w
At the bottom of the circle, w acts away from the center of the circle, and so
T = F c + w

Fig. 9-2

SOLVED PROBLEM 9.8
A string 0.5 m long is used to whirl a 1-kg stone in a verticle circle at a uniform velocity of 5 m/s. What
is the tension in the string (a) when the stone is at the top of the circle and (b) when the stone is at the
bottom of the circle?
(a) The centripetal force needed to keep the stone moving at 5 m/sis
mv 2 (1kg)(5 m/s) 2
F c = = = 50 N
r 0.5m
2
The weight of the stone is w = mg = (1kg)(9.8 m/s ) = 9.8N. At the top of the circle,
T = F c − w = 50 N − 9.8N = 40.2N
(b) At the bottom of the circle,

T = F c + w = 59.8N