Page 120 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 9] CIRCULAR MOTION AND GRAVITATION 105

SOLVED PROBLEM 9.9
An airplane pulls out of a dive in a circular arc whose radius is 1200 m. The airplane’s velocity is a
constant 200 m/s. Find the force with which the 80-kg pilot presses down on his seat at the bottom of
the arc.

The downward force F the pilot exerts on his seat is the equal and opposite reaction to the upward force of
the seat on him. This upward force both supports his weight w and provides the centripetal force F c that keeps him
in a circular path. The upward force of the seat on the pilot plays the same role here as the tension in the string
of Fig. 9-2(b) does in the motion of a body moving in a vertical circle when the body is at the bottom of its path.
Hence

mv 2 (80 kg)(200 m/s) 2 2
F = F c + w = + mg = + (80 kg)(9.8 m/s ) = (2667 + 784) N = 3451 N
r 1200 m
The pilot presses down on his seat with a force of 3451 N, which is 4.4 times his weight.

SOLVED PROBLEM 9.10
A ball at the end of an 80-cm string is being whirled in a vertical circle. At what critical velocity v 0 will
the string begin to go slack at the top of the ball’s path?
The situation here corresponds to that shown in Fig. 9-2(a), with T = F c − w. When the string goes slack, the
tension in it is T = 0, and

F c = w
mv 2 0 = mg
r

√ 2
v 0 = rg = (0.8m)(9.8 m/s ) = 2.8 m/s
For velocities less than v 0 , the top of the ball’s path will flatten out and will not be circular.

SOLVED PROBLEM 9.11
A ball is being whirled vertically at constant energy at the end of an 80-cm string. If the ball’s speed at
the top of the circle is 3 m/s, what is its speed at the bottom of the circle?
At the top of the circle, the ball has a potential energy relative to the bottom of the circle of

PE 1 = mgh = mg(2r) = 2mgr

At the bottom of the circle, PE 2 = 0. Because the total energy of the ball is constant,

KE 2 + PE 2 = KE 1 + PE 1
where KE 1 is the ball’s kinetic energy at the top and KE 2 is its kinetic energy at the bottom. Therefore

1 mv + 0 = mv + 2mgr
2
2
1
2 2 2 1
2
2
v = v + 4gr
2 1

2 2 2
v 2 = v + 4gr = (3 m/s) + (4)(9.8 m/s )(0.8m) = 6.35 m/s
1
The ball’s velocity at the bottom of its path is more than twice its velocity at the top. The difference is due to the
conversion of the ball’s potential energy at the top into additional kinetic energy at the bottom.