Question

An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the rate of 8.13 rev/s when the length of the chain is 0.600 m. When he increases the length to 0.900 m, he is able to rotate the ball only 6.04 rev/s. (a) Which rate of rotation gives the greater speed for the ball? 6.04 rev/s 8.13 rev/s (b) What is the centripetal acceleration of the ball at 8.13 rev/s? m/s2 (c) What is the centripetal acceleration at 6.04 rev/s? m/s2

Answer 1

(a) 6.04 rev/s

The speed of the ball is given by:

$v=\omega r$

where

$\omega$ is the angular speed

r is the distance of the ball from the centre of the circle

In situation 1), we have

$\omega=8.13 rev/s \cdot 2\pi = 51.0 rad/s$

r = 0.600 m

So the speed of the ball is

$v=(51.0 rad/s)(0.600 m)=30.6 m/s$

In situation 2), we have

$\omega=6.04 rev/s \cdot 2\pi = 37.9 rad/s$

r = 0.900 m

So the speed of the ball is

$v=(37.9 rad/s)(0.900 m)=34.1 m/s$

So, the ball has greater speed when rotating at 6.04 rev/s.

(b) $1561 m/s^2$

The centripetal acceleration of the ball is given by

$a=\frac{v^2}{r}$

where

v is the speed

r is the distance of the ball from the centre of the trajectory

For situation 1),

v = 30.6 m/s

r = 0.600 m

So the centripetal acceleration is

$a=\frac{(30.6 m/s)^2}{0.600 m}=1561 m/s^2$

(c) $1292 m/s^2$

For situation 2 we have

v = 34.1 m/s

r = 0.900 m

So the centripetal acceleration is

$a=\frac{v^2}{r}=\frac{(34.1 m/s)^2}{0.900 m}=1292 m/s^2$