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 8.54 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 6.31 rev/s.
What is the centripetal acceleration of the ball at 8.54 rev/s?

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 8.54 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 6.31 rev/s. 
 What is the centripetal acceleration of the ball at 8.54 rev/s?

jamesj · Answer

What's the formula for centripetal acceleration for uniform circular motion?
If
v = speed of ball (meters/second)
r = radius of motion (meters)
w (omega) = angular velocity (radians/second)

then the acceleration, a, is given by

$$ a = \frac{v^2}{r} = \omega^2 r $$
So either of the two expressions can be used.  Remember also that $ v = \omega r $.

anonymous · Answer

we know  basically, that arch lenghth=angle subtend by this at centre * radius of the circle.
                                           L=$$\theta \times radius$$=>$$v(velocity)=\omega \times radius$$ after differentiate on both side w.r.to time. and centripetal acceleration of the body is $$a(centripetal acceleration)=v(linear velocity)^2\div radius  OR  \omega(angular velocity)^2\times radius$$.apply the formula you get your answer.......