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 8.16 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 6.35 rev/s.(a) Which rate of rotation gives the greater speed for the ball?6.35 8.16 (b) What is the centripetal acceleration of the ball at 8.16 rev/s?m/s2(c) What is the centripetal acceleration at 6.35 rev/s?m/s2

Answer 1

Answer:

(a) Ball with 6.35 rev/sec gives greater speed

(b) So centripetal acceleration of ball with rotation 8.16 m/sec

$a_c=\frac{v^2}{r}=\frac{4.896^2}{0.6}=39.95m/sec^2$

So centripetal acceleration of ball with rotation 6.35 m/sec

$a_c=\frac{v^2}{r}=\frac{5.75^2}{0.9}=36.73m/sec^2$

Explanation:

(a) In first case angular speed $\omega =8.16rev/sec$rev/sec and length of the chain = 0.8 m

So velocity $v=\omega r=8.16\times 0.6=4.896 m/sec$

In second case angular speed angular speed $\omega =6.35rev/sec$ and length of the chain that is r = 0.6 m

So velocity $v=\omega r=6.35\times 0.9=5.715m/sec$

So 6.35 rev/sec gives greater speed

(b) Centripetal acceleration is given by $a_c=\frac{v^2}{r}$

So centripetal acceleration of ball with rotation 8.16 m/sec

$a_c=\frac{v^2}{r}=\frac{4.896^2}{0.6}=39.95m/sec^2$

So centripetal acceleration of ball with rotation 6.35 m/sec

$a_c=\frac{v^2}{r}=\frac{5.75^2}{0.9}=36.73m/sec^2$