Answer to Question #94822 in Mechanics | Relativity for stefanie

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 8.06 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 6.01 rev/s.
(a) Which rate of rotation gives the greater speed for the ball?
6.01 rev/s or
8.06 rev/s

(b) What is the centripetal acceleration of the ball at 8.06 rev/s?
m/s2

(c) What is the centripetal acceleration at 6.01 rev/s?
m/s2

Answer 1

Solution.

(a) The linear velocity is proportional to the angular velocity and the radius. Hence

$$v_1=\omega_1\times R, v_2=\omega_2\times R,$$

Therefore, we obtain a greater speed at 8.06 rev/s .

(b) Find the centripetal acceleration of the ball at 8.06 rev/s using formula

$$a=\omega^2 \times R$$

where R=0.6m is radius rotation; w=8.06 rev/s. As result

$$a=(2\pi \times 8.06)^2 \times 0.6=1539 \frac {m} {s^2}$$

(c) Find the centripetal acceleration of the ball at 6.01 rev/s using formula

$$a=\omega^2 \times R$$

where R=0.6m is radius rotation; w=6.01 rev/s. As result

$$a=(2\pi \times 6.01)^2 \times 0.6=856 \frac {m} {s^2}$$

Answer. (a) 8.06 rev/s  (b) 1539 m/s^2 (c) 856 m/s^2.