Question

A child swings a tennis ball attached to a 0.626-m string in a horizontal circle above his head at a rate of 4.50 rev/s.

a. What is the centripetal acceleration of the tennis ball?
b. The child now increases the length of the string to 1.00m but has to decrease the rate of rotation to 4.00 rev/s. Is the speed of the ball greater now or when the string was shorter?
c. What is the centripetal accleration of the tennis ball when the string is 1.00 m in length?

Answer 1

Explanation:

Length of a string, l = 0.626 m

A tennis ball revolves in a horizontal circle above his head at a rate of 4.50 rev/s.

(a) The centripetal acceleration of the tennis ball is given by :

$a=\omega^2r\\\\a=(4.5\times 2\pi)^2\times 0.626\\\\=500.44\ m/s^2$

(b) When r = 0.626 m and ω = 4.50 rev/s

Speed,

$v=r\omega\\\\=0.626\times 4.50 \times 2\pi\\\\=17.69\ m/s$

When r = 1 m and ω = 4 rev/s

Speed,

$v=r\omega\\\\=1\times 4 \times 2\pi\\\\=25.13\ m/s$

Speed is more in second case when child now increases the length of the string to 1.00m but has to decrease the rate of rotation to 4.00 rev/s.

(c) $a=\omega^2r\\\\a=(4\times 2\pi)^2\times 1\\\\=631.65\ m/s^2$

Hence, this is the required solution.