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.

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.