Answer to Question #271801 in Mechanics | Relativity for Happpy

Question

In a Bangladesh-Australia cricket match, Sakib Al Hasan throws a ball towards the batsman. The ball starts to spin with 18.0 rev/s. The radius of the ball is 7.1cm. (a) Find the tangential speed of the outer periphery of the ball as it spins. (b) What is the centripetal acceleration of the cricket ball? (c) Let’s consider, due to air friction, the spin of the ball decays at the rate of 0.6 rev/s2 . Now find how long it will take for the ball to stop spinning and the total angle through which it will rotate during this time. (Consider the rate as constant)

Answer 1

$\omega = 18\frac{rev}{s}$

$r = 7.1cm = 0.071m$

$a)\omega = 18\frac{rev}{s}=18*2*\pi\frac{rad}{s}\approx113.01\frac{rad}{s}$

$v_\tau= rw=113.01*0.071=8.02\frac{m}{s}$

$b) a_c = \omega^2 r=113.01^2*0.071=906.76\frac{m}{s^2}$

$c)\epsilon= -0.6\frac{rev}{s^2}= -3.77\frac{rad}{s^2}$

$\omega_0 =113.01\frac{rad}{s}$

$\omega_1=0$

$\epsilon=\frac{\omega_1-\omega_0}{t}$

$t = \frac{-\omega_0}{\epsilon}=\frac{113.01}{3.77}=29.97s$

$\phi= \omega_0t+\frac{\epsilon t^2}{2}=113.01*29.97-\frac{3.77 *29.97^2}{2}=1693.8 rad$

$\text{Answer:}$

$a)v_\tau=8.02\frac{m}{s}$

$b) a_c = 906.76\frac{m}{s^2}$

$c)\phi= 1693.8 rad$