Question

A small steel sphere of mass $m$ is tied to a string of length $r$ and is whirled in a horizontal circle with a uniform angular velocity $2 \, \omega $ The string is suddenly pulled, so that radius of the circle is halved. The new angular velocity will be

• A. 2 $ \omega $
• B. 4 $ \omega $
• C. 6 $ \omega $
• D. 8 $ \omega $

Answer 1

Answer: (D)

When no external torque is acting upon a body rotating about an axis, then angular momentum of the body remains constant.

Therefore,
$J=I \omega=$ constant
where $ I=m r^{2} $ (moment of inertia)
$\therefore m_{1} r_{1}^{2} \omega_{1}=m_{2} r_{2}^{2} \omega_{2}$
$\Rightarrow \omega_{2}=\frac{m_{1} r_{1}^{2} \omega_{1}}{m_{2} r_{2}^{2}}$
Given, $ m_{1}=m, r_{1}=r, \omega_{1}=2 \omega$
$m_{2}=m, r_{2}=\frac{r}{2}$
$\therefore \omega_{2}=\frac{m r^{2}(2 \omega)}{m(r / 2)^{2}}=\frac{m r^{2}}{m r^{2}} 8 \omega$
$\Rightarrow \omega_{2}=8 \omega$