Q. A dancer is standing on a stool rotating about the vertical axis passing through its centre. She pulls her arms towards the body reducing her moment of inertia by factor of $n$. The new angular speed of turn table is proportional to
A) $n$
B) $n^{- 1}$
C) $n^{0}$
D) $n^{2}$
Explanation
(a) Since, no external torque is acting the angular $(J)$ is conserved.
$J = I\omega =$constant
Where $I( = mr^{2})$ is moment of inertia and $\omega$ the angular velocity.
Given, $I_{1} = I,\ I_{2} = \frac{I}{n},\ \omega_{1} = \omega$
$\therefore$ $J = I_{1}\omega_{1} = I_{2}\omega_{2}$
$I\omega = \frac{I}{n}\omega_{2}$
$\Rightarrow$ $\omega_{2} = n\omega$
Hence, angular velocity increases by a factor of $n$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.