Q 11-06-036NEETJEE MainTop questionMedium
A man stands at the centre of a turntable rotating freely at $30$ rpm, holding two dumbbells at arm's length. The moment of inertia of the system is $6\ \text{kg m}^2$. When he pulls the dumbbells in, it becomes $2\ \text{kg m}^2$. The new rate of rotation and the ratio of final to initial kinetic energy are
Answer: (B) $90$ rpm, $3$
No external torque, so $I_1\omega_1 = I_2\omega_2$: $\omega_2 = \dfrac{6}{2} \times 30 = 90$ rpm.
$K = \dfrac{L^2}{2I}$ with $L$ constant, so $\dfrac{K_2}{K_1} = \dfrac{I_1}{I_2} = 3$. The extra energy comes from the work the man does pulling his arms in.
Solution by Sreeraj P, M.Sc Physics