Question

A child is standing with his two arms outstretched at the centre of a turntable that is rotating about its central axis with an angular speed $\omega_0$ . Now, the child folds his hands back so that moment of inertia becomes $3$ times the initial value. The new angular speed is

• A. $3\omega_0$
• B. $\frac {\omega_0}{3}$
• C. $6\omega_0$
• D. $\frac {\omega_0}{6}$

Answer 1

Answer: (B)

$\frac {\omega_0}{3}$

Answer 2

Here, Initial angular speed, $\omega_i = \omega_0$ Initial moment of inertia $= I_i$ Final moment of inertia $I_f = 3I_i$ According to the law of conservation of angular momentum, we get $L_i = L_f$ or $I_i\omega_i = I_f \omega_f$ $\omega_{f} = \frac{I_{i}\omega_{i}}{I_{f}} = \left(\frac{I_{i}}{I_{f}}\right)\omega_{i}$ $= \left(\frac{I_{i}}{3I_{i}}\right)\omega_{0} = \frac{\omega_{0}}{3}$