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)

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 $I_i{\omega }_i=I_f{\omega }_f$
${\omega }_f=\frac{I_i{\omega }_i}{I_f}=\left(\frac{I_i}{I_f}\right){\omega }_i$
$=\left(\frac{I_i}{3I_i}\right){\omega }_0=\frac{{\omega }_0}{3}$