Question

A dancer is rotating on a smooth horizontal floor with an angular momentum $L$. The dancer folds her hands so that her moment of inertia decreases by $25\%$. The new angular momentum is

• A. $\frac{L}{4}$
• B. $\frac{L}{8}$
• C. $\frac{L}{2}$
• D. $L$

Answer 1

The correct option is D $L$

Since, there is no external torque acting on the dancer. The total angular momentum remains conserved.
$\therefore I_i{\omega }_i=I_f{\omega }_f$
So, when dancer fold her hands, the moment of inertia decreases which means the angular speed increases such that the angular momentum remains constant. Therefore, $L_i=L_f=L$.
Hence, the new angular momentum will be $L$.