Question

A ballet dancer is rotating about his own vertical axis at an angular velocity $100rpm$ on smooth horizontal floor. The ballet dancer folds himself close to his axis of rotation by which is moment of inertia decrease to half of initial moment of inertia then his final angular velocity is:

• A. $50$ rpm
• B. $100$ rpm
• C. $150$ rpm
• D. $200$rpm

Answer 1

The correct option is D $200$rpm

Let the initial and final angular velocities be ${\omega }_1$ and ${\omega }_2$ respectively and the initial and final moment of inertia be $I_1$

and $I_2$ respectively.

By conservation of angular momentum ,

$I_1{\omega }_1$ = $I_2{\omega }_2$

Now , $I_1$ = $2I_2$

Therefore , $2{\omega }_1$ = ${\omega }_2$

Hence , ${\omega }_2$ = $200$ $rpm$