Question

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

• A. $\frac{3L}{4}$
• B. $\frac{L}{4}$
• C. $\frac{L}{2}$
• D. $L$

Answer 1

The correct option is D $L$

The law of conservation of angular momentum states that , in an isolated system , total angular momentum remains constant . A rotating dancer is an isolated system therefore when she folds hands , her moment of inertia decreases but her angular speed increases such that angular momentum remains constant as L i.e.

$L=I\omega$

when $I$ decreases , $\omega$ increases .

Hence , total angular momentum would be L .