A ballet dancer spins about a vertical axis at `60` rpm with arms outstretched. When her arms folded the angular frequency increases to `90` rpm. Find the change in her moment of inertia.

Explanation

By the principle of conservation of angular momentum
`Ixx60=I_(2)xx90`
Final moment of inertia, `I_(2)=(2I)/(3)`
Change in moment of inertia `=I-(2I)/(3)=(I)/(3)`
Assuming the initial `I` to be `100%` change in moment of inertia `=(100)/(3)%=33(1)/(3)%`