Question

A figure skater is spinning slowly with arms outstretched. She brings her arms in close to her body and her moment of inertia decreases by 12. By what factor does her rotational Kinetic energy change?

Answer 1

Given is:-

Initial moment of inertial  

Final moment of inertia $I_2 = \frac{I}{2}$

Now,

The rotational kinetic energy is given by

$KE_{rotational } = \frac{1}{2}Iw^2$

thus initial rotational kinetic energy is

$KE_{rot,1 } = \frac{1}{2}Iw^2$

and the final rotational kinetic energy is

$KE_{rot,2 } = \frac{1}{2}(\frac{I}{2})w^2$

or

$KE_{rot,2 } = \frac{1}{2}KE_{rot,1}$

Hence,

The rotational kinetic energy is change by a factor of   $\boxed{\frac{1}{2}}$

thus option - 1 is correct answer.