Rotational kinetic energy: At any angular speed, a certain uniform solid sphere of diameter D has half as much rotational kinetic energy as a certain uniform thin-walled hollow sphere of the same diameter when both are spinning about an axis through their centers. If the mass of the solid sphere is M, the mass of the hollow sphere is
A) 3/5 M.
B) 5/3 M.
C) 5/6 M.
D) 6/5 M.
E) 2 M.

Question

Quizplus · Accepted Answer

The rotational kinetic energy is given by $ KE = \frac{1}{2} I \omega^2 $. For a solid sphere, $ I = \frac{2}{5} MR^2 $, and for a hollow sphere, $ I = \frac{2}{3} MR^2 $. Given that the solid sphere has half the rotational kinetic energy of the hollow sphere, we set up the equation $ \frac{1}{2} \left(\frac{1}{2} \cdot \frac{2}{5} MR^2 \cdot \omega^2\right) = \frac{1}{2} \cdot \frac{2}{3} M_h R^2 \cdot \omega^2 $, solving for $ M_h $ gives $ M_h = \frac{6}{5} M $.

Rotational Kinetic Energy: at Any Angular Speed, a Certain Uniform