Answer to Question #118122 in Classical Mechanics for Lizwi

Question

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
1 : 3/5 M.
2 : 5/3 M.
3 : 5/6 M.
4 : 6/5 M.
5 : 2 M.

Answer 1

Correct option is 4.

Reason:

Let $\omega$ be the speed at which both sphere rotates,$I_s,I_h$ be the moment of inertia of solid and hollow sphere ,passing through center respectively.

$D$ is diameter of both the sphere, $M,M'$ be the mass of solid and hollow sphere respectively.

We have given that the rotational kinetic energy of solid sphere is half of the rotational kinetic energy of hollow sphere,thus

$$K_s=\frac{1}{2}K_h\\ \implies \frac{1}{2}I_s\omega^2=\frac{1}{4}I_h\omega^2 \hspace{1cm}(1)$$

Since,

$$I_s=\frac{2}{5}M\bigg(\frac{D}{2}\bigg)^2$$

And

$$I_h=\frac{2}{3}M'\bigg(\frac{D}{2}\bigg)^2$$

Thus, on plugin these values to $(1)$ we get,

$$\frac{2}{5}M=\frac{1}{3}M'\implies M'=\frac{6}{5}M$$

Therefore, mass of hollow sphere is $\frac{6}{5}M$ .