Physics - Rotation Question with Solution | TestHub

Let $M$ be the mass and $\rho$ be the density of the material for both spheres.

The volume of the material for the solid sphere is $V_s=\frac{4}{3}\pi R_s^3$.

The volume of the material for the hollow sphere is $V_h=\frac{4}{3}\pi (R_{out}^3-R_{in}^3)$.

Since they have the same mass and same material, their volumes of material must be equal, so $V_s=V_h$.

$\frac{4}{3}\pi R_s^3=\frac{4}{3}\pi (R_{out}^3-R_{in}^3)\Rightarrow R_s^3=R_{out}^3-R_{in}^3$

This implies $R_{out}>R_s$. For a thin-walled hollow sphere, its effective radius $R_h\approx R_{out}$, so $R_h>R_s$.

The moment of inertia of the solid sphere about its diameter is $I_s=\frac{2}{5}M{R}_s^2=0.4M{R}_s^2$.

The moment of inertia of the thin-walled hollow sphere about its diameter is $I_h=\frac{2}{3}M{R}_h^2\approx 0.67M{R}_h^2$.

Since $R_h>R_s$ and $\frac{2}{3}>\frac{2}{5}$, it follows that $I_h>I_s$.

Answer: ${\mathrm{I}}_{\mathrm{h}}>{\mathrm{I}}_{\mathrm{s}}$