The moment of inertia of a cube of mass $m$ and side $a$ about one of its edges is equal to

Question

Aicharya · Accepted Answer

From theorem of perpendicular axes, we have $\begin{aligned} I & =I_C+m\left(\frac{a}{\sqrt{2}}ight)^2 \ & =\left[\frac{m a^2}{12}+\frac{m a^2}{12}ight]+\frac{m a^2}{2} \ & =\frac{2}{3} m a^2\end{aligned}$

The moment of inertia of a cube of mass $m$ and side $a$ about one of its edges is equal to

Solution