Question

Four spheres of mass $M$ and radius $R$ each are touching each other as shown. Calculate the MI about the axis which passes through centre of the square and perpendicular to the plane.

• A. $\frac{12}{5}M{R}^2$
• B. $\frac{24}{5}M{R}^2$
• C. $\frac{48}{5}M{R}^2$
• D. $\frac{M{R}^2}{28}$

Answer 1

The correct option is C $\frac{48}{5}M{R}^2$

Using parallel axis theorem,
Moment of inertia about an axis is $I=I_c+M{d}^2$
Where,
$I$ is moment of inertia due of centre of square,
$I_c$ is moment of inertia due of centre of sphere
$d$ is distance between center of sphere of square
$\Rightarrow d=\sqrt{2}R$
Hence, $I=(\frac{2}{5}M{R}^2+M(\sqrt{2}R{)}^2)$
For four spheres,
$4I=(\frac{2}{5}M{R}^2+M(\sqrt{2}R{)}^2)\times 4$
$=\frac{48}{5}M{R}^2$