Question

Two identical solid spheres each of mass $M$ and radius $R$ are joined at the two ends of a light rod of length $2R$ as shown in the figure. The moment of inertia of the system about an axis perpendicular to the length of rod and passing through centre of mass of the two spheres is:

• A. $\frac{22M{R}^2}{5}$
• B. $\frac{44M{R}^2}{5}$
• C. $\frac{11M{R}^2}{5}$
• D. $\frac{33M{R}^2}{5}$

Answer 1

Answer: (B)

$\frac{44M{R}^2}{5}$

Moment of inertia of the sphere about its center of mass is $\frac{2}{5}M{R}^2$

Using the parallel axis theorem we get the moment of inertia about the rotational axis as $\frac{2}{5}M{R}^2+M(2R{)}^2=\frac{22}{5}M{R}^2$

As there are two sphere we get the total moment of inertia as $2\times \frac{22}{5}M{R}^2=\frac{44}{5}M{R}^2$