Question

Three identical solid spheres each of mass $M$ and radius $R$ are fixed at three corners of light equilateral triangular frame of side length $3R$ such that centre of spheres coincide with corners of the frame. The moment of inertia of these three spheres about an axis perpendicular to the plane of frame and passing through centre of any sphere is:

• A. $\frac{34M{R}^2}{5}$
• B. $\frac{96M{R}^2}{5}$
• C. $\frac{51M{R}^2}{5}$
• D. $17M{R}^2$

Answer 1

Answer: (B)

$\frac{96M{R}^2}{5}$

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

Using parallel axis theorem we get the moment of inertia about the rotational axis for the two remaining spheres as $\frac{2}{5}M{R}^2+(3R{)}^2=\frac{47}{5}M{R}^2$

Thus we get the total moment of inertia as $2\times \frac{47}{5}M{R}^2+\frac{2}{5}M{R}^2=\frac{96}{5}M{R}^2$