Question

Four identical solid spheres each of mass $M$ and radius $R$ are fixed at four corners of a light square frame of side length $4R$ such that centres of spheres coincide with corners of square. The moment of inertia of $4$ spheres about an axis perpendicular to the plane of frame and passing through its centre is:

• A. $\frac{21M{R}^2}{5}$
• B. $\frac{42M{R}^2}{5}$
• C. $\frac{84M{R}^2}{5}$
• D. $\frac{168M{R}^2}{5}$

Answer 1

Answer: (D)

$\frac{168M{R}^2}{5}$

Distance of the center of the spheres from the axis is $d=2\sqrt{2}R$

Now the MI of a solid sphere about its axis is $2M{R}^2/5$

Using the parallel axis theorem, the MI about the given axis is $2M{R}^2/5+M{d}^2=2M{R}^2/5+M(2\sqrt{2}R{)}^2=2M{R}^2/5+8M{R}^2=42M{R}^2/5$

So total MI of system is $4\times 42M{R}^2/5=168M{R}^2/5$