Question

Moment of inertia of solid sphere about its diameter is $I$. If that sphere is recast into $8$ identical small spheres, then the moment of inertia of a small sphere about its diameter is:

• A. $\frac{I}{8}$
• B. $\frac{I}{16}$
• C. $\frac{I}{24}$
• D. $\frac{I}{32}$

Answer 1

Answer: (D)

$\frac{I}{32}$

Let Mass and radius of bigger sphere be $M$ and $R$.

So MI is $I=2M{R}^2/5$

As this sphere is recast into $8$ smaller spheres, the mass of smaller spheres is $M/8$ and radius is $(\frac{4}{3}\pi R^3/\frac{4}{3}\pi 8{)}^{1/3}=R/2$

So MI of the smaller sphere is $2\left(M/8\right)(R/2{)}^2/5=2M{R}^2/5\times 32=I/32$