Question

Find the moment of inertia $I$ of hollow sphere of outer radius $R=20\text{cm}$ and mass $M=1\text{kg}$ about its tangent on outer surface.

• A. $\frac{1}{15}{\text{kg-m}}^2$
• B. $\frac{1}{45}{\text{kg-m}}^2$
• C. $\frac{1}{30}{\text{kg-m}}^2$
• D. $\frac{1}{60}{\text{kg-m}}^2$

Answer 1

The correct option is A $\frac{1}{15}{\text{kg-m}}^2$


$I$ is the moment of inertia of hollow sphere about its tangent.

We know, moment of inertia about its center of mass $I_{\text{com}}=\frac{2}{3}M{R}^2$
M.O.I about tangent of hollow sphere is $I$

Applying parallel axes theorem,
$I=I_{com}+M{d}^2$ [$\because$ here $d=R$]
$=\frac{2}{3}M{R}^2+M{R}^2$
$=\frac{5}{3}M{R}^2$
$=\frac{5}{3}\times 1\times (0.20{)}^2{\text{kg-m}}^2$
$=\frac{1}{15}{\text{kg-m}}^2$
So, M.O.I of hollow sphere about its tangent is $\frac{1}{15}{\text{kg-m}}^2$