Question

The centres of four spheres each of mass $m$ and diameter $2a$ are at four corners of a square of side $b$. The moment of inertia of the system about one side of the square will be

• A. $\frac{2}{5}m[4a^2+5b^2]$
• B. $\frac{2}{5}m[5a^2+4b^2]$
• C. $\frac{2}{5}m[a^2+b^2]$
• D. $m[\frac{8}{5}{a}^2+b^2]$

Answer 1

The correct option is A $\frac{2}{5}m[4a^2+5b^2]$

Two of the spheres rotate about their own axis thus there moment of inertia is given as $2\times \frac{2}{5}m{a}^2$ and the moment of inertia of other two are found out using parallel axis theorem as $2\times (\frac{2}{5}m{a}^2+m{b}^2)$
Thus we get total moment of inertia as $\frac{2}{5}m[4a^2+5b^2]$