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 the diagonal of square will be

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

Answer 1

Answer: (A)

$\frac{2}{5}m[4a^2+\frac{5}{2}{b}^2]$

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 $2\times (\frac{2}{5}m{a}^2+m\frac{b}{(\sqrt{2}{)}^2})$
Thus we get total moment of inertia as $m[\frac{8}{5}{a}^2+\frac{b^2}{2}]=\frac{2}{5}m[4a^2+\frac{5}{2}{b}^2]$