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 an axis passing through one corner of the square and perpendicular to its plane will be

• A. $\frac{4}{5}m[2a^2+5b^2]$
• B. $\frac{5}{4}m[a^2+2b^2]$
• C. $\frac{2}{5}m[3a^2+4b^2]$
• D. $\frac{3}{4}m[2a^2+4b^2]$

Answer 1

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

One of the sphere rotate about its own axis, thus its moment of inertia is given as $\frac{2}{5}m{a}^2$ and the moment of inertia of other three are found out using parallel axis theorem. Two of them are at a distance of b from the rotational axis the sum of their moment of inertia is $2\times (\frac{2}{5}m{a}^2+m{b}^2)$ and the third one is at a diagonal distance of $\sqrt{2}b$, thus its moment of inertia is $\frac{2}{5}m{a}^2+2m{b}^2$
Thus we get total moment of inertia as $\frac{4}{5}m[2a^2+5b^2]$