Physics - Rotation Question with Solution | TestHub

Mass of each sphere, $M=0.5$ kg

Radius of each sphere, $R=\frac{\sqrt{5}}{2}$ cm $=\frac{\sqrt{5}}{2}\times 10^{-2}$ m

Side of the square, $a=4$ cm $=4\times 10^{-2}$ m

The diagonal of the square passes through the centres of two spheres. The perpendicular distance of the centres of the other two spheres from this diagonal is $r=\frac{a}{\sqrt{2}}=\frac{4}{\sqrt{2}}=2\sqrt{2}$ cm $=2\sqrt{2}\times 10^{-2}$ m.

The moment of inertia of the system about the diagonal is the sum of the moments of inertia of all four spheres. Using the parallel axis theorem for the two spheres not on the diagonal:

$I=2\times \left(\frac{2}{5}M{R}^2\right)+2\times \left(\frac{2}{5}M{R}^2+M{r}^2\right)$

$I=\frac{8}{5}M{R}^2+2M{r}^2$

Substituting the given values:

$I=\frac{8}{5}\times 0.5\times {\left(\frac{\sqrt{5}}{2}\times 10^{-2}\right)}^2+2\times 0.5\times {\left(2\sqrt{2}\times 10^{-2}\right)}^2$

$I=\frac{8}{5}\times \frac{1}{2}\times \frac{5}{4}\times 10^{-4}+1\times 8\times 10^{-4}$

$I=1\times 10^{-4}+8\times 10^{-4}$

$I=9\times 10^{-4}$ kg-m${}^2$

Comparing this with $N\times 10^{-4}$ kg-m${}^2$, we get $N=9$.

Answer: $9$