Question

Four solid spheres each of diameter $\sqrt{5}$ m and mass $0.5kg$ are placed with their centres at the corners of a square of side $4cm$. The moment of inertia of the system about the diagonal of the square is $N\times {10}^{-4}kg{m}^2$ then N is

Answer 1

Moment of inertia of sphere about its center

$I_1=2m{r}^2/5=m{d}^2/10=0.25\times {10}^{-4}kg{m}^2$

no using parallel axis theorem ,distance of sphere center to square center is $4/\sqrt{2}\times {10}^{-2}m$

so $I=I_1+m\times \left(\frac{4}{\sqrt{2}}\right)\times {10}^{-4}$

$I=2.25\times {10}^{-4}$

since we have 4 sphere ,so total moment of inertia =$4I=9\times {10}^{-4}kg{m}^2$