Question

Four solid spheres each of diameter $\sqrt{5}cm$ and mass $0.5kg$ are placed with their centers 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:

• A. $6$
• B. $7$
• C. $8$
• D. $9$

Answer 1

Answer: (D)

$9$

MI of the system $I=I_1+I_2+I_3+I_4$ where $I_i$ is the MI of the ith sphere about the diagonal XY where i goes from 1 to 4.
$I_1=I_3$ by symmetry
$I_2=I_4$ by symmetry
$I_1=\frac{2}{5}m{R}^2+m(\frac{\sqrt{2}a}{2}{)}^2$ using Parallel Axis Theorem since $\frac{2}{5}m{R}^2$ is the $MI$ of the sphere about the Center of Mass of the sphere. $\frac{\sqrt{2}a}{2}$ is the distance between the center of the square and sphere.
$I_2=\frac{2}{5}m{R}^2$
$\therefore I=2I_1+2I_2$

$=2\ast (\frac{2}{5}m{R}^2+m(\frac{\sqrt{2}a}{2}{)}^2)+2\ast (\frac{2}{5}m{R}^2)$ $=\frac{8}{5}m{R}^2+m{a}^2$

$=\frac{8}{5}\ast 0.5\ast (\frac{\sqrt{5}}{2}{)}^2\ast {10}^{-4}+0.5\ast 16\ast {10}^{-2}=9\ast {10}^{-4}kg{m}^2$
$\therefore$ comparing I with $N\times {10}^{-4}kg{m}^2$, we get $N=9$