Four identical uniform solid spheres each of same mass ' $M$ ' and radius ' $R$ ' are placed touching each…
Four identical uniform solid spheres each of same mass ' $M$ ' and radius ' $R$ ' are placed touching each other as shown in figure, with centres $A, B, C, D . I_A, I_B, I_C$ and $I_D$ are the moment of inertia of these spheres respectively about an axis passing through centre and perpendicular to the plane. The difference in $\mathrm{I}_{\mathrm{A}}$, and $\mathrm{I}_{\mathrm{B}}$ is
$24 \mathrm{MR}^2$
$32 \mathrm{MR}^2$
$56 \mathrm{MR}^2$
$80 \mathrm{MR}^2$
Solution
Using the parallel axes theorem, the M.I. of the system about the axis passing through the centre of the sphere $\mathrm{A}$ is
$\begin{aligned}
\mathrm{I}_{\mathrm{A}}=\mathrm{I}_{\mathrm{A}}{ }^{\prime}+\mathrm{I}_{\mathrm{B}}{ }^{\prime}+\mathrm{I}_{\mathrm{C}^{\prime}}+\mathrm{I}_{\mathrm{D}}{ }^{\prime} & \\
\mathrm{I}_{\mathrm{A}}=\frac{2}{5} \mathrm{MR}^2+\left(\frac{2}{5} \mathrm{MR}^2+4 \mathrm{MR}^2\right) & +\left(\frac{2}{5} \mathrm{MR}^2+16 \mathrm{MR}^2\right) \\
& +\left(\frac{2}{5} \mathrm{MR}^2+36 \mathrm{MR}^2\right)
\end{aligned}$
$\therefore \quad \mathrm{I}_{\mathrm{A}}=57.6 \mathrm{MR}^2$
The M.I. about sphere B is,
$\begin{aligned}
\mathrm{I}_{\mathrm{B}}=\frac{2}{5} \mathrm{MR}^2+\left(\frac{2}{5} \mathrm{MR}^2+4 \mathrm{MR}^2\right) & +\left(\frac{2}{5} \mathrm{MR}^2+16 \mathrm{MR}^2\right) \\
& +\left(\frac{2}{5} \mathrm{MR}^2+4 \mathrm{MR}^2\right) \\
\therefore \quad \mathrm{I}_{\mathrm{B}}=25.6 \mathrm{MR}^2 & \mathrm{I}_{\mathrm{A}}-\mathrm{I}_{\mathrm{B}}=57.6 \mathrm{MR}^2-25.6 \mathrm{MR}^2=32 \mathrm{MR}^2
\end{aligned}$
.
