Question 545: Centre of Mass and Rotational Motion

Question

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. $\mathrm{I}_{\mathrm{A}}, \mathrm{I}_{\mathrm{B}}, \mathrm{I}_{\mathrm{C}}$ and $\mathrm{I}_{\mathrm{D}}$ 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

Accepted Answer

Using the parallel axes theorem, the M.I. of the system about the axis passing through the centre of the sphere $A$ is
$\begin{aligned}
& \mathrm{I}_{\mathrm{A}}=\mathrm{I}_{\mathrm{A}}{ }^{\prime}+\mathrm{I}_{\mathrm{B}}{ }^{\prime}+\mathrm{I}_{\mathrm{C}^{\prime}}+\mathrm{I}_{\mathrm{D}}{ }^{\prime} \
& \mathrm{I}_{\mathrm{A}}=\frac{2}{5} \mathrm{MR}^2+\left(\frac{2}{5} \mathrm{MR}^2+4 \mathrm{MR}^2ight)+\left(\frac{2}{5} \mathrm{MR}^2+16 \mathrm{MR}^2ight)  +\left(\frac{2}{5} M^2+36 M R^2ight) \
&
\end{aligned}$
$	herefore \quad \mathrm{I}_{\mathrm{A}}=57.6 \mathrm{MR}^2$
The M.I. about sphere B is,
$\begin{aligned}
I_B=\frac{2}{5} M R^2+\left(\frac{2}{5} M R^2+4 M R^2ight)+ & \left(\frac{2}{5} M R^2+16 M R^2ight) \
& +\left(\frac{2}{5} M R^2+4 M R^2ight)
\end{aligned}$
$\begin{aligned}
& \mathrm{I}_{\mathrm{B}}=25.6 \mathrm{MR}^2 \
	herefore \quad & \mathrm{I}_{\mathrm{A}}-\mathrm{I}_{\mathrm{B}}=57.6 \mathrm{MR}^2-25.6 \mathrm{MR}^2=32 \mathrm{MR}^2
\end{aligned}$

Answer

24 MR$^2$

Answer

32 MR$^2$

Answer

56 MR$^2$

Answer

80 MR$^2$

Centre of Mass and Rotational Motion

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