Seven identical discs each of mass $M$ and radius $\mathrm{R}$ are arranged in a hexagonal plane pattern so…

Seven identical discs each of mass $M$ and radius $\mathrm{R}$ are arranged in a hexagonal plane pattern so as to touch each neighbour disc as shown in the figure. The moment of inertia of the system of seven discs about an axis passing through the centre of central disc and normal to the plane of all discs is

1. $\frac{7}{2} \mathrm{MR}^2$
2. $\frac{13}{2} \mathrm{MR}^2$
3. $\frac{29}{2} \mathrm{MR}^2$
4. $\frac{55}{2} \mathrm{MR}^2$

Solution

M.I of a circular disc $=\frac{\mathrm{MR}^2}{2}$ Using parallel axis theorem, M.I. about origin $\mathrm{I}=\mathrm{I}_{\mathrm{cm}}+6 \mathrm{I}$ where, $\mathrm{I}_{\mathrm{cm}}=\mathrm{M}$. I of the central disc $\mathrm{I}^{\prime}=\mathrm{M}$. I of the each disc about the given axis. $\therefore \quad \mathrm{I}=\frac{\mathrm{MR}^2}{2}+6\left(\mathrm{I}_{\mathrm{cm}}+\mathrm{MD}^2\right)$ $=\frac{\mathrm{MR}^2}{2}+6\left(\frac{\mathrm{MR}^2}{2}+4 \mathrm{MR}^2\right) \quad \ldots(\because \mathrm{D}=2 \mathrm{R})$ $\begin{aligned} & =\frac{\mathrm{MR}^2}{2}+6\left(\frac{\mathrm{MR}^2+8 \mathrm{MR}^2}{2}\right) \\ & =\frac{55 \mathrm{MR}^2}{2} \end{aligned}$ .

Asked in: MHT CET 2023 (10 May Shift 1)

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