Three point masses, each of mass ' $m$ ' are placed at the corners of an equilateral triangle of side ' $L$…

Three point masses, each of mass ' $m$ ' are placed at the corners of an equilateral triangle of side ' $L$ '. The moment of inertia of the system about an axis passing through one of the vertices and parallel to the side joining other two vertices will be
  1. $\frac{3 \mathrm{~mL}^2}{4}$
  2. $\frac{\mathrm{mL}^2}{4}$
  3. $\frac{3 \mathrm{~mL}^2}{2}$
  4. $\frac{\mathrm{mL}^2}{2}$

Solution

Consider the mass at vertex $A$ as shown in figure. Hence, M.I. about the line through A is, $\mathrm{I}=2 \mathrm{mh}^2$
From figure, $\begin{aligned} & \mathrm{h}=\mathrm{L} \sin 60^{\circ}=\mathrm{L} \times \frac{\sqrt{3}}{2} \\ \therefore \quad & \mathrm{I}=2 \mathrm{~m} \times \frac{3}{4} \mathrm{~L}^2=\frac{3}{2} \mathrm{~mL}^2 \end{aligned}$

Asked in: MHT CET 2024 (16 May Shift 2)

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