MHT CET · Physics · Rotational Motion
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
- A \(\frac{3 \mathrm{~mL}^2}{4}\)
- B \(\frac{\mathrm{mL}^2}{4}\)
- C \(\frac{3 \mathrm{~mL}^2}{2}\)
- D \(\frac{\mathrm{mL}^2}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{3 \mathrm{~mL}^2}{2}\)
Step-by-step Solution
Detailed explanation
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,
\(\begin{aligned}
& \mathrm{h}=\mathrm{L} \sin 60^{\circ}=\mathrm{L} \times \frac{\sqrt{3}}{2} \\
\therefore \quad & \mathrm{I}=2 \mathrm{~m} \times \frac{3}{4} \mathrm{~L}^2=\frac{3}{2} \mathrm{~mL}^2
\end{aligned}\)
Hence, M.I. about the line through A is,
\(\mathrm{I}=2 \mathrm{mh}^2\)
From figure,
\(\begin{aligned}
& \mathrm{h}=\mathrm{L} \sin 60^{\circ}=\mathrm{L} \times \frac{\sqrt{3}}{2} \\
\therefore \quad & \mathrm{I}=2 \mathrm{~m} \times \frac{3}{4} \mathrm{~L}^2=\frac{3}{2} \mathrm{~mL}^2
\end{aligned}\)
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