Question 511: Centre of Mass and Rotational Motion

Question

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

Accepted Answer

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,
$$\begin{aligned}
& \mathrm{h}=\mathrm{L} \sin 60^{\circ}=\mathrm{L} 	imes \frac{\sqrt{3}}{2} \
	herefore\quad& \mathrm{I}=2 \mathrm{~m} 	imes \frac{3}{4} \mathrm{~L}^2=\frac{3}{2} \mathrm{~mL}^2
\end{aligned}$$

Answer

$\frac{3 \mathrm{~mL}^2}{4}$

Answer

$\frac{\mathrm{mL}^2}{4}$

Answer

$\frac{3 \mathrm{~mL}^2}{2}$

Answer

$\frac{\mathrm{mL}^2}{2}$

Centre of Mass and 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

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