Three point masses each of mass ' m ' are kept at the corners of an… | MHT CET 2019 (2 May, Shift 2)

Q156MHT CET 2019 · 2 MAY · SHIFT 2Easy

Moment of Inertia & Radius of Gyration

Three point masses each of mass ' $m$ ' are kept at the corners of an equilateral triangle of side ' $L$ '. The system rotates about the center of the triangle without any change in the separation of masses during rotation. The period of rotation is directly proportional to $\left(\cos 30^\circ = \sin 60^\circ = \frac{\sqrt{3}}{2}\right)$

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Answer: (a)

The separation stays fixed because mutual gravity supplies the centripetal force. Each mass is at $r = \frac{L}{\sqrt{3}}$ from the centre.
Net pull on one mass toward the centre: $2\frac{Gm^2}{L^2}\cos 30^\circ = \frac{\sqrt{3}Gm^2}{L^2}$
$m\omega^2 \frac{L}{\sqrt{3}} = \frac{\sqrt{3}Gm^2}{L^2} \Rightarrow \omega^2 = \frac{3Gm}{L^3}$
$T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{L^3}{3Gm}} \Rightarrow T \propto L^{3/2}$
So the answer is (b) $L^{3/2}$ .

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Moment of Inertia & Radius of Gyration question from MHT CET 2019 (2 May, Shift 2)

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