Three, point masses each of mass ' $\mathrm{m}$ ' 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

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Aicharya · Accepted Answer

Consider mass $\mathrm{m}$ at $\mathrm{A}$. The forces exerted on it by the other two masses are given by $\mathrm{F}_1=\mathrm{G} \frac{\mathrm{m}^2}{\mathrm{~L}^2}=\mathrm{F}_2$ The angel between the two forces is $60^{\circ}$. Hence the resultant force The angel between the two forces is $60^{\circ}$. Hence the resultant force $\begin{aligned} & \mathrm{F}=\sqrt{\mathrm{F}_1^2+\mathrm{F}_1^2+2 \mathrm{~F}_1^2 \cos 60^{\circ}}=\sqrt{3} \mathrm{~F}_1 \ & \mathrm{~F}=\sqrt{3} \cdot \mathrm{G} \frac{\mathrm{m}^2}{\mathrm{~L}^2}\end{aligned}$ radius $r=\frac{L}{\sqrt{3}}$ For uniform circular motion, the gravitational force provides the centripetal force. $\begin{aligned} & 	herefore \mathrm{mr} \omega^2=\mathrm{F} \ & 	herefore \mathrm{m} \frac{\mathrm{L}}{\sqrt{3}} \cdot \omega^2=\sqrt{3} \mathrm{G} \frac{\mathrm{m}^2}{\mathrm{~L}^2} \ & 	herefore \omega^2=3 \mathrm{G} \frac{\mathrm{m}}{\mathrm{L}^3}\end{aligned}$ $\begin{aligned} & 	herefore \omega=\left(3 G \frac{\mathrm{m}}{\mathrm{L}^3}ight)^{1 / 2} \ & 	herefore \frac{2 \pi}{\mathrm{T}}=\left(\frac{3 \mathrm{Gm}}{\mathrm{L}^3}ight)^{1 / 2}\end{aligned}$ $	herefore \mathrm{T}=2 \pi\left(\frac{\mathrm{L}^3}{3Gm}ight)^{1 / 2}$ $	herefore \mathrm{T} \propto \mathrm{L}^{3 / 2}$ .

Three, point masses each of mass ' $\mathrm{m}$ ' are kept at the corners of an equilateral triangle of side…

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