Question 349: Gravitation

Question

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

Accepted Answer

Consider mass $$\mathrm{m}$$ at $$\mathrm{A}$$.
The forces exerted on it by the other two masses are given by
$$F_1=G \frac{\mathrm{m}^2}{\mathrm{~L}^2}=\mathrm{F}_2$$
The angle between the two forces is $$60^{\circ}$$.
Hence the resultant force
$$\begin{aligned}
& \quad F=\sqrt{\mathrm{F}_1^2+\mathrm{F}_1^2+2 \mathrm{~F}_1^2 \cos 60^{\circ}}=\sqrt{3} \mathrm{~F}_1 \
& 	herefore \quad F=\sqrt{3} \cdot G \frac{\mathrm{m}^2}{\mathrm{~L}^2}
\end{aligned}$$
Mass $$\mathrm{m}$$ rotates around the centre $$\mathrm{G}$$. The radius of the circular motion is $$\mathrm{AG}$$.
$$\begin{aligned}
\mathrm{AG} & =\frac{2}{3} \mathrm{AD} \
\mathrm{AD} & =\mathrm{AC} \sin 60^{\circ}=\mathrm{L} \sin 60^{\circ}=\frac{\sqrt{3}}{2} \mathrm{~L} \
	herefore \quad \mathrm{AG} & =\frac{2}{3} \cdot \frac{\sqrt{3}}{2} \mathrm{~L}=\frac{\mathrm{L}}{\sqrt{3}} \
	herefore 	ext { radius } \mathrm{r} & =\frac{\mathrm{L}}{\sqrt{3}}
\end{aligned}$$
For uniform circular motion, the gravitational force provides the centripetal force.
$$\begin{aligned}
& 	herefore \mathrm{mr} \frac{\mathrm{L}}{\sqrt{3}} \cdot \omega^2=\sqrt{3} \mathrm{G} \frac{\mathrm{m}^2}{\mathrm{~L}^2} \
& 	herefore \omega^2=3 \mathrm{G} \frac{\mathrm{m}}{\mathrm{L}^3} \
& 	herefore \quad \omega=\left(3 \mathrm{G} \frac{\mathrm{m}}{\mathrm{L}^3}ight)^{1 / 2} \
& 	herefore \frac{2 \pi}{T}=\left(\frac{3 G \mathrm{~m}}{\mathrm{~L}^3}ight)^{1 / 2} \
& 	herefore T=2 \pi\left(\frac{\mathrm{L}^3}{3 \mathrm{Gm}}ight)^{1 / 2} \
& 	herefore T \propto \mathrm{L}^{3 / 2}
\end{aligned}$$

Answer

L

Answer

$$\mathrm{L}^{1 / 2}$$

Answer

$$\mathrm{L}^{3 / 2}$$

Answer

$$\mathrm{L}^{-2}$$

Gravitation

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