Q.
Three particles each of mass $m$ rotate in a circle of radius $r$ with uniform angular speed $\omega$ under their mutual gravitational attraction. If at any instant the points are on the vertex of an equilateral of side $L$, then angular velocity $\omega$ is
A) $\sqrt{\frac{2Gm}{L^{3}}}$
B) $\sqrt{\frac{3Gm}{L^{3}}}$
C) $\sqrt{\frac{5Gm}{L^{3}}}$
D) $\sqrt{\frac{Gm}{L^{3}}}$
Explanation
(b) $a\frac{Gm^{2}}{L^{2}}\cos{30{^\circ} = m\omega^{2}r = \frac{m\omega^{2}L}{\sqrt{3}}}\ \ \therefore r = \frac{L}{\sqrt{3}}\therefore\omega = \sqrt{\frac{3Gm}{L^{3}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.