Questions

Three point masses each of mass ‘m’ rotate in a circle of radius r with constant angular velocity $ω$ due to their mutual gravitational attraction. If at any instant, the masses are on the vertex of an equilateral triangle of side ‘a’, then the value of $ω$ is

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detailed solution

Correct option is B

F=F12+F22+2F1F2cosθ ⇒If F1=F2 then Fnet=2Fcosθ2θ=60∴ Fnet=2Fcos602=2Fcos30=2Gmma232=3Gmma2Fnet provides centripetal force3GMma2=mrω23GMa2=a3ω2ω2=3GMa3ω=3GMa3

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