Question

Three particles, each of mass m, are situated at the vertices of equilateral triangle of side length a. The only forces. It is desired that each particle moves in a circle while maintaining the original mutual speration a. Find the intial velocity that should be given to each particle and also the time period of the circular motion.

Answer 1

Correct Answer - A::B::C
The radius of the circle `r = (2)/(3) sqrt(a^2 - (a^2)/(4)) = (alpha)/(sqrt3)`
Let v be the velocity given . The centripetal force is provied by the resultant gravitational attraction of the two masses.
`F_F = sqrt(F^2 +F^2 +2F^2 cos 60^@)`
`=sqrt3 F =sqrt3G (mxxm)/(a^2)`
`:. sqrt3G (m^2)/(a^2) = (mv^2)/(r )`
`((mv^2)/(r ) = centripetal force)`
`v^2 = (sqrt3Gmr)/(a^2) = sqrt(3Gma)/(a^2xxsqrt3) rArr v = sqrt((Gm)/(a))`
Time period of circular motion
`T = (2pir)/(v) = (2pia//sqrt3)/(sqrt(Gm)/(a)) = 2pisqrt((a^3)/(3Gm))`