Two point masses `m` are connected the light rod of length `l` and it is free to rotate in vertical plane as shown in figure. Calculate the minimum horizontal velocity is given to mass so that it completes the circular motion in vertical lane.

Answer

`u=sqrt((48)/(5)gl)`

Explanation

Let the initial given to the mass at `A` be `u`. Then the velocity of mass at `B` is `u//2`
As the system moves from initial the final position Increase in potential energy is `4 ml+ 2 mgl` Decrease in kinetic energy.
`=(1)/(2) m u^(2)+(1)/(2)m((u)/(2))^(2)=(5)/(2)m u^(2)`
From conservation of energy
`(5)/(8) m u^(2)=6 m l or u = sqrt((48)/(5) gl)`