A weightless rod of length `2l` carries two equal masses `'m'` , one secured at lower end `A` and the other at the middle of the rod at `B`. The rod can rotate in vertical plane about a fixed horizontal axis passing through `C`. What horizontal velocity must be imparted to the mass at `A` so that it just completes the vertical circle.

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)`