Two beads of mass `2m` and `m`, connected by a rod of length `l` and of negligible mass are free to move in a smooth vertical circular wire frame of radius `l` as shown. Initially the system is held in horizontal position `(` Refer figure `)`

If the rod is replaced by a massless string of length `l` and the system is released when the string is horizontal then `:`

Options

A. Mass `2m` will arrive earlier at the bottom
B. Mass `m` will arrive earlier at the bottom.
C. Both the masses will arrive together but different speeds.
D. Both the masses will arrive together with same speeds.

Answer

D

Explanation

The speeds given to `2m` will also be possessed by `m`
`:. KE` in horizontal position gets converted in `PE` in vertical position.
`(1)/(2)2mv^(2)+(1)/(2)mv^(2)=` change in `PE` in vertical position.

`DeltaPE=2mg[l cos 30^(@)-l cos 60^(@)]+mg[l cos 30^(@)+(l)/(2)]`
`2mg[(lsqrt(3))/(2)-(l)/(2)]+mg[(lsqrt(3))/(2)+(l)/(2)]`
`:. mgl[sqrt(3)-1]-mgl[(sqrt(3)+1)/(2)]`
`=mgl[sqrt(3)-1+(sqrt(3))/(2)+(1)/(2)]=mgl[(3sqrt(3))/(2)-(1)/(2)]`
`K.E.=(1)/(2)3mv^(2)=mgl[(3sqrt(3)-1)/(2)]`
`:. v=sqrt((3sqrt(3)-1)/(3)gl)` Ans.
Both the masses will have same magnitude of acceleration all the time.
`:.` Their velocities and distance covered will same.
Hence `(D)`.