A small body `A` is fixed to the inside of a thin rigid hoop of radius `R` and mass equal to that of the body `A`. The hoop rolls without slipping over a horizontal plane, at the moments when the body `A` gets into the lower position, the centre of the hoop moves with velocity `v_(0)`. At what values of `v_(0)` will the hoop move without bouncing?

Answer

`sqrt(8gR)`

Explanation

One the hoop, the grond is giving a normal force `R`. Again the attached mass `A` is also giving a normal force `N`. Now value of `R` is changing as the attached body is changing its position. Value of `R` is minimum when the attached body is at the highest position in its circular motion. but this minimum value of `R` should also the greater than zero to prevent the hoop from bouncing.
Now consider `FBD` of the attached body
`N+mg=(mv^(2))/R`.i
[where `v=`velocity of cenntre of mass of the hoop at tis moment. Again velocity of the body at its moment with respect to the hooop centre of mass `=2v-v=v`]
For the hoop
`N+R=mg`.....ii
From i and ii
`(mv^(2))/R-mg+R=mgimpliesR=2mg=(mv^(2))/R`..........iii
`Rge0, implies v^(2)le2ghr`........iv
Now conserving energying the reference rame of the ground:
`mv_(0)^(2)1/2 mv_(0)^(2)+1/2MR^(2)omega_(0)^(2)`
`=1/2mv^(2)+1/2(mR^(2))omega^(2)+1/2m(2v)^(2)+2mgR`
`implies` Putting `v_(0)=Romega_(0)` and `v=Romega:`
`v_(0)=sqrt(8gR)`