A semicircular track of radius `R=62.5cm` is cut in a block. Mass of block having track, is `M=1kg` and rests over a smooth horizontal floor. A cylinder of radius `r=10`cm and mass `m=0.5kg` is hanging by thread such that axes of cylinder and track are in same level and surface of cylinder is in contact with the track as shown in figure When the thread is burnt, cylinder starts to move down the track. Sufficient friction exists between surface of cylinder and track, so that cylinder does not slip.
Calculate velocity (in `m//s`) of the block when it reaches bottom of the track. `(g=10m//s^(2))`

Answer

2

Explanation

Since block having semi circular track is over a smooth horizontal floor, therefore no horizontal reaction is provided by the floor or horizontal momentum of system of cylinder and block remains constant. But system is initially at rest or its initial horizontal momentum is zero. Therefore, horizontal momentum of the system always remains zero.
According to law of conservation of momentum
`MV=mv` or `V=v/2`
Hence velocity of cylinder axis, relative to the block is
`(v+V)=(3v)/2`(Right ward)
Angular velocity of cylinder is
`omega((1.5v))/r=15v` (clockwise)
Moment of inertia of cylindder about its own axis is
`I=(mr^(2))/2=2.5xx10^(-3) kgm^(2)`
Since cylinder does nopt slip over the track, therefore, there is no loss of energy due to friction.
Hence sum of `KE` of cylinder and block `=` Loss of `PE` of cylinder from top to bottom of the track
`{1/2Iomega^(2)+1/2mv^(2)}+1/2MV^(2)=mg(R-r)`
`v=2m//s`