A wheel of radius r and moment of inertia I about its axis is fixed at top of an inclined plane of inclination `theta` as shown in figure. A string is wrapped round the wheel and its free end supports a block of mass M which can slide on the plane. Initially, the wheel is rotating at a speed `omega` in direction such that the block slides up the plane. How far will the block move before stopping?

Explanation

Suppose the deceleration of the block is `a`. The linear deceleration of the rim of the wheel is also `a`. the angular deceleration of the wheel is `alpha=a/r`. If the tension in the string is `T`,the equations of motion are as follows:

`Mgsintheta=T=Ma ` and `Tr=Ialpha=(Ia)/r`
Elinating `T` from these equations
`Mgsintheta-I a/r^(2)=Ma`
`implies a=(Mgr^(2)sitheta)/((I+Mr^(2)))`
The initial velocity of the block up the incline is `v=omegar`. Thus the distance moved by the block before stopping is
`x=(v^(2))/(2a)=(omega^(2)r^(2)(I+Mr^(2)))/(2Mgr^(2)sitheta)-((I+Mr^(2))omega^(2))/(2Mgsintheta)`