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

A wheel of rdius r and moment of inertia I about its axis is fixed at

1.	A wheel of rdius r and moment of inertia I about its axis is fixed at top of an inclined plane of inclination `theta` as shownin figure. A string is wrapped round the wheel and its free end supports a block of mass M which can slide on the plane. INitialy, tehwheel is rotating at a speed `omega` in direction such that the block slides pu the plane. How far wil the block move before stopping?
Answer» Suppose the deceleration of the block is a. The linear decelerationof the rim of the wheel is also a. The angular decelratinof the wheel is `alpha=a/r`. I the tension in the string is T, the equations of motion are as follows: `Mgsintheta-T=Ma` and `Tr=Ialpha=Ia/r` ElimiN/Ating T from these equations `Mgsintheta-Ia/r^2=Ma` giving `a=(Mgr^2sintheta)/(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^2r^2(I+Mr^2))/(2Mgr^2sintheta)=((I+Mr^2)omega^2)/(2Mgsintheta)`

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