An extensible string is wound over a rough pulley of mass `M_(1)` and radius `R` and a cylinder of mass `M_(2)` and radius `R` such that as the cylinder rolls down. The string un wounds over the pulley as well the cylinder. Find the acceleration of cylinder `M_(2)`.

Explanation

Through this illustration we will learn the application of torque equation and constraint relation in more complex case here. We have change the block with cylinder `M_(2)` Pulley `M_(1)` will have oly pure rotation while cylinder `M_(2)` will have rotatioin and translation combined. Let us analyse step by step in the same way as the previous illustration.
Step I: Analyse the motion of the pulley and the cyinder. Pulley: One rotational acceleration `alpha_(1)` (clockwise)
Cylinder: One rotational acceleration `alpha_(2)` (clockwise) andn a linear acceleration `a_(2)` (downward)
Step II: Equation of motion for `M_(2)`
`M_(2)g-T=M_(2)a_(2)` ..........i
Step III: Torque equation for pulley `tau_(c)=I_(c)alpha`
`TR=((M_(1)R^(2))/2)alpha_(1)implies(2T)/(M_(1)R)`.......ii
Torquue equation for the cylinder (about centre of mass of cylinder `tau_(c)=I_(c)alpha`
`implies TR=((M_(2)R^(2))/2)alpha_(2)`
`alpha_(2)=(2T)/(M_(2)R)`.........iii
Step IV: The acceleration of `P` and `Q` should be equal as both are connected with the same inextensible string ltbr. Accceleration of `P, a_(M)=alpha_(1)R` (downward) .........iv
Acceleration of `Q, a_(N)=a_(2)-alpha_(2)R` (downwards) .............. v
Hence, constraint relation `alpha_(1)R=a_(2)-alpha_(2)R` .............vi
Step V: Solving equations
After solving eqn i, ii, iii and iv, we get
` a_(2)=[((2M_(1)+M_(2)))/(3M_(1)+2M_(2))]g`