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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 unwounds over the pulley as well the cylinder. Findthe acceleration of cylinder M_(2). |
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Answer» Solution :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` |
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