Question

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. Find the acceleration of cylinder $M_2$

Answer 1

Through this illustration, we will learn the application of torque equation and constraint relation in more complex case here. We have changed the block with cylinder $M_2$. Pulley $M_1$ will have only pure rotation while cylinder $M_2$ will have rotation and translation combined. Let us analyse step by step in the same way as the previous illustration.

Step I: Analyse the motions of the pulley and the cylinder Pulley. One rotational acceleration ${\alpha }_1$ (clockwise) cylinder. One rotational acceleration ${\alpha }_2$ (cloclwise) and a linear acceleration $a_2$ (downward)

Step II: Equation of motion for $M_2$

$M_{2}g-T=M_2{a}_2...\left(i\right)$

Step III: Torque equation for pulley: ${\tau }_C=I_C\alpha$

$TR=\left(\frac{M_1{R}^2}{2}\right){\alpha }_1\Rightarrow {\alpha }_1=\frac{2T}{M_{1}R}...\left(ii\right)$

Torque equation for the cylinder (About centre of mass of cylinder) ${\tau }_C=I_C\alpha$

$\Rightarrow TR=\left(\frac{M_2{R}^2}{2}\right){\alpha }_2\Rightarrow {\alpha }_2=\frac{2T}{M_{2}R}...\left(iii\right)$

Step IV: The acceleration of $P$ and $Q$ should be equal as both are connected with the same inextensible string.

Acceleration of $P,a_M={\alpha }_{1}R.....\left(iv\right)$

Acceleration of $Q,a_N=a_2-{\alpha }_{2}R....\left(v\right)$ (Downward)

Hence, constraint relation: ${\alpha }_{1}R=a_2-{\alpha }_{2}R...\left(vi\right)$

Step V: Solving equations

After solving Eqs. (i), (ii), (iii) and (iv) we get

$a_2=\left[\frac{2(M_1+M_2)}{{3M}_1+{2M}_2}\right]g$