Question

If figure mass $m_1$ slides without friction on the horizontal surface, the frictionless pulley is in the form of a cylinder of mass $M$ and radius $R$, and a string turns the pulley without slipping. Find the acceleration of each mass, and tension in each part of the string.

Answer 1

Step I: Analysis of motion: In this case the string rotates the pulley i.e., the string does not slide. The pulley will have angular acceleration. Hence, the tension on the string on both the sides of the pulley will not be equal. The motion of $m_1$ and $m_2$ will be translational. Positive direction of angular acceleration $\alpha$ is taken along the corresponding acceleration $a_1$ which is taken positive in the direction of net force.

Step II: Equation of motion: Let these be $T_1$ and $T_2$. The equation of motion respectively, for masses $m_1$ and $m_2$ is

$T_1=m_{1}a...\left(i\right)$

$m_{2}g-T_2=m_{2}a....\left(ii\right)$

Step III: Application of torque equation:

The torque equation for the pulley

$T_{2}R-T_{1}R=I\alpha ....\left(iii\right)$

($N_2$ constitutes no torque about the axis of rotation)

There is no slipping of the string over the pulley

Step IV: Constraint relation

$a=\alpha R\Rightarrow \alpha =\frac{a}{R}...\left(iv\right)$

From Eqs. (iii) and (iv) we get

$\therefore T_{2}R-T_{1}R=I\frac{a}{R}or{T}_2-T_1=I\frac{a}{R^2}...\left(v\right)$

Step V: Solving equations:

Solving the above equations, we get $a=\frac{m_{2}g}{m_1+m_2+\frac{I}{R^2}}$

Here $I=\frac{M{R}^2}{2}$

$T_1=\frac{m_1{m}_{2}g}{(m_1+m_2+\frac{I}{R^2})};T_2=\frac{(m_1+\frac{I}{R^2})m_{2}g}{m_1+m_2+\frac{I}{R^2}}$