Question

Find the tension in the string between the pulleys and the acceleration of the blocks as shown in the figure. Assume that there is no slipping between the string and the pulleys.
[Given, mass of bigger pulley $=2m$, radius of bigger pulley $=R$, mass of smaller pulley $=m$ and radius of smaller pulley $=\frac{R}{2}$]

• A. $\frac{4}{3}mg,\frac{2g}{9}$
• B. $\frac{9}{4}mg,\frac{9}{2}g$
• C. $mg,g$
• D. $mg,\frac{g}{2}$

Answer 1

The correct option is A $\frac{4}{3}mg,\frac{2g}{9}$

Let $a$ be the acceleration of pulley and block.
From the free body diagram of the blocks,


$2mg-T_1=2ma---\left(1\right)$
$T_2-mg=ma---\left(2\right)$

Let the tension in the string between the pulleys be $T_3$.
FBD of the pulleys:


For pulley of mass $2m$:
Torque about centre of pulley ${\tau }_0=T_{1}R-T_{3}R$
${\tau }_0=I\alpha$
$\Rightarrow I_1{\alpha }_1=T_{1}R-T_{3}R$
$\Rightarrow \frac{2m{R}^2}{2}\left(\frac{a}{R}\right)=R(T_1-T_3)$
$\Rightarrow ma=T_1-T_3.......\left(3\right)$

For pulley of mass $m$ :
$I_2{\alpha }_2=T_3\frac{R}{2}-T_2\frac{R}{2}$
$\Rightarrow \frac{m}{2}{\left(\frac{R}{2}\right)}^2\left(\frac{a}{\frac{R}{2}}\right)=\frac{R}{2}(T_3-T_2)$
$\Rightarrow \frac{ma}{2}=T_3-T_2---\left(4\right)$
From eq. (2) and (4),
$\frac{ma}{2}=T_3-ma-mg$
$\Rightarrow \frac{3ma}{2}=T_3-mg--\left(5\right)$
and from eq. (1) & (3),
$ma=2mg-2ma-T_3$
$\Rightarrow T_3=2mg-3ma--\left(6\right)$
Lastly, from eq. (5) & (6)
$\frac{3}{2}ma+mg=2mg-3ma$
$\Rightarrow a=\frac{2}{9}g$ is the acceleration of the system.
From eq. (6),
$T_3=2mg-3m\times \frac{2}{9}g=\frac{4mg}{3}$
$\therefore T_3=\frac{4}{3}mg$ is the tension in the string between the pulleys.