Question

Find the tension in the string between the pulleys and the acceleration of the system:-
[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 $T_0=T_{1}R-T_{3}R$
$T_0=I\alpha$
$\Rightarrow I\alpha =T_{1}R-T_{3}R$
$\Rightarrow \frac{2m{R}^2}{2}\left(\frac{a}{R}\right)=R(T_1-T_3)$
$\Rightarrow ma=T_1-T_3.......\left(3\right)$

For pulley of mass $m$ :
$I\alpha =T_3\frac{R}{2}-T_2\frac{R}{2}$
$\Rightarrow \frac{m}{2}{\left(\frac{R}{2}\right)}^2\left(\frac{a}{\frac{R}{2}}\right)=\frac{R}{2}(T_3-T_2)$
$\Rightarrow \frac{ma}{2}=T_3-T_2---\left(4\right)$
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),
$T_3=2mg-3m\frac{2}{9}g=\frac{4mg}{3}$
$\therefore T_3=\frac{4}{3}mg$ is the tension in the string between the pulleys.