Question

Two blocks of masses 1 kg and 2 kg are suspended at the end of a light string passing over a frictionless pulley of mass 4 kg and radius 10 cm. When the masses are released, the acceleration of the masses is:

• A. $\frac{g}{9}$
• B. $\frac{g}{7}$
• C. $\frac{g}{5}$
• D. $\frac{g}{8}$

Answer 1

The correct option is C $\frac{g}{5}$

Refer to the diagram.
For the two blocks the equations of motion are:
$T_1-m_{1}g=m_{1}a$ ......(1)
$m_{2}g-T_2=m_{2}a$ ......(2)
For the pulley, equation of motion for rotation is( since the pulley only rotates)
Net torque ${\tau }_{net}=I\alpha$

$\therefore T_{2}R-T_{1}R=\frac{1}{2}M{R}^2\alpha$ ....(3)

Dividing (3) by R and adding (1) and (2) to it we get,

$m_{2}g-m_{1}g=m_{2}a+m_{1}a+\frac{M}{2}R\alpha$

$\therefore m_{2}g-m_{1}g=m_{2}a+m_{1}a+\frac{M}{2}a$

$\Rightarrow a=\frac{m_2-m_1}{m_2+m_1+\frac{M}{2}}g$ .....(4)

Substituting $m_1=1$, $m_2=2$, and $M=4$ and $R=10cm=0.1m$ in eqn(4), we get:

$a=\frac{2-1}{2+1+\frac{4}{2}}g=\frac{g}{5}$