Question

A circular pulley of mass ${}^{\prime }{M}^{\prime }$ and radius ${}^{\prime }{R}^{\prime }$, is hinged at the centre. A long string is wound over this disc and two bodies of mass $m_1$ and $m_2$ are attached at it's free ends. Now the bodies are released. Find the accceleration of each body.

• A. $\frac{2(m_1-m_2)g}{M+(m_1+m_2)}$
• B. $\frac{(m_1-m_2)g}{M+2(m_1+m_2)}$
• C. $\frac{2(m_1-m_2)g}{M+2(m_1+m_2)}$
• D. $\frac{(m_1-m_2)g}{M+(m_1+m_2)}$

Answer 1

The correct option is C $\frac{2(m_1-m_2)g}{M+2(m_1+m_2)}$


Pulley:
$\tau =T_{1}R-T_{2}R$
$I\alpha =(T_1-T_2)R$
$\frac{1}{2}M{R}^2\times \frac{a}{R}=(T_1-T_2)R$
$\frac{Ma}{2}=T_1-T_2$
$m_1:$

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

$m_2:$

$T_2-m_{2}g=m_{2}a.....\left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$
$T_1-T_2=m_{1}g-m_{1}a-m_{2}g-m_{2}a$
$\frac{Ma}{2}+(m_1+m_2)a=(m_1-m_2)g$
$a=\frac{2(m_1-m_2)g}{M+2(m_1+m_2)}$