Question

Pulley of atwood machine has mass $m_1$ and is free to rotate about its geometrical center without friction. Centre of mass of the pulley is at a distance $R/2$ from geometrical centre, where $R$ is the radius of the pulley. There is no slipping between the string and pulley. The ratio of maximum acceleration to minimum acceleration of block of mass $m_1$ is (Take $m_1{R}^2$ as moment of inertia of pulley about axis of rotation, $\frac{m_1}{m_2}=3$ )

Answer 1

Answer: 7

$m_{1}g-T_1=m_{1}a$ .......(i)
$T_2-m_{2}g=m_{2}a$ ....(ii)
$T_1-T_2\pm \frac{m_{1}g}{2}=m_{1}a$
$\left(m_1-m_2\right)g\pm \frac{m_{1}g}{2}=\left(2m_1+m_2\right)a$
$a=\frac{2m_{2}g\pm \frac{3g{m}_2}{2}}{7m_2}$
$a_{max}=\frac{g}{2}$
$a_{min}=\frac{g}{14}$
$\frac{a_{max}}{a_{min}}=7$