Consider two masses with m_1 > m_2 connected by a light inextensi… - Vidyadip

MCQ

Consider two masses with $m_1 > m_2$ connected by a light inextensible string that passes over a pulley of radius $R$ and moment of inertia $I$ about its axis of rotation. The string does not slip on the pulley and the pulley turns without friction. The two masses are released from rest separated by a vertical distance $2 h$. When the two masses pass each other, the speed of the masses is proportional to

✓
$\sqrt{\frac{m_1-m_2}{m_1+m_2+\frac{I}{R^2}}}$
B
$\sqrt{\frac{\left(m_1+m_2\right)\left(m_1-m_2\right)}{m_1+m_2+\frac{1}{R^2}}}$
C
$\sqrt{\frac{m_1+m_2+\frac{I}{R^2}}{m_1-m_2}}$
D
$\sqrt{\frac{1}{R^2}}$

✓

Answer

Correct option: A.

$\sqrt{\frac{m_1-m_2}{m_1+m_2+\frac{I}{R^2}}}$

a
(a) Loss of potential energy of $m_1$ appears as kinetic energies of $m_1, m_2$ and pulley and also as potential energies of $m_1$ and $m_2$. So, by energy conservation, we get

$m_1 g h=\frac{1}{2} m_1 v^2+\frac{1}{2} m_2 v^2+\frac{1}{2} I \omega^2+m_2 g h$

Here, $m_1$ falls by distance $h, m_2$ rises by distance $h , v=$ speed of $m_1=$ speed of $m_2$ as they passes each other and $\omega=$ angular speed of pulley $=\frac{v}{R}$.

$\text { So, }\left(m_1-m_2\right) g h=\frac{v^2}{2}\left(m_1+m_2+\frac{I}{R^2}\right)$

$\therefore \quad v=\frac{2 g h\left(m_1-m_2\right)}{\left(m_1+m_2+\frac{I}{R^2}\right)}$

$\text { or } \quad v \propto \sqrt{\left(\frac{m_1-m_2}{m_1+m_2}+\frac{I}{R^2}\right)}$

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