Consider two masses with m_1 m_2 connect | vedclass

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Question

(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 e

Vedclass · Accepted Answer

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

Vedclass · Answer

(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 $=\frac{v}{R}$.

$	ext { So, }\left(m_1-m_2ight) g h=\frac{v^2}{2}\left(m_1+m_2+\frac{I}{R^2}ight)$

$	herefore \quad v=\frac{2 g h\left(m_1-m_2ight)}{\left(m_1+m_2+\frac{I}{R^2}ight)}$

$	ext { or } \quad v \propto \sqrt{\left(\frac{m_1-m_2}{m_1+m_2}+\frac{I}{R^2}ight)}$

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