Consider two masses with m_1 m_2 connected | vedclass

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(A) The loss in potential energy of mass $m_1$ is converted into the kinetic energy of $m_1$,$m_2$,and the rotational kinetic energy of the pulley,plu

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The loss in potential energy of mass $m_1$ is converted into the kinetic energy of $m_1$,$m_2$,and the rotational kinetic energy of the pulley,plus the gain in potential energy of $m_2$.By the principle of conservation of energy:$m_1 g h = m_2 g h + \frac{1}{2} m_1 v^2 + \frac{1}{2} m_2 v^2 + \frac{1}{2} I \omega^2$Here,$m_1$ falls by distance $h$,$m_2$ rises by distance $h$,$v$ is the speed of the masses when they pass each other,and $\omega = \frac{v}{R}$ is the angular speed of the pulley.Substituting $\omega = \frac{v}{R}$ into the equation:$(m_1 - m_2) g h = \frac{1}{2} (m_1 + m_2) v^2 + \frac{1}{2} I \left(\frac{v}{R}\right)^2$$(m_1 - m_2) g h = \frac{1}{2} v^2 \left(m_1 + m_2 + \frac{I}{R^2}\right)$Solving for $v$:$v^2 = \frac{2 g h (m_1 - m_2)}{m_1 + m_2 + \frac{I}{R^2}}$$v = \sqrt{\frac{2 g h (m_1 - m_2)}{m_1 + m_2 + \frac{I}{R^2}}}$Thus,the speed $v$ is proportional to $\sqrt{\frac{m_1 - m_2}{m_1 + m_2 + \frac{I}{R^2}}}$.

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