First slide

Dynamics of rotational motion about fixed axis

Question

Consider two objects with $m_{1} > m_{2}$ connected by a light string that passes over a pulley having a moment of inertia of I about its axis of rotation as shown in figure. The string does not slip on the pulley or stretch. The pulley tums without friction. The two objects are released from rest separated by a vertical distance 2h.The translational speeds of the objects as they pass each other is

Moderate

A

$\sqrt{\frac{2 (m_{1} + m_{2}) gh}{m_{1} + m_{2} + \frac{I}{R^{2}}}}$
B

$\sqrt{\frac{2 (m_{1} - m_{2}) gh}{m_{1} + m_{2} + \frac{I}{R^{2}}}}$
C

$\sqrt{\frac{(m_{1} - m_{2}) gh}{m_{1} + m_{2} + \frac{I}{R^{2}}}}$
D

$\sqrt{\frac{(m_{1} + m_{2}) gh}{m_{1} + m_{2} + \frac{I}{R^{2}}}}$

Solution

Take the two objects, pulley, and Earth as the system. If we neglect friction in the system, then mechanical energy is conserved and we can state that the increase in kinetic energy of the system equals the decrease in potential energy. Since $K_{i} = 0$ (the system is initially at rest), we have
$∆ K = K_{f} - K_{i}$
= $\frac{1}{2} m_{1} v^{2} + \frac{1}{2} m_{2} v^{2} + \frac{1}{2} {Iω}^{2}$
where $m_{1} and m_{2}$ have a common speed. But
$v = Rω so that ∆ K = \frac{1}{2} (m_{1} + m_{2} + \frac{I}{R^{2}}) V^{2}$
From FIG. we see that the system loses potential energy because of the motion of $m_{1}$ and gains potential energy because of the motion of $m_{2}$ . Applying the law of conservation of energy,
$∆ K + ∆ U = 0, gives$
$\frac{1}{2} (m_{1} + m_{2} + \frac{I}{R^{2}}) v^{2} + m_{2_{}} gh - m_{1} gh = 0$
$v = \sqrt{\frac{2 (m_{1} - m_{2} gh)}{m_{1} + m_{2} + \frac{I}{R^{2}}}}$

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