Question
Two masses 400 g and 350 g are suspended from the ends of a light string passing over a heavy pulley of radius 2 cm . When released from rest the heavier mass is observed to fall 81 cm in 9 s . The rotational inertia of the pulley is $\_\_\_\_$ $kg.m^{2}$.
$$\left( g = 9.8\text{ }m/s^{2} \right) $$
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Solution
$${S = ut + \frac{1}{2}{at}^{2} }{a = \frac{2\text{ }s}{t^{2}} = \frac{2 \times 0.81}{81} = 0.02\text{ }m/s^{2} }{m_{1}\text{ }g - T_{1} = m_{1}a }{T_{2} - m_{2}\text{ }g = m_{2}a }{\left( T_{1} - T_{2} \right)R = I \cdot \frac{a}{R} }{\therefore a = \frac{\left( m_{1} - m_{2} \right)g}{m_{1} + m_{2} + \frac{I}{R^{2}}}}$$
$$\begin{matrix} & 0.02 = \frac{(400 - 350)\left( 10^{- 3} \right)g}{(400 + 350)\left( 10^{- 3} \right) + \frac{I}{R^{2}}} \\ & \frac{I}{R^{2}} = \frac{50 \times 10^{- 3}\text{ }g}{0.02} - 750 \times 10^{- 3} = 23.75 \\ & I = 23.75 \times 4 \times 10^{- 4} = 9.5 \times 10^{- 3}\text{ }kg - m^{2} \end{matrix}$$
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