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Newton's Second Law: Rotatory Version

A circular pu...

Question

A circular pulley of mass $^{'} M^{'}$ and radius $^{'} R^{'}$ , is hinged at the centre. A long string is wound over this disc and two bodies of mass $m_{1}$ and $m_{2}$ are attached at its free ends. Now the bodies are released. Find the acceleration of each body.

\frac{2 [m_{1} - m_{2}] g}{M + (m_{1} + m_{2})}

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\frac{[m_{1} - m_{2}] g}{M + 2 (m_{1} + m_{2})}

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\frac{2 [m_{1} - m_{2}] g}{M + 2 (m_{1} + m_{2})}

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\frac{[m_{1} - m_{2}] g}{M + (m_{1} + m_{2})}

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Solution

The correct option is C $\frac{2 [m_{1} - m_{2}] g}{M + 2 (m_{1} + m_{2})}$
Pulley:
$τ = T_{1} R - T_{2} R$
$I α = (T_{1} - T_{2}) R$
$\frac{1}{2} M R^{2} \times \frac{a}{R} = (T_{1} - T_{2}) R$
$\frac{M a}{2} = T_{1} - T_{2}$
From (i) and (ii)
$T_{1} - T_{2} = m_{1} g - m_{1} a - m_{2} g - m_{2} a$
$\frac{M a}{2} + (m_{1} + m_{2}) a = (m_{1} - m_{2}) g$
$a = \frac{2 (m_{1} - m_{2}) g}{M + 2 (m_{1} + m_{2})}$

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