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Pulley Problem

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 it's free ends. Now the bodies are released. Find the accceleration 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:
$τ = T_{1} R - T_{2} R$
$I α = (T_{1} - T_{2}) R$
$\frac{1}{2} M R^{2} \times \frac{a}{R} = (T_{1} - T_{2}) R$
$\frac{M a}{2} = T_{1} - T_{2}$
$m_{1} :$

$m_{1} g - T_{1} = m_{1} a$
$m_{1} g - m_{1} a = T_{1} . . . (i)$

$m_{2} :$

$T_{2} - m_{2} g = m_{2} a . . . . . (i i)$
From $(i)$ and $(i i)$
$T_{1} - T_{2} = m_{1} g - m_{1} a - m_{2} g - m_{2} a$
$\frac{M a}{2} + (m_{1} + m_{2}) a = (m_{1} - m_{2}) g$
$a = \frac{2 (m_{1} - m_{2}) g}{M + 2 (m_{1} + m_{2})}$

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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 m1 and m2 are attached at it's free ends. Now the bodies are released. Find the accceleration of each body.

The correct option is C 2(m1−m2)gM+2(m1+m2) Pulley: τ=T1R−T2R Iα=(T1−T2)R 12MR2×aR=(T1−T2)R Ma2=T1−T2 m1: m1g−T1=m1a m1g−m1a=T1 ...(i) m2: T2−m2g=m2a .....(ii) From (i) and (ii) T1−T2=m1g−m1a−m2g−m2a Ma2+(m1+m2)a=(m1−m2)g a=2(m1−m2)gM+2(m1+m2)

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 it's free ends. Now the bodies are released. Find the accceleration of each body.