A uniform cylinder of radius ;r' and mass 'm' can rotate freely about a fixed horizontal axis. A thin cord of length 'l' and mass 'm_{0} is would on the cylinder in a single layer. Find the angular acceleration of the cylinder as a function of the length of x of the hanging part of the cord. The would part of the cord is supposed to have a centre of gravity on the cylinder axis as shown in fig

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Solution

As shown in the fig the free-body diagram of the two bodies, viz., cylinder and overhanging portion of the cord. If $α$ is the angular accleration of the cylinder, then from $τ$ = $I α$
we have $T r$ = $[\frac{m r^{2}}{2} + \frac{m_{0}}{l} (l - x) r^{2}] α$
$T$ = $[\frac{m}{2} + m_{0} - \frac{m_{0} x}{l}] r α$ (i)
If $α$ be the acceleration of the hanging part, then from $F = m a$ ,
$\frac{m_{0} x g}{l} - T$ = $\frac{m_{0} x}{l} a$ = $\frac{m_{0} x}{l} (α r)$ (ii)
[Linear acceleration of the rim of the cylinder is same as that of the cord]
Adding Eq. (i) and (ii), we get
$\frac{m_{(} 0) x}{l} g$ = $α [\frac{m}{2} + m_{(} 0)] r$ $\Rightarrow$ $α$ $\frac{m_{(} 0) x g}{r l [\frac{m}{2} + m_{(} 0)]}$

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