Dynamics of rotational motion about fixed axis

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Question

A block of mass m is attached to the end of an inextensible string which is wound over a rough pulley p of mass m and radius R. Assuming no slipping between the string and the pulley, find the acceleration of the block when released

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Solution

For the block, mg-T=ma ……..(1)
For the pulley,
$T . R = (\frac{1}{2} {mR}^{2}) . α \Rightarrow T . R = (\frac{1}{2} {mR}^{2}) . (\frac{a}{R}) T = \frac{1}{2} ma . . . . . . . (2)$
$Solving, a = \frac{2 g}{3}$

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Dynamics of rotational motion about fixed axis

Remember concepts with our Masterclasses.

A block of mass m is attached to the end of an inextensible string which is wound over a rough pulley p of mass m and radius R. Assuming no slipping between the string and the pulley, find the acceleration of the block when released

For the block, mg-T=ma ……..(1)For the pulley,T.R=12mR2.α ⇒T.R=12mR2.aR T=12ma.......(2)Solving, a=2g3

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For the block, mg-T=ma ……..(1)
For the pulley,
$T . R = (\frac{1}{2} {mR}^{2}) . α \Rightarrow T . R = (\frac{1}{2} {mR}^{2}) . (\frac{a}{R}) T = \frac{1}{2} ma . . . . . . . (2)$
$Solving, a = \frac{2 g}{3}$