In the following figure, a body of mass m&nbs | vedclass

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Question

We know the tangential acceleration

$a = \frac{g}{{1 + \frac{I}{{m{R^2}}}}}$$ = \frac{g}{{1 + \frac{{1/2M{R^2}}}{{m{R^2}}}}} = \frac{{2mg}}{{2m + M

Vedclass · Accepted Answer

$\frac{{2mgt}}{{R(M + 2m)}}$

Vedclass · Answer

We know the tangential acceleration

$a = \frac{g}{{1 + \frac{I}{{m{R^2}}}}}$$ = \frac{g}{{1 + \frac{{1/2M{R^2}}}{{m{R^2}}}}} = \frac{{2mg}}{{2m + M}}$   

[As$I = \frac{1}{2}M{R^2}$ for cylinder]

After time $t$, linear velocity of mass $m$,

$v = u + at$ $ = 0 + \frac{{2mgt}}{{2m + M}}$

So angular velocity of the cylinder ,

$\omega  = \frac{v}{R} = $$\frac{{2mgt}}{{R(M + 2\,m)}}$.

In the following figure, a body of mass $m$ is tied at one end of a light string and this string is wrapped around the solid cylinder of mass $M$ and radius $R$. At the moment $t = 0$ the system starts moving. If the friction is negligible, angular velocity at time $t$ would be

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