A block of mass m is attached to a pulley dis | vedclass

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Question

For block,

$-\mathrm{Tdt}=\mathrm{mv}-\mathrm{mv}_{0}$

For disc, $\mathrm{rTdt}=\mathrm{I} \omega$

or $\quad \mathrm{Tdt}=\frac{\mathrm{I} \o

Vedclass · Accepted Answer

$10/3\, m/s$

Vedclass · Answer

For block,

$-\mathrm{Tdt}=\mathrm{mv}-\mathrm{mv}_{0}$

For disc, $\mathrm{rTdt}=\mathrm{I} \omega$

or $\quad \mathrm{Tdt}=\frac{\mathrm{I} \omega}{\mathrm{r}}=\frac{\mathrm{Iv}}{\mathrm{r}^{2}}$

or $\quad \mathrm{mv}_{0}-\mathrm{mv}=\frac{\mathrm{mv}}{2}$

or $\quad \mathrm{v}_{0}=\mathrm{v}+\frac{\mathrm{v}}{2}=\frac{3}{2} \mathrm{v}$

$	herefore \mathrm{v}=\frac{2}{3} \mathrm{v}_{0}=\frac{2}{3} 	imes 5=\frac{10}{3} \mathrm{m} / \mathrm{s}$

$A$ block of mass $m$ is attached to a pulley disc of equal mass $m$, radius $r$ by means of a slack string as shown. The pulley is hinged about its centre on a horizontal table and the block is projected with an initial velocity of $5\, m/s$. Its velocity when the string becomes taut will be

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