A uniform cylinder of mass m can rotate freel | vedclass

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Question

$\mathrm{m}_{0} \mathrm{g}-\mathrm{T}=\mathrm{m}_{0} \mathrm{a}$

$\mathrm{Tr}=\mathrm{I} \alpha$

or $\operatorname{Tr}=\frac{\mathrm{mr}^{2}}{2}

Vedclass · Accepted Answer

$\frac{{2{m_o}g}}{{m + 2{m_o}}}$

Vedclass · Answer

$\mathrm{m}_{0} \mathrm{g}-\mathrm{T}=\mathrm{m}_{0} \mathrm{a}$

$\mathrm{Tr}=\mathrm{I} \alpha$

or $\operatorname{Tr}=\frac{\mathrm{mr}^{2}}{2} \alpha$

or $T=\frac{m r \alpha}{2}=\frac{m a}{2}$

or $\mathrm{m}_{0} \mathrm{g}-\frac{\mathrm{ma}}{2}=\mathrm{m}_{0} \mathrm{a}$

or $\mathrm{m}_{0} \mathrm{g}=\left(\mathrm{m}_{0}+\frac{\mathrm{m}}{2}ight) \mathrm{a}$

$	herefore \quad \mathrm{a}=\frac{\mathrm{m}_{0} \mathrm{g}}{\mathrm{m}_{0}+\frac{\mathrm{m}}{2}}=\frac{2 \mathrm{m}_{0} \mathrm{g}}{\mathrm{m}+2 \mathrm{m}_{0}}$

$A$ uniform cylinder of mass $m$ can rotate freely about its own axis which is horizontal.$A$ particle of mass mo hangs from the end of a light string wound round the cylinder which does not slip over it. When the system is allowed to move, the acceleration of the descending mass will be

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