In the given figure linear acceleration of so | vedclass

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Question

$\mathrm{m}_{2} \mathrm{g}-\mathrm{T}=\mathrm{m}_{2} \mathrm{a}_{2} \ldots .(1)$

$\mathrm{TR}=\frac{\mathrm{m}_{2} \mathrm{R}^{2}}{2} \alpha_{2} \l

Vedclass · Accepted Answer

$\frac{{2({a_2} + g)}}{R}$

Vedclass · Answer

$\mathrm{m}_{2} \mathrm{g}-\mathrm{T}=\mathrm{m}_{2} \mathrm{a}_{2} \ldots .(1)$

$\mathrm{TR}=\frac{\mathrm{m}_{2} \mathrm{R}^{2}}{2} \alpha_{2} \ldots \ldots(2)$

$\alpha_{1} \mathrm{R}=\mathrm{a}_{2}-\alpha_{2} \mathrm{R} \ldots .(3)$

$\mathrm{TR}=\left(\frac{\mathrm{m}_{1} \mathrm{R}^{2}}{2}\right) \alpha_{1} \ldots .(4)$

$\alpha_{2}=\frac{2 \mathrm{T}}{\mathrm{m}_{2} \mathrm{R}}=\frac{2}{\mathrm{m}_{2} \mathrm{R}}\left(\mathrm{m}_{2} \mathrm{a}_{2}+\mathrm{m}_{2} \mathrm{g}\right)=\frac{2\left(\mathrm{a}_{2}+\mathrm{g}\right)}{\mathrm{R}}$

In the given figure linear acceleration of solid cylinder of mass $m_2$ is $a_2$. Then angular acceleration $\alpha_2$ is (given that there is no slipping).

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