There is a rod of lenght l, mass m lying on&n | vedclass

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Question

$\mathrm{m}_{1} \mathrm{g}-\mathrm{T}=\mathrm{m}_{1} \mathrm{a}$

$\mathrm{T}=\mathrm{m} \mathrm{a}_{\mathrm{cm}}$

$\mathrm{T}(\ell / 2)=\frac{\m

Vedclass · Accepted Answer

$4$

Vedclass · Answer

$\mathrm{m}_{1} \mathrm{g}-\mathrm{T}=\mathrm{m}_{1} \mathrm{a}$

$\mathrm{T}=\mathrm{m} \mathrm{a}_{\mathrm{cm}}$

$\mathrm{T}(\ell / 2)=\frac{\mathrm{mL}^{2}}{12} \alpha$

$\mathrm{T}=\mathrm{m} \frac{\mathrm{L}}{6} \alpha ; \quad \mathrm{a}_{\mathrm{cm}}=\frac{\mathrm{L}}{6} \mathrm{\alpha}$

$a=a_{e m}+\frac{L}{2} \alpha$

$=a_{\infty}+3 a_{\mathrm{cm}}=4 \mathrm{a}_{\mathrm{cm}}$

$\mathrm{m}_{1} \mathrm{g}=\mathrm{m}_{1} \mathrm{a}+\mathrm{ma}_{\mathrm{em}}$

$\mathrm{m}_{1} \mathrm{g}=4 \mathrm{m}_{1} \mathrm{a}_{\mathrm{cm}}+\mathrm{cm}_{\mathrm{cm}}$

$\frac{m_{1}}{4 m_{1}+m} g=a_{c m}$

$\frac{1}{4+\frac{m}{m_{1}}} g=a_{c m}$ $=g/4$

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