A small block slides down from rest at point | vedclass

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Question

$\mathrm{h}=\mathrm{R}\left(\cos 	heta_{1}-\cos 	heta_{2}ight)$

$\frac{1}{2} m v^{2}=m g h$

$\frac{\mathrm{mv}^{2}}{\mathrm{R}}=\mathrm{mg}

Vedclass · Accepted Answer

$cos\ 	heta_2 = \frac{2}{3}cos\ 	heta_1$

Vedclass · Answer

$\mathrm{h}=\mathrm{R}\left(\cos 	heta_{1}-\cos 	heta_{2}ight)$

$\frac{1}{2} m v^{2}=m g h$

$\frac{\mathrm{mv}^{2}}{\mathrm{R}}=\mathrm{mg} \cos 	heta_{2} \Rightarrow \mathrm{v}=\sqrt{\mathrm{Rg} \cos 	heta_{2}}$

$\frac{1}{2} \mathrm{m} \mathrm{Rg} \cos 	heta_{2}=\mathrm{mgR}\left(\cos 	heta_{1}-\cos 	heta_{2}ight)$

$3 \cos 	heta_{2}=2 \cos 	heta_{1}$

A small block slides down from rest at point $A$ on the surface of a smooth cylinder, as shown. At point $B$, the block falls off (leaves) the cylinder. The equation relating the angles $\theta_1$ and $\theta_2$ is given by

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