A small block slides down from rest at point | vedclass

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(C) Let $R$ be the radius of the cylinder. The vertical height fallen by the block from point $A$ to point $B$ is $h = R \cos 	heta_1 - R \cos 	heta

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $R$ be the radius of the cylinder. The vertical height fallen by the block from point $A$ to point $B$ is $h = R \cos 	heta_1 - R \cos 	heta_2 = R(\cos 	heta_1 - \cos 	heta_2)$.By the law of conservation of energy,the kinetic energy gained is equal to the potential energy lost: $\frac{1}{2}mv^2 = mgh = mgR(\cos 	heta_1 - \cos 	heta_2)$.At point $B$,the normal force $N$ becomes zero as the block leaves the surface. The radial equation of motion is $mg \cos 	heta_2 - N = \frac{mv^2}{R}$.Setting $N = 0$,we get $v^2 = Rg \cos 	heta_2$.Substituting this into the energy equation: $\frac{1}{2}m(Rg \cos 	heta_2) = mgR(\cos 	heta_1 - \cos 	heta_2)$.Dividing by $mgR$: $\frac{1}{2} \cos 	heta_2 = \cos 	heta_1 - \cos 	heta_2$.Rearranging gives: $\frac{3}{2} \cos 	heta_2 = \cos 	heta_1$,or $\cos 	heta_2 = \frac{2}{3} \cos 	heta_1$.

$A$ 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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