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(B) Let $R$ be the radius of the circle. The particle starts from rest at the highest point.By the law of conservation of energy,the kinetic energy at

Vedclass · Accepted Answer

$h = R/3$

Vedclass · Answer

(B) Let $R$ be the radius of the circle. The particle starts from rest at the highest point.By the law of conservation of energy,the kinetic energy at a vertical distance $h$ below the highest point is equal to the loss in potential energy:$\frac{1}{2} m v^2 = m g h \Rightarrow v^2 = 2 g h$.From the geometry of the circle,the vertical distance $h$ from the top is $h = R - R \cos 	heta$,where $	heta$ is the angle with the vertical.Thus,$v^2 = 2 g R(1 - \cos 	heta)$.The forces acting on the particle are gravity $(m g)$ and the normal reaction $(N)$. The radial equation of motion is:$m g \cos 	heta - N = \frac{m v^2}{R}$.When the particle leaves the circle,the normal reaction $N$ becomes zero:$m g \cos 	heta = \frac{m v^2}{R} \Rightarrow g \cos 	heta = \frac{2 g R(1 - \cos 	heta)}{R}$.$\cos 	heta = 2(1 - \cos 	heta) \Rightarrow \cos 	heta = 2 - 2 \cos 	heta \Rightarrow 3 \cos 	heta = 2 \Rightarrow \cos 	heta = \frac{2}{3}$.Substituting this into the expression for $h$:$h = R - R \cos 	heta = R - R(\frac{2}{3}) = \frac{R}{3}$.

$A$ particle originally at rest at the highest point of a smooth vertical circle is slightly displaced. It will leave the circle at a vertical distance $h$ below the highest point,such that

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