A particle originally at a rest at the highest point of a smooth circle in a vertical plane, is gently pushed and starts sliding along the circle. It will leave the circle at a vertical distance $h$ below the highest point such that [image:4_image50.jpeg]

Question

Dr. Abubacker · Accepted Answer

(d) From law of conservation of energy, potential energy of fall gets converted to kinetic energy.
[image:4_image138.emf]
$	herefore\ \ PE = KE$
$mgh = \frac{1}{2}{mv}^{2}$
$or\ \ v = \sqrt{2gh}\ \ \ldots.(i)$
Also, the horizontal component of force is equal centrifugal force.
$	herefore\ \ \ mg\cos{	heta = \frac{{mv}^{2}}{R}\ \ldots(ii)}$
From Eq. (i)
$	herefore\ \ \ mg\cos{	heta = \frac{2mgh}{R}\ \ldots(iii)}$
From $\mathrm{\Delta}\ AOB,$
$\ \cos{	heta = \frac{(R - h}{R}\ }$
$\Longrightarrow \ \ mg\left( \frac{(R - h)}{R} ight) = \frac{2mgh}{R}$
$\Longrightarrow \ \ 3h = R$
$\Longrightarrow h = \frac{R}{3}$

Q. A particle originally at a rest at the highest point of a smooth circle in a vertical plane, is gently pushed and starts sliding along the circle. It will leave the circle at a vertical distance $h$ below the highest point such that

Explanation

Reviewed by Dr. Abubacker

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