Question

A particle originally at 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

• A. $h=2R$
• B. $h=\frac{R}{2}$
• C. $h=R$
• D. $h=\frac{R}{3}$

Answer 1

Answer: (D)

$h=\frac{R}{3}$

Answer 2

From law of conservation of energy, potential energy of fall gets converted to kinetic energy.
$PE = KE$
$mgh =\frac{1}{2} m v^{2}$
$v =\sqrt{2 g h}\,\,\,$ ...(i)
Also, the horizontal component of force is equal centrifugal force.
$\therefore m g \cos \theta=\frac{m v^{2}}{R} \,\,\,$.,.(ii)
From E (i),
$v=\sqrt{2 g h}$
$\therefore m g \cos \theta=\frac{2 m g h}{R} \,\,\,\,$...(iii)
From $\triangle A O B$
$\cos \theta=\frac{2 R-h}{R}$
$\Rightarrow m g\left(\frac{R-h}{R}\right)=\frac{2 m g h}{R}$
$\Rightarrow 3 h=R$
$\Rightarrow h=\frac{R}{3}$