question_answer

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:

A) \[h=R\]             

B) \[h=\frac{R}{2}\]

C) \[h=\frac{R}{3}\,\]                    

D) \[h=\,2R\]

Correct Answer: C

Solution :

[c]    At p let the velocity gained by v \[{{v}^{2}}=2gh\]    \[[where\,\frac{R-h}{R}=cos\theta ,h=R[1-cos\theta ]\] \[mgCos\theta -N=\frac{m{{v}^{2}}}{R}\] It leaves when N = 0 \[mg\frac{[R-h]}{R}-0=\frac{m}{R}2gh\left[ Cos\theta =\frac{R-h}{R} \right]\] \[\frac{R-h}{R}=\frac{2R}{R}\,\,[1-cos\theta ]\left[ h=[1-\cos \theta ] \right]\] \[cos\theta =2\,[cos\theta ]\] \[cos\theta =2-2\text{ }cos\theta \] \[3cos\,\theta =\frac{2}{3}\] use this in             \[h=R\text{  }\!\![\!\!\text{ }1-Cos\theta ]\]             \[=F\left[ 1-\frac{2}{3} \right]\] \[h=\frac{R}{3}\]