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 \[h=R\text{ }\!\![\!\!\text{ }1-Cos\theta ]\] \[=F\left[ 1-\frac{2}{3} \right]\] \[h=\frac{R}{3}\]