A particle originally at rest at the highest point of 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)\[\operatorname{h}=\frac{R}{2}\]
C)h = R
D)\[\operatorname{h}=\frac{R}{3}\]
Correct Answer:
D
Solution :
Apply law of conversation of energy \[{{\operatorname{K}}_{A}}+{{P}_{A}}={{K}_{P}}+{{P}_{P}}\] \[\operatorname{O}+mgh=\frac{1}{2}n{{v}^{2}}_{p}+O\,\,\,\,\,\,[assume\,\,PE=0\,\,at\,\,P]\] \[{{\operatorname{V}}^{2}}_{p}=2gh\] \[{{\operatorname{V}}^{2}}_{p}=2g\,\,\operatorname{R}\left[ 1-\cos \theta \right]\] ?..(i) \[\operatorname{mg}\,\,\cos \theta -\operatorname{N}=\frac{m{{V}^{2}}_{p}}{R}\] Body to lose contact N=0 \[\cos \theta =\frac{m}{R}\times 2gh\left[ 1-cos\theta \right]\] \[\cos \theta =2-2cos\theta \] \[\cos \theta =\frac{2}{3}\] We have to find valve of ?h? \[\operatorname{h}=R\left[ 1-cos\theta \right]\] \[R=\left[ 1-\frac{2}{3} \right]\] \[\operatorname{h}=\frac{R}{3}\]