question_answer

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? \[\operatorname{h}=R\left[ 1-cos\theta  \right]\] \[R=\left[ 1-\frac{2}{3} \right]\] \[\operatorname{h}=\frac{R}{3}\]