question_answer

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)  \[h=\frac{R}{3}\]

Correct Answer: D

Solution :

From law of conservation of energy, potential energy of fall gets converted to kinetic energy. \[\therefore \] \[PE=KE\] \[mgh=\frac{1}{2}m{{v}^{2}}\] \[\Rightarrow \] \[v=\sqrt{2gh}\] ?..(i) Also, the horizontal component of force is equal centrifugal force. \[\therefore \] \[mg\,\,\cos \theta =\frac{m{{v}^{2}}}{R}\] ?..(ii) From Eq. (i), \[v=\sqrt{2gh}\] \[\therefore \] \[mg\,\,\cos \theta =\frac{2mgh}{R}\] ?..(iii) From \[\Delta AOB,\] \[\cos \theta =\frac{2R-h}{R}\] \[\Rightarrow \] \[mg\left( \frac{R-h}{R} \right)=\frac{2mgh}{R}\] \[\Rightarrow \] \[mg\left( \frac{R-h}{R} \right)=\frac{2mgh}{R}\] \[\Rightarrow \] \[3h=R\] \[\Rightarrow \] \[h=\frac{R}{3}\]