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 \[\Delta \,AOB,\]                 \[\cos \theta =\frac{2R-h}{R}\] \[\Rightarrow \]               \[Mg\left( \frac{R-h}{R} \right)=\frac{2mgh}{R}\] \[\Rightarrow \]               \[Mg\left( \frac{R-h}{R} \right)=\frac{2mgh}{R}\] \[\Rightarrow \]               \[3h=R\] \[\Rightarrow \]               \[h=\frac{R}{3}\]