question_answer

The height at which body leaves the vertical circle is:

A) \[\frac{2}{3}r\] 

B) \[\frac{3}{2}r\]

C)  \[3r\]                                   

D) r

Correct Answer: A

Solution :

                 When body leaves the vertical circle its reaction is zero. The various forces acting on the body are as shown. Let body leave the path at point P, then                 \[R=0=mg\sin \theta -\frac{m{{v}^{2}}}{r}\] where\[\frac{m{{v}^{2}}}{r}\]is the centripetal force. \[\Rightarrow \]               \[{{v}^{2}}=rg\sin \theta \]                          ?.. (i) Also, from law of conservation of energy Potential energy of fall from = Kinetic energy Q to P. \[\therefore \]  \[mgr-mgr\sin \theta =\frac{1}{2}m{{V}^{2}}\]   ?. (ii) From Eqs. (i) and (ii), we get                 \[2-2\sin \theta =\sin \theta \] \[\Rightarrow \]               \[\sin \theta =\frac{2}{3}\] Also,      \[\sin \theta =\frac{h}{r}=\frac{2}{3}\] \[\Rightarrow \]               \[h=\frac{2}{3}r\]