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, \[\sin \theta =\frac{h}{r}=\frac{2}{3}\] \[\Rightarrow \] \[h=\frac{2}{3}r\]You need to login to perform this action.
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