question_answer

A bob of mass m attached to one end of a light rod hangs vertically. The rod is turned through \[\text{9}0{}^\circ \] so that it becomes horizontal and then released. Calculate the angle between the rod and vertical position at which the tension in the rod is equal to weight of the bob.  

Answer:

                As the rod moves from horizontal position OA to vertical position OB, suppose T= mg at position P,          where Z.BOP = 9, Fig. 4(HT).2.  If \[\upsilon \]is velocity of bob at P, then \[\frac{1}{2}m{{\upsilon }^{2}}=mg\left( L\cos \theta  \right)\]                 \[{{\upsilon }^{2}}=2\,g\,L\,\cos \theta \]                                                      ?.. (i) Net force acting on the bob along \[PO=T-mg\,\cos \theta \] This must provide the necessary centripetal force      \[T-mg\,\cos \theta =\frac{m{{\upsilon }^{2}}}{L}=\frac{m}{L}\left( 2g\,L\,\cos \theta  \right)\]                         ?? using (i) \[=2\,mg\,\cos \theta \] \[T=3mg\,cos\theta \] When \[T=mg\]; \[mg=3mg\,\cos \theta \] \[\cos \theta =\frac{1}{3}=0.3333\]    \[\therefore \]\[\theta ={{\cos }^{-1}}\left( 0.3333 \right)={{70}^{o}}30'\]