question_answer

A rigid uniform rod of mass M and length L is resting on a smooth horizontal table, Two marbles each of mass m and travelling with uniform speed v collide with two ends of the rod simultaneously and inelastically as shown. The marbles get struck to the rod after the collision and continue to move with the rod. If \[m=\frac{M}{6}\] and v = L ms-1, then the time taken by the rod to rotate through \[\frac{\pi }{2}\] is

A)                  1 s                                         

B)  2\[\pi \] s

C)  \[\pi \] s      

D)  \[\frac{\pi }{2}\] s

Correct Answer: C

Solution :

           Since, mass of rigid uniform rod = M kg Mass of particle (marbles) \[=m=\frac{M}{6}kg\] (given) Speed of marbles \[v=L\,m{{s}^{-1}}\] Both marbles give equal momentum in opposite direction to the rod. After striking rod does not move straight but rotate about point C. Since, no external torque acting on rod, therefore applying conservation of angular momentum. Angular momentum \[{{J}_{i}}=mv.\frac{L}{2}+mv.\frac{L}{2}\]                                 \[=\frac{M}{6}.\frac{{{L}^{2}}}{2}+\frac{M}{6}.\frac{{{L}^{2}}}{2}\]                                 \[=\left( \frac{2M{{L}^{2}}}{12} \right)kg.{{m}^{2}}{{s}^{-1}}\] Since, collision is inelastic. Angular momentum after striking \[{{J}_{f}}=\left( \frac{M{{L}^{2}}}{2}+\frac{m{{L}^{2}}}{4}+\frac{m{{L}^{2}}}{4} \right)\omega \] Where \[\omega \]is angular velocity of rod. \[{{J}_{f}}=\left( \frac{M{{L}^{2}}}{12}+\frac{M}{6}.\frac{{{L}^{2}}}{4}+\frac{M}{6}.\frac{{{L}^{2}}}{4} \right)\omega \] \[=\left( \frac{M{{L}^{2}}}{12}+\frac{2M.{{L}^{2}}}{24} \right)\omega \] \[{{J}_{f}}=\left( \frac{2M{{L}^{2}}}{12} \right)\omega w\]                 By law of conservation of momentum \[{{J}_{f}}={{J}_{i}}\] \[\frac{2M{{L}^{2}}}{12}\omega =\frac{2M{{L}^{2}}}{12}\]                 By initial angular velocity \[{{\omega }_{0}}=0\,rad\,{{s}^{-1}}\]                 \[\therefore \]Angular acceleration after time t                 \[\alpha =\left( \frac{\omega -{{\omega }_{0}}}{t} \right)=\frac{1-0}{t}=\left( \frac{1}{t} \right)\] \[\therefore \]From IInd equation of angular motion \[\theta ={{\omega }_{0}}t+\frac{1}{2}\alpha {{t}^{2}}\] \[\therefore \]                  \[\frac{\pi }{2}=0+\frac{1}{2}\times \frac{1}{t}.{{t}^{2}}\] or            \[\frac{\pi }{2}=\frac{1}{2}.t\]     \[\Rightarrow \]\[t=\pi s\]