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A ball of radius R=20 cm has mass m=0.75 kg and moment of inertia (about its diameter ) I=0.0125 kgm^(2). The ball rolls without sliding over a rough horizontal floor wilth velocity v_(0)=10 ms^(-1) towards a smooth vertical wall if coefficient of resutitution between the wall and the ball is e=0.7, calculate velocity v of the ball long after the collision. (g=10ms^(-2)) |
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Answer» Solution :Since the ball is rolling without sliding, therefore, its angular velocity `(omega_(0))` just before collision with WALL is, `(v_(0))/R=10/0.2=50 rads^(-1)` Translation velocity of centre of ball, just after collision. `ev_(0)=0.7xx10=7ms^(-1)`(Right wards) Since the wall is smooth, therefore no TANGENTIAL force is applied by the wall. Hence angular velocity `w_(0)` of ball remains UNCHANGED during collision. Now surface of ball slides on flor to the right as shown in the figure. Let coefficient of friction between ball and the floor be m. Considering free body diagram of ball while sliding. for verticla forces `N=mg` For horizontal forces `muN=ma` Taking moment (about `O`) of forces acting on the ball, `muNR=Ialpha` or `alpha=120 MU`  Long after the collision there will be no sliding or it wil be pure rolling Let sliding stop after a time `t` after collision, then final translational velocity `v=(7-at)` or `v=7-10mut` and final angular velocity. `omega=(-omega_(0))+alphat` `omega=(120mut-50)rads^(-1)` (clockwise) But at that instant `v=Romega` `(7-10mut)=0.2(120mut-50)` `mut=0.5` `v=2ms^(-1)` |
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