Two particles `A` and `B` of mass `m` each hit the ends of a rigid bar of mass `M` and length `l` in-elastically from opposite sides the speeds `v` and perpendicular to the rod. The bar is kept on a smooth horizontal plane (as shown). Find the linear and angular speed of the system (bar`+`particle).
Text Solution
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Conservation of linear momentum. There is not external force acting on the system along `x`-axis. There fore the linear momentum of the system before and after the system remains constant. The momentum of the system before impact. `P_(i)=mv-mv=0` The linear momentum of the system after the impact must be zero. therefore, the centre of mass of the system does not translate. `implies v_(cm)=0` Conservation of angular momentum Since the impact forces are internal force, the net torque about a pperpendicular axis passing through `CM` of the system is zero. Thefore the angular momentum of the system before and after the impact will be equal. `implies L_("initial")=L_("final") where L_("initial")|(ml)/2v+(ml)/2v|=mvl` Let the system rotate about is `CMO` with an angular speed `omega` `implies L_("final")=(I_("system"))omega` where `I_("system")=MI` of the system about `0=(I_("rod"))_(0)+(I_(A))+(I_(B))_(0)` `implies I_("system")=(Ml^(2))/12+m(l/2)^(2)+m(l/2)^(2)=((M+6m)/12)l^(2)` Using the above equations we get `F_(1)=1/2mralpha=1/3masintheta=mvlimpliesomega=(12mv)/((M+6m)l)`
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