Figure 11-53 is an overhead view of a thin uniform rod of length0.803 m and mass

Question

Figure 11-53 is an overhead view of a thin uniform rod of length0.803 m and mass M rotating horizontally at angular speed 21.8rad/s about an axis through its center. A particle of mass M/3.00initially attached to one end is ejected from the rod and travelsalong a path that is perpendicular to the rod at the instant ofejection. If the particle's speed vp is 4.83m/s greater than the speed of the rod end just after ejection, whatis the value of vp?

I cant understand the submitted solution to this problem. Couldsomebody help clarify how to work this? Thanks.

Explanation / Answer

               Given thatlength of rod is L = 0.803 m                massis M                massof particle is M/3               Initial angular velocity of the system is = 21.8rad/s                        -----------------------------------------------------------------------                      Letthe final velocity of the rod is V then from given dataVp = V + 4.83 m/s                                                                                                                     Since thereis no external force on the system then the angualr momentumis conserved                               Initial total angular momentum  = final angularmomentum                                                                     Ii* = (M/3)Vp*L/2 + I1*1                                                               ( Irod +Imass) = (M/3)*Vp*L/2 +Irod*1                                                                    (Irod + (M/3)(L/2)2 ) = (M/3)Vp*L/2 +Irod*1               where 1 = V / (L/2) = [Vp - 4.83m/s ]/ (L/2)                                                        Themoment of inertia of the rod about the center ofthe axis is Irod = ML2 /12        slove for Vp =--------m/s                                                                                                                                                             

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