Figure 11-53 is an overhead view of a thin uniform rod of length0.803 m and mass
ID: 1745416 • Letter: F
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/sRelated Questions
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