A thin uniform rod of length 0.754 m and mass M is rotating horizontally at angu

Question

A thin uniform rod of length 0.754 m and mass M is rotating horizontally at angular speed 14.7 rad/s about an axis through its center. A particle of mass M/3.00 initially attached to one end is ejected from the rod and travels along a path that is perpendicular to the rod at the instant of ejection. If the particle's speed vp is 5.59 m/s greater than the speed of the rod end just after ejection, what is the value of vp? I'm getting stuck somewhere but I'm not sure why my calculation is off. Thanks!

Explanation / Answer

By conservation of angular momentum. the total angular momentum after the ejection must be equal to before the ejection:
Lp'+Lr' =Lp+Lr
(L/2)mVp +(1/12)ML2' =IP+(1/12)ML2
where
Ip=m(L/2)2
' =(Vp-5.59)/(L/2)
(L/2)mVp +(1/12)ML2(Vp-5.59)/(L/2) =m(L/2)2+(1/12)ML2
0.377*(M/3)Vp +(1/12)M*0.7542[(Vp-5.59)/0.377]=(M/3)0.3772+(1/12)M*0.7542*14.7
(0.377/3)Vp +(0.57/4.524)Vp -3.178/4.524=0.14/3+(8.36/12)
0.251Vp =1.446
VP =5.75m/s

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