How Does Ejecting a Particle Affect the Angular Momentum of a Rotating Rod?

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The discussion centers on the effect of ejecting a particle from a rotating rod on the system's angular momentum. A thin, uniform rod of length L = 0.850 m and mass M rotates at an angular velocity of 18.5 rad/s. When a particle with mass M/3.00 is ejected, the conservation of momentum principle is applied to determine the speed of the ejected particle. The correct approach involves recognizing that the explosion is an internal force, thus conserving the total angular momentum of the system.

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A thin, uniform rod of length L = 0.850 m and mass M rotates horizontally about an axis through its center, at an angular velocity 18.5 rad/s. A particle with mass M/3.00 is attached to one end. The particle is then ejected from the rod by a small explosion. The explosion sends the particle along a path that is perpendicular to the rod at the instant of the explosion, with a speed that is 6.00 m/s greater than the speed of the end of the rod after the explosion. What is the speed of the ejected particle?

I attempted to find the speed by solving for v in the following equation:
[(1/12)ml^2]*(w) = (M/3)(v)(l/2)

then i added 6. this gives me 8.62, which is wrong.

Please tell me what I'm doing wrong! Thanks!
 
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On what principles did you base your equation on? The conservation of momentum principle states that the momenum of a system cannot change with no external forces acting on it. Hint: an internal explosion is not considered an external force.
 

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