Angular momentum of a system of a rotating rod and sliding rings

  • #1
Homework Statement
A uniform, 0.0300-kg rod of length 0.400 m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.0200 kg, are mounted so that they can slide along the rod. They are initially held by catches at positions 0.0500 m on each side of the center of the rod, and the system is rotating at With no other changes in the system, the catches are released, and the rings slide outward along the rod and fly off at the ends. (a) What is the angular speed of the system at the instant when the rings reach the ends of the rod? (b) What is the angular speed of the rod after the rings leave it?
Relevant Equations
Conservation of angular momentum
I got the correct answer for the first part but I'm not sure why the answer for (b) is the same for (a). Wouldn't the rings falling off mean that [itex] I_f = \frac{1}{12}M_L L^2[/itex] only where [itex]I_F, M_L, L[/itex] are the final moment of inertia, mass of the rod and length of the rod as opposed to [itex] I_f = \frac{1}{12}M_L L^2 + 2M_Rd^2[/itex] where [itex]M_R[/itex] is the mass of the rings and [itex]d_L[/itex] is their distance from the axis of rotation?

Answers and Replies

  • #2
The angular momentum of an isolated system is conserved. When you calculate the final angular momentum (after the rings have left the rod) you must still use the same system as you did for the initial angular momentum. You can't neglect the angular momentum of the rings once they have left the rod.

Draw a picture that shows the motion of the rings after they leave the rod. Neglect gravity, it's irrelevant.
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