Rings on a rod: angular momentum conceptual question

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Homework Statement


A uniform rod of mass 3.15×10−2kg and length 0.380m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.250kg , are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 5.20×10^−2m on each side from the center of the rod, and the system is rotating at an angular velocity 33.0rev/min. Without otherwise changing 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?

Homework Equations


[itex]I_1 \omega_1 = I_2 \omega_2[/itex]

The Attempt at a Solution


I got the problem right but want to understand what's going on. I used conservation of angular momentum to calculate the angular velocity when the rings reached the end, but apparently the system has the same angular velocity after the rings leave. How does that work? Doesn't the mass and moment of inertia go down since it's only the rod in the case b?
 

Answers and Replies

  • #2
haruspex
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the system has the same angular velocity after the rings leave. How does that work? Doesn't the mass and moment of inertia go down since it's only the rod in the case b?
When the rings depart, they carry away mass and angular momentum. These are the same mass and angular momentum they contributed to the system total the instant before departure. Therefore the mass and angular momentum that remain are those the rod had the instant before departure. Why would the angular speed of the rod change?
 
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Thank you! I understand now. Sounds dumb, but I neglected that angular velocity. Under my assumptions, the rings would have just fell straight down after they reached the ends of the rod. That defies inertia.
 

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