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anonymous
 one year ago
Conceptual physics question: I got the problem right but want to understand what's going on. In the problem, a rod was spinning with a given angular velocity. Two rings were held in place on the rod, and released so that they flew outwards towards the ends. 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?
anonymous
 one year ago
Conceptual physics question: I got the problem right but want to understand what's going on. In the problem, a rod was spinning with a given angular velocity. Two rings were held in place on the rod, and released so that they flew outwards towards the ends. 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?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Continue using conservation of momentum and you'll see that the angular velocity remains the same. You can also use conservation of energy. The rings will take some kinetic energy with them when they fly off the rod. Let's look at the moment when the rings fly off the rod. Just before, they increase the moment of inertia of the system, thus decreasing the angular velocity. The next instant, they are no longer actually attached to the rod. They still have angular momentum, however. They have the same angular momentum they had just an instant before. The rings would continue onward on their path, tangential to the point where they came off the rod. They'd maintain that amount of angular momentum. Since they still have that angular momentum, and the angular momentum of the rodring system is a constant value, the rod must maintain that final angular velocity.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for the clear explanation! I think I understand now.
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