Angular Momentum Problem of rod

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The problem involves a uniform rod rotating about a fixed axis with two small rings sliding outward before detaching. The initial angular speed is denoted as omega, and the moment of inertia for the rod and rings is calculated using the formula I(rod)=(1/12)mL^2 and I(ring)=mr^2. The final angular speed (wf) can be determined using the conservation of angular momentum, expressed as wf=wi*((I_initial)/(I_final)). The discussion highlights the analogy of the problem to sand falling from a moving truck, indicating a similar conservation principle at play. The solution emphasizes understanding the changes in moment of inertia as the rings leave the system.
nateshoe
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Homework Statement



A uniform rod of mass m_1 and length L rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass m_2, are mounted so that they can slide along the rod. They are initially held by catches at positions a distance r on each side from the center of the rod, and the system is rotating at an angular velocity omega. Without otherwise changing the system, the catches are released, and the rings slide outward along the rod and fly off at the ends.

What is the angular speed of the rod after the rings leave it?

Homework Equations


I(rod)=(1/12)mL^2

I(ring)=mr^2

wf=wi*(Ii/If)

The Attempt at a Solution



wf=wi*((1/12*m1*L^2+2*m2*r^2)/(1/12*m1*L^2)Nevermind I figured it out, it is the same concept as sand falling out of a moving truck.
 
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Hmm I'm having trouble with this one. Could you lead me in the right direction?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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