Angular Momentum Problem of rod

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SUMMARY

The discussion revolves around the angular momentum problem involving a uniform rod of mass m1 and length L, rotating about a fixed axis. Two small rings, each with mass m2, slide outward from positions r on either side of the rod's center when released. The final angular speed (ωf) of the rod after the rings leave is calculated using the conservation of angular momentum formula: ωf = ωi * (Ii / If), where Ii is the initial moment of inertia and If is the final moment of inertia. The moment of inertia for the rod is given by I(rod) = (1/12)mL2 and for the rings by I(ring) = m2r2.

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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.
 
Last edited:
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Hmm I'm having trouble with this one. Could you lead me in the right direction?
 

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