Conservation of Angular Momentum in a Rotating System

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The discussion centers on the conservation of angular momentum in a rotating system involving a uniform rod and two sliding rings. The initial moment of inertia is calculated using the rod's mass and length, along with the rings' mass and their initial positions. Upon releasing the rings, they slide to the ends of the rod, changing the system's moment of inertia. The final angular velocity is determined by applying the conservation principle, but an error in calculating the radius of the rings leads to an incorrect result. The realization that the rings only extend to half the rod's length corrects the misunderstanding.
redribbbon
A uniform rod of mass 3.15×10^−2 kg and length 0.380 m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.250 kg, are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 5.20×10^−2 m on each side from the center of the rod, and the system is rotating at an angular velocity 34.0 rev/min. Without otherwise changing the system, the catches are released, and the rings slide outward along the rod and fly off at the ends.

now
I think that the its just pure conservation of momentum

Iwinitial=Iwfinal

so i do

((1/12)(3.5*10^-2)(.38)^2+2*.25*(5.2x10^-2)^2))
divided by
((1/12)(3.5*10^-2)(.38)^2+2*.25*(.38^2))
and then multiply the result by 34
but i get 0.8109
which is the wrong answer
what i am doing wrong?
 
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What is the question you are trying to answer? It would be good if you would identify all those numbers by the symbols they replaced, and units would go a long way to helping in that identification.
 
What is the angular speed of the system at the instant when the rings reach the ends of the rod?

the numbers represent

(1/12)(mass(rod))(Length)^2+2*mass(ring)*radius(ring)^2
Which is just the moment of inertia of the system before it starts
divided by the moment of inertia of the system after the rings get to the edge
the only thing that changes is the radius of the ring which goes from (5.2*10^-2)m to .38m

and then multipy by the initial angular velocity to
get the final angular velocity

i just happen to end up with .8109, which is not right
 
Last edited by a moderator:
redribbbon said:
What is the angular speed of the system at the instant when the rings reach the ends of the rod?

the numbers represent

(1/12)(mass(rod))(Length)^2+2*mass(ring)*radius(ring)^2
Which is just the moment of inertia of the system before it starts
divided by the moment of inertia of the system after the rings get to the edge
the only thing that changes is the radius of the ring which goes from (5.2*10^-2)m to .38m

and then multipy by the initial angular velocity to
get the final angular velocity

i just happen to end up with .8109, which is not right
Each ring only goes out to half the length of the rod
 
wow

i am an idiot

thanks a lot for pointing that out
 
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