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  1. #1
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    Physics Question Angular Momentum (2700 reps)

    A uniform rod of mass 2.60 x 10^-2 kg and length 0.440m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.150kg, are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 4.7 x 10^-2 m on each side from the center of the rod, and the system is rotating at an angular velocity 35.0rev/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 when the rings are at the end of the rod?

    b) What is the angular speed after rings leave the rod?





    I got part A simply enough, by applying I1w1=I2w2.

    Generic expression for I (not specifying whether the rings are the end or not)

    I = (1/12)ML^2 +2md^2

    Now going onto part B, where the rings leave the rod, wouldn't the moment of inertia decrease to merely

    I = (1/12)ML^2 because "d" = 0?

    Naturally when moment of inertia decreases in a system with no net external torque (angular momentum is conserved), the angular speed should increase right? but for some reason it stays the same.
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  2. #2
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    too bad i did crap in first year physics otherwise i would have helped you
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  3. #3
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    bump. anyone?
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    idk much about physics, but keep bumping it, there are a good amount of miscers familiar with this stuff.
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  5. #5
    Registered User KillahJon's Avatar
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    i gotchu bro. give me 5 minutes plz.
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  6. #6
    Registered User KillahJon's Avatar
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    wait, wtf. of course it stays the same lol, it's conservation of angular momentum. You aren't adding any extra torque to the system.
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    Originally Posted by KillahJon View Post
    wait, wtf. of course it stays the same lol,
    How? If I = (1/12)ML^2 +md^2+md^2 initially
    and the rings fall off the rod then those md^2 parts disappear, so the moment of inertia decreases, and since angular momentum is conserved, the angular speed should go up
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  8. #8
    Registered User KillahJon's Avatar
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    Originally Posted by NephilimRising View Post
    How? If I = (1/12)ML^2 +md^2+md^2 initially
    and the rings fall off the rod then those md^2 parts disappear, so the moment of inertia decreases, and since angular momentum is conserved, the angular speed should go up
    nah, the rings still have angular momentum about the rotation axis.

    thinkg of it this way.

    if you reverse the process.
    say u throw the rings into the rod, and the rod is stationary. then the rod begins to rotate.
    you wouldn't say that the md^2 parts have disappeared in the beginning now would u?
    Just because it isn't connected to the rod doesn't mean it's not part of the system.
    That is the point of angular momentum conservation.
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  9. #9
    Registered User NephilimRising's Avatar
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    Originally Posted by KillahJon View Post
    wait, wtf. of course it stays the same lol, it's conservation of angular momentum. You aren't adding any extra torque to the system.
    Yes, angular momentum stays the same in the absence of a net external torque but L = Iw, I and w can change.
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  10. #10
    Registered User NephilimRising's Avatar
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    Originally Posted by KillahJon View Post
    nah, the rings still have angular momentum about the rotation axis.

    thinkg of it this way.

    if you reverse the process.
    say u throw the rings into the rod, and the rod is stationary. then the rod begins to rotate.
    you wouldn't say that the md^2 parts have disappeared in the beginning now would u?
    Just because it isn't connected to the rod doesn't mean it's not part of the system.
    That is the point of angular momentum conservation.
    Oh I see. This is presumably right after the rings leave the rod? The moment of inertia of the whole system is (1/12)ML^2 +md^2+md^2 same as before. But now the system consists of 3 separate entities instead of 1?

    Edit: will rep on recharge.
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  11. #11
    Registered User KillahJon's Avatar
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    Originally Posted by NephilimRising View Post
    Oh I see. This is presumably right after the rings leave the rod? The moment of inertia of the whole system is (1/12)ML^2 +md^2+md^2 same as before. But now the system consists of 3 separate entities instead of 1?

    Edit: will rep on recharge.
    yes whatever you include in our system from your initial state has to be in the final state.
    initial angular momentum = final angular momentum.
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