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Rotating and Nonrotating Rods Superposed

  1. Mar 24, 2012 #1
    1. The problem statement, all variables and given/known data

    A uniform disk turns at 3.7 rev/s around a frictionless spindle. A nonrotating rod, of the same mass as the disk and length equal to the disk's diameter, is dropped onto the freely spinning disk . They then both turn around the spindle with their centers superposed. What is the angular frequency in rev/s of the combination?


    2. Relevant equations

    L = Iω

    3. The attempt at a solution

    I was trying to relate the angular velocity to this but I don't think that's the correct approach. For example, on my second attempt I had:

    ω = L / (1/6)(M)(l)^2

    (3.7 rev/s)(6) = 22.2 rev/s

    I'm having trouble finding a connection.
     

    Attached Files:

  2. jcsd
  3. Mar 24, 2012 #2

    gneill

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    Staff: Mentor

    Think inelastic collision angular style!
     
  4. Mar 24, 2012 #3
    I'll try to explain this without it turning into a complete mess. I still thought of L = Iω and assumed it would be conserved. I made two equations:

    L = (1/12)Mr^2 * 3.7 rev./s

    L = ((1/12)Mr^2 + (1/12)Mr^2) * ω

    So then I tried to make a proportion.

    L1 / L2 = 0.30833 rev./s * Mr^2
    ---------------------
    (1/6 Mr^2) * ω


    6 Mr^2 * ω L1 = L2 * 0.30833 rev./s * Mr^2

    ω = L2 * 0.30833 rev./s * Mr^2
    ---------------------------
    L1 * 6 Mr^2

    I assumed momentum was conserved which I why I though L2 and L1 canceled. For ω, I got 5.138 * 10 ^-2 rev./s which doesn't make a lot of sense.
     
    Last edited: Mar 24, 2012
  5. Mar 24, 2012 #4

    gneill

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    Staff: Mentor

    One problem here: the expression for the moment of inertia of a disk is not the same as that of a thin rod. Use the expression for a disk to determine the system's angular momentum. Use a common measure for the radius of the disk and the length of the rod (so if the disk's radius is r, then the rod's length is 2r).
    You won't need to make a proportion if you simply equate the expression for the angular momentum after collision to the expression before. You should find that in solving for the new angular velocity that a LOT of stuff is going to cancel out :wink:
     
  6. Mar 24, 2012 #5
    Ok. I think I have this figured out.

    The initial moment of inertia is just the disk. 1/2 Mr^2.
    Once the disk is added, the final moment of inertia is 1/12 M (2r)^2 + 1/2 Mr^2 = 5/6 Mr^2.

    So now I can make an equation relating the two:

    3.7 rev./s (1/2 Mr^2) = 5/6 Mr^2 * ω
    ω final ≈ 2.2 rev./s
     
  7. Mar 24, 2012 #6

    gneill

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    Staff: Mentor

    That looks much better! :smile:
     
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