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Two rotating disks, angular speed and kinetic energy question.

  1. Nov 2, 2011 #1
    1. The problem statement, all variables and given/known data
    A uniform disk of mass 9.00m and radius 2.00r can rotate freely about its fixed center like a merry-go-round. A smaller uniform disk of mass m and radius r lies on top of the larger disk, concentric with it. Initially the two disks rotate together with an angular velocity of 21.4 rad/s. Then a slight disturbance causes the smaller disk to slide outward across the larger disk, until the outer edge of the smaller disk catches on the outer edge of the larger disk. Afterward, the two disks again rotate together (without further sliding). (a) What then is their angular velocity about the center of the larger disk? (b) What is the ratio K/K0 of the new kinetic energy of the two-disk system to the system's initial kinetic energy?

    3. The attempt at a solution
    To be honest, I do not know where to start with this one, can anybody help me out?
     
  2. jcsd
  3. Nov 2, 2011 #2
    Use conservation of angular momentum of the whole system.
     
  4. Nov 2, 2011 #3
    Ok so i want you to look at the formulas for velocity:
    v=vi(i=initial not a variable) + at
    x = xi + vit + 1/2at^2
    v^2 = vi^2 +2aΔx
    Now circular motion formulas use the same variables only written differently:
    (ω=angular velocity)similar to v
    (α=angular acceleration)similar to a
    (θ=angular displacement)similar to x
    Now look at circular motion equations:
    ω = ωi(i=initial) + αt
    Δθ = wit + 1/2αt^2
    ω^2 = ωi^2 + 2αΔθ

    Simply plug in the givens and there you go!
    Also a side note, some questions might ask for the velocity in revoltions and in radians.
    conversion is 1 revolution = 2pi radians
    hope that helps! (I would solve the question for you but the rules say I can't...sorry!)
     
  5. Nov 2, 2011 #4

    Filip Larsen

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    Unfortunately those equations are not going to help the OP find his answers.

    The key concept to use, as Grzz already pointed out, is conservation of angular momentum and how it relates to angular velocity and rotational energy of a rigid disc.
     
  6. Nov 2, 2011 #5
    Ok, I think I know how to solve for the angular velocity and KE of the system before sliding, but How do I do it after the other disk moved to the edge. Maybe we haven't gone over this in class yet, but I have another similar problem and I do not know how to do it, so I will attempt it, and wait till Friday's class to look into further it.
     
  7. Nov 2, 2011 #6

    Filip Larsen

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    If you list the equations and work you've done so far we should be able to provide better help. In order to solve this problem you do need to know how angular momentum relates to torque and angular velocity and you also need to be able to calculate (or look up) the moment of inertia around the central axis of the two solid discs both before and after the slide.
     
  8. Nov 2, 2011 #7
    Ok so for the initial KE, I did KE=(1/2)(I)(w^2) so KE=(1/2)(10m)(21.4rad/s) and
    KE=107 m x rad/s

    Now I re-read the problem and I realized I am unsure how to calculate the Angular velocity after the slide. Would I have to use L=Iw to find initial angular momentum and set the initial L= to the final L and find w that way? and use that to find the final KE? Am I on the right track. Thanks for the help!
     
  9. Nov 3, 2011 #8

    Filip Larsen

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    While it is correct that [itex]K = \frac{1}{2}I \omega^2[/itex] you should think about what I in that equation is and how you can calculate it from the knowledge of the shape, mass and size of the two discs. You should end up with a moment of inertia about the center axis that is expressed as a function of m and r (since m and r is given only as variables you cannot insert any number for them).

    And while you are at it, you may also want to think about if I after the slide is different and if so how to calculate that. This is probably the most tricky part of the solution (hint: one disc had its center of mass moved some distance relative to the center axis of rotation).

    Try it and see what it brings :smile:
     
  10. Nov 3, 2011 #9
    Alright so since it is a disk, then I=(1/2)MR^2 so since there are two disks, should it be I=(1/2)(M)(R^2)+(1/2)(m)(r^2) where M is the mass of the large disk and m is the mass of the other disk, and R is radius of large disk and r is radius of small disk? So if that is the case then it is I=(1/2)(9m)(2r)^2+(1/2)(m)(r^2) so with much simplification I get I=18.5m(r^2) so since we know I, I can plug that into the KE=(1/2)(I)(w^2) and KE=(1/2)(18.5m(r^2))(21.4rad/s)^2 so (with simplification) KE=4236.13mr^2

    I Can't think how I would treat this, I am a little worn out now. So I will think more later and see how my above attempt worked out and try again later.
     
  11. Nov 4, 2011 #10

    Filip Larsen

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    Spot on :smile:

    For calculating the moment of inertia after the slide you can do the same, but with a correction for the fact that the small disc no longer rotates around an axis that goes through its center of mass as before the slide but now instead rotate around an axis parallel to that (assuming here that the small disc is keeping itself flat against the bigger disc after slide).
     
  12. Nov 4, 2011 #11
    Ok, so if I used I=(x)MR^2 the only difference would be the x in front of MR^2, so instead of (1/2) it would be 1/some number. So After consulting my book and a brief web search, I cannot find a rotational inertia for a disk with the axis on the rim. So either a) how do i find this, or b) what is it?

    And after I have that I, then I get KE, like I did above, and for the angular velocity, do I use L=Iw and set initial L equal to final L and use the separate I's and calculate for w. I don't have time to crunch numbers but I'll be back in a hour or so to put this to the test.
     
  13. Nov 4, 2011 #12
    Just found out for the disk with axis at rim, it is (3/2)(M)(R^2) so I=(1/2)(9m)(2r)^2+(3/2)(m)(r)^2 and with simplification I=19.5mr^2 so for initial L=Iw so L=(18.5mr^2)(21.4) and L=395.9mr^2 so if Li=Lf then 395.9mr^2=(19.5mr^2)w and w=20.3025641
    so the KE=(1/2)(19.5mr^2)(20.3025641) and KE=197.95mr^2
    so final answers are w=20.3025641rad/s
    and k/k0 then 197.95mr^2/4236.13mr^2 and k/k0=0.046728972
    Thanks for the help!
     
  14. Nov 4, 2011 #13
    hmmm, the k/k0 is wrong, any suggestions?
     
  15. Nov 4, 2011 #14

    Filip Larsen

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    Check your calculation of the final kinetic energy.

    You may also find it instructive to solve the problem without inserting a numeric value for the angular speed until at the very end. If you carry through with a symbolic angular speed you will end up with some very simple expressions for the ratio of angular speeds and for the ratio of kinetic energies.
     
  16. Nov 4, 2011 #15
    Agh I didn't square w; the KEf=4018.892563 so the ratio is 0.948, correct?
     
  17. Nov 4, 2011 #16

    Filip Larsen

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    Correct.

    Notice, that from [itex]I_1\omega_1 = I_0 \omega_0[/itex], [itex]K_0 = I_0\omega_0^2[/itex] and [itex]K_1 = I_1\omega_1^2[/itex] alone you end up with the rather nice result that

    [tex]
    \frac{\omega_1}{\omega_0} = \frac{K_1}{K_0} = \frac{I_0}{I_1}
    [/tex]
     
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