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Rotating on a small platform attached by massless rods to a pivot

  1. May 6, 2006 #1
    I recall my teacher explaining this concept in class, but he did not go over how we would solve these types of problems:

    Bob the monkey (20 kg) is on a small platform (2 kg), which is attached by massless rods (R = 3m) to a frictionless pivot, and initially at rest.
    [​IMG]
    Bob would like to get the banana shown, which he could reach-- if only the platform were rotated a half-circle from its current location. Being a smart monkey, he starts spinning around his center of mass on the platform, and soon has his snack.

    a. What is Bob's rotational inertia about his center of mass? (You make approximate him as a sphere of radius 0.5 m).
    b. What is the rotational inertia of the system around the frictionless pivot? (Remember that Bob also moves with the platform. In this calculation, treat him as a point mass.)
    c. If Bob spins himself with ana ngular speed of 70 rad/s, how fast do Bob+platform rotate about the frictionless pivot? How long does it take for Bob to get his snack?

    For A, I used I = (20)(2.5)^2 and (22)(2.5)^2, and both answers were wrong. I don't know what else I can use to figure out the rotational inertia around Bob's center of mass.

    For B, I tried I = (2)(3)^2, but this answer was wrong.

    I haven't tried C, since I felt that I would again get a wrong answer, since the previous 2 parts are incorrect. Any help on this problem would be greatly appreciated! Thanks!
     
    Last edited: May 6, 2006
  2. jcsd
  3. May 6, 2006 #2

    nrqed

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    That's a fun problem! :biggrin: And a smart monkey!

    You need the moment of inertia of a *sphere* (with respect to an axis going through its center)! Do you know the expression?

    why 2??
    You will have 20 kg * (3 m)^2 plus the moment of inertia of the platform (you will need the moment of inertia of a full disk with respect to its center)
     
  4. May 6, 2006 #3
    So I = 7/5MR^2?
     
  5. May 6, 2006 #4

    nrqed

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    ?? I thought it was 2/5 MR^2 for a sphere!
     
  6. May 6, 2006 #5
    Okay, for that I got 2 for Bob's rotational inertia about his center of mass.

    I then used (20)(3)^2 + 2, and got 182, but got the question wrong. Did I do something wrong?
     
  7. May 6, 2006 #6

    nrqed

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    Ok for Bob's moment of inertia.

    But in the next calculation, where did you get 2?? The 20*3^2 is the moment of inertia of Bob (now treated as a point particle) . To that you must add the moment of inertia of the disk. Do you know the formula for the I of a disk rotated with respect to its axis? Compute that (using the mass of the disk and its radius) and add that to the I of Bob (the 20*3^2).
     
  8. May 6, 2006 #7
    So I will use (20)(3)^2 + (0.5)(2)(3)?
     
  9. May 6, 2006 #8

    nrqed

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    well, it is 1/2 M R^2 (so it is squared).

    But wait.. are you given the radius of the disk?? If not, we are stuck. (unless one treats the disk as having a negligible radius, in which case it contributes nothing to the total moment of inertia).

    So, are you sure you gave all the information provided??
     
  10. May 6, 2006 #9
    There is no radius for the disk.
     
  11. May 6, 2006 #10

    nrqed

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    oh, my mistake!! Sorry!:frown:
    When I saw ''2k of the platform'' I thought it was the mass of the swivel!!! Sorry!

    Well, then the platform can be treated as point mass also so you would add to the I of the monkey 2kg *(3 m)^2. (or you could have combined the monkey and the platform in one object of 22k and use (22kg) (3m)^2

    my apologies..
     
  12. May 6, 2006 #11
    So (22)(3)^2 = 198?
     
  13. May 6, 2006 #12

    nrqed

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    yes..don't forget to include the units
     
  14. May 6, 2006 #13
    Okay cool, and what about the final part?
     
  15. May 7, 2006 #14
    Bump! Anyone able to help me finish off this problem?
     
  16. May 8, 2006 #15

    Hootenanny

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    I think C can be solved if you consider conservation of angular momentum.

    ~H
     
  17. May 8, 2006 #16
    How so? Would you be able to elaborate more on that?
     
  18. May 8, 2006 #17

    Doc Al

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    Since no external torques act about the vertical axis, the total angular momentum of the monkey-platform system about that axis remains constant.
     
  19. May 8, 2006 #18
    Okay, and how would I be able to incorporate this into what I have now...how should I set my equation up?
     
  20. May 8, 2006 #19

    Doc Al

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    Find the total angular momentum: The monkey rotates about his axis, what's his angular momentum? The platform/monkey rotates about the central axis, what's its angular momentum?

    What's the total angular momentum?
     
  21. May 9, 2006 #20
    What is the equation for total angular momentum? Is it L=Iw? I don't know L or w, so I have 2 unknowns?
     
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