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Problems With Fleas and Angular Momentum

  1. Nov 8, 2004 #1
    A straight uniform rigid hair of mass M lies on a smooth table; at each end of the hair sits a flea of mass m. For what values of m can the fleas simultaneously jump with the same speed and take-off angle, so that they exchange ends without colliding?

    All i know is that this occurs when m is above some fraction of M. lol I need the fraction though. Any help or ideas?
     
    Last edited: Nov 8, 2004
  2. jcsd
  3. Nov 8, 2004 #2
    help....I know it has to do with conservation of angular momentum
     
  4. Nov 8, 2004 #3
    Would you clarify the question statement a little? As it's written, the question makes no sense. Particularly, what's rotating?

    --J
     
  5. Nov 8, 2004 #4
    That's exactly how the question was given to me lol
     
  6. Nov 8, 2004 #5
    Yeap this was a question written by a grad student and it just puzzles my classmates and I.
     
  7. Nov 8, 2004 #6
    And it doesn't strike you as odd that you've been told you need to use conservation of angular momentum, but you have nothing in the problem that actually has any angular momentum?

    --J
     
  8. Nov 8, 2004 #7
    I know, and I agree, but that's what we were given.
     
  9. Nov 8, 2004 #8

    AKG

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    If I understand it right, imagine that it is 9:15 so the hour hand is pointing straight to the left, and the minute hand is straight to the right (I know that's not how a real clock is, never mind that). Assume that the hr and minute hands form one rigid rod. Now, there is a flea at each end, so one is at "3" and the other at "9". Now, something like this happens : the flea at "9" jumps to "7" and the flea at "3" jumps to "1". In jump, each flea pushes the rod in the clockwise direction. In the time flea at "3" gets to "1", the end of the rod that was at "9" spins clockwise to "1", and in this they exchange ends.

    I don't really care to do the work, but I'm sure it's a simple matter of plugging values into the right equations and solving.
     
  10. Nov 8, 2004 #9
    Ah. The hair's not fixed. Clever.

    --J
     
  11. Nov 8, 2004 #10
    thank you , your explanation is helpful, but i'm at a loss to what equations and values to actually use. I figured the hair was fixed since it's lying on a table.
     
  12. Nov 8, 2004 #11
    Start with the first one you cited, conservation of momentum.

    --J
     
  13. Nov 8, 2004 #12

    AKG

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    Let l be the length of the hair. Let v be the speed of the fleas, and h be the angle above the table with which they leap. Their total angular momentum is L = mvl*cos(h). Let I be the moment of inertia of the hair about it's center. It's angular velocity is w = mvl*cos(h)/I. The fleas will spend time t in the air, where:
    a = (v2 - v1)/t in the vertical direction. We get:

    t = v*sin(h)/4.9

    If the hair rotates H radians, and H = wt = 12mv²sin(2h)/9.8Ml. The fleas must go pi - H, and the amount by which the fleas "rotate" is a function of the linear distance that they travel (since they have to end up on some point on the circle of radius l/2), and that in turn is a function of v and h. If D is "angle" through which the fleas "rotate", then:

    D = pi - H
    H = 12mv²sin(2h)/9.8Ml
    D = f(v, h)
    0 < pi < H

    So, you have M and l fixed, and D, H, m, v and h as (5) variables, with 3 equations and 1 inequality. You should be able to express them all in terms of 2 of the variables, get everything expressed in terms of m and H. You should end up with m expressed in terms of m and H such that you can't isolate m. At any rate, you know the range of values for H, so you should be able to find the range of values for m. I think this should work, let us know if it does.
     
  14. Nov 8, 2004 #13

    AKG

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    Put a pencil on a table, and push the eraser end in one direction, and the lead end in the other. See if it is "fixed". Of course, I believe you'll have to assume the table is frictionless, otherwise the problem gets even uglier.
     
  15. Nov 8, 2004 #14
    I'm sorry it's just not clicking, I'm not getting any kind of answer. And i'm assuming by smooth table it means frictionless lol
     
  16. Nov 8, 2004 #15

    AKG

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    Oh, didn't even notice that it siad smooth. Where are you getting stuck? If you know how to use LaTeX, show us what you have so far; if not, learn it and show us what you have so far.

    [tex]\sin (x) = \frac{\pi}{17}[/tex]

    There's an example of LaTeX. You can click on it to see how it was done, and when you click on it, there's a link in the window that pops up to a .pdf file that explains how to use it.
     
  17. Nov 8, 2004 #16
    ok I'm doing Li=Lf, and after substituting L with mvl(cos(h)), and then h with 12mv²sin(2h)/9.8Ml, I get an equation in which both sides are the same so obviously something is missing, and i don't knwo how to incorporate a and t into this.
     
  18. Nov 9, 2004 #17

    AKG

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    You don't need to do Li = Lf. In fact, I have no idea why you're doing that. I gave you a list of 3 equations (and an inequality). One of the "equations" was simply D = f(v, h). You actually need to figure out what f is. Basically, you've already figured out the time spent in the air by the flea, t. You also already know it's horizontal speed, v*cos(h). The distance travelled is vt*cos(h), which is:

    D = v*cos(h)*t = v²sin(2h)/9.8

    So, we have:

    D = (Ml/12m)H

    D = pi - H
    (Ml/12m + 1)H = pi
    Ml/12m = (pi - H)/H
    m = HMl/12(pi - H)

    dm/dH = (1/12)(Ml(pi - H) + HMl)/(pi - H)²
    dm/dH = Ml(pi)/12(pi - H)², so m is a strictly positive function of H, and it can clearly be seen that it is strictly increasing. Therefore, m takes on it's minimum value as H approaches 0, and it's max as H approaches pi. Plug in these values into:

    m = HMl/12(pi - H), and you get m approaching zero, and m approaching infinity. Well there's a problem :eek:. I think there's some other restriction on H which need to be found. That being, if you have a given D, then there are only certain values for v and h where that will work, and so you will only get certain values for angular momentum of the hair, and so it will only go through certain angular displacements H. Since H is restricted in THAT way by D, but also that D + H = pi, you should be able to limit H to some tigher interval than [0, pi] and so you shouldn't get m approaching infinity. However, figuring out how to restrict H in terms of D "That way" is the ugliest part, and I'll leave that to you.
     
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