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Bug Jumps off a rod; what is the angular velocity of the rod

  1. Apr 10, 2013 #1
    1. The problem statement, all variables and given/known data

    A small 13.0g bug stands at one end of a thin uniform bar that is initially at rest on a smooth horizontal table. The other end of the bar pivots about a nail driven into the table and can rotate freely, without friction. The bar has mass 65.0g and is 120cm in length. The bug jumps off in the horizontal direction, perpendicular to the bar, with a speed of 25.0cm/s relative to the table.

    mass bug = 0.013 kg
    mass bar = 0.065 kg
    distance from axis of rotation = 1.2 m
    velocity bug = 0.025 m/s

    Find
    What is the angular speed of the bar just after the frisky insect leaps?

    2. Relevant equations

    L = Iω
    I(bug) = Mr^2
    ω(bug) = v/r
    L(bug) = Iω = Mr^2ω = Mrv
    I(bar) = Mr^2
    ω(bar) = v/r
    L(bar) = Iω = Mr^2ω

    3. The attempt at a solution

    Conservation of Angular Momentum

    L(bug) = L(bar)
    M*r*v = M*r^2*ω
    0.013*0.12*0.025 = 0.065*0.12^2*ω
    3.9*10^5 = 9.36*10^4ω
    ω = 0.204

    Conservation of Kinetic Energy

    KE = 1/2Iω^2

    KE(bug) = 1/2Iω^2
    KE(bug) = 1/2(mr^2)ω^2
    KE(bug) = 1/2(mr^2)(v/r)^2
    KE(bug) = 1/2mv^2

    KE(bar) = 1/2Iω^2
    KE(bar) = 1/2(1/3mr^2)ω^2
    KE(bar) = 1/6mr^2ω^2

    KE(bug) = KE(bar)
    1/2mv^2 = 1/6mr^2ω^2
    4.0625E-6 = 0.0156ω^2
    ω^2 = 2.6042E-4
    ω = 0.01614

    So far none are correct. Please Help!
     
  2. jcsd
  3. Apr 10, 2013 #2

    mfb

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    Staff: Mentor

    Kinetic energy is not conserved, the bug has to use some source of energy to jump. There is no particular reason to expect the same kinetic energy for both bug and bar.
    Conservation of angular momentum works - but you have to consider that the mass of the bar is not at its end only, it is distributed over the whole bar.

    Do you know the right answer? If you divide your value (for conservation of angular momentum) by the right answer, you should note some relation between the two.
     
  4. Apr 10, 2013 #3
    Also, you said the moment of inertia of the bar is mr^2 - I don't think that's right. If you integrate r^2*dm for a uniform thin bar, you get something slightly different.

    Edit: think mfb and I are saying the same thing: re-check your moment of inertia for the bar.
     
    Last edited: Apr 10, 2013
  5. Apr 10, 2013 #4
    So if I change the moment of inertia of the bar to 1/3mr^3 then I should get the correct answer?
     
  6. Apr 10, 2013 #5
    1/3mr2 should do it :)
     
  7. Apr 10, 2013 #6
    Why is it r2 and not r3?

    because I thought you would take the integral of mr2 and in order to get it outside of the [itex]\Sigma[/itex] you had to take the anti-derivative which would be 1/3mr3?
     
  8. Apr 10, 2013 #7
    L = Iω
    I(bug) = Mr^2
    ω(bug) = v/r
    L(bug) = Iω = Mr^2ω = Mrv
    I(bar) = 1/3*Mr^2
    ω(bar) = v/r
    L(bar) = Iω = 1/3Mr^2ω

    L(bug) = L(bar)
    M*r*v = (1/3)*M*r^2*ω
    0.013*0.12*0.025 = 0.02167*0.12^2*ω
    3.9*10^-5 = 3.12*10^-4ω
    ω = 0.125
     
  9. Apr 10, 2013 #8
    First, r2 is dimensionally correct, but the reason also comes from

    ∫r2dm.

    dm = (m/L)*dx, and r = x, so we have

    (m/L) * ∫x2dx. Integrate from 0 to L.

    What does that work out to?
     
  10. Apr 10, 2013 #9
    (m/L) * (1/3)L3 - (1/3)03

    which when you simplify is M*(1/3)L2

    Thank you so much for your help. I will post if the answer is correct or not shortly
     
  11. Apr 10, 2013 #10
    (m/L) * (1/3)x3

    Use the limit - evaluate that between x=0 and x=L. Notice one of the L's cancels.
     
  12. Apr 10, 2013 #11
    Yeah I noticed that when I made the reply. I edited my previous reply! Thank you so much!
     
  13. Apr 10, 2013 #12
    The answer was correct!

    Thank you very much!
     
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