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Conservation of Angular Momentum of Bug Problem

  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 = 0.12 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

    This is the incorrect answer. I am not quite sure where I made my mistake?
    All help would be appreciated!!
     
  2. jcsd
  3. Apr 10, 2013 #2
    Three problems that I can see.

    The first is that the length of the bar needs to be in terms of meters. You've changed 120cm into 0.120 meters which is not right. 120cm=1.20m

    The second is that the change is the fly's linear momentum transfers to the bars angular momentum. Find the fly's change in linear momentum (easy) and set it equal to the bar's angular momentum (slightly more work) and then solve for the angular speed of the bar.

    The third is that the moment of inertia for a bar which rotates about its end is I=(1/3)ML^2. You forgot the 1/3

    See if that helps
     
    Last edited: Apr 10, 2013
  4. Apr 10, 2013 #3
    ΔLinear Momentum (fly) = ΔAngular Momentum (bar)

    mv(final) - mv(initial) = Iω(final) - Iω(initial)
    = mr^2ω(final) - mr^2ω(initial)

    0.013*0.025 - 0 = 0.065*1.2^2ω(final) - 0
    3.25E-4 = 0.0936ω(final)
    ω(final) = 0.00347

    Does this look better?
     
  5. Apr 10, 2013 #4
    I apologize, i just made an edit on my last post. Look at what I wrote about the moment of inertia for the bar.

    Also, it's not as straight forward as I thought. You need to figure out the torque that results as the fly jumps off of the rod. It's related to the force that the fly applies to the bar as it jumps off as well at the distance down the bar in which the fly jumped.

    I will keep looking at this and get back to you although I'm sure somewhat will read this and correct me before I get another chance.
     
    Last edited: Apr 10, 2013
  6. Apr 10, 2013 #5
    So when I calculate the torque that the fly applies to the bar how would I integrate it into finding the angular velocity? Im just confused how the torque would be added to the angular velocity
     
  7. Apr 10, 2013 #6
    [itex]\tau[/itex] = rFsin(theta)

    since here we know that theta = 90...

    [itex]\tau[/itex] = rF

    would F here be = ma(centripetal) = (mv^2)/r

    **All for the bug**
     
  8. Apr 10, 2013 #7
    Try conservation of energy. KE of fly = KE of bar

    If this doesn't work out I promise to stop trying to help
     
  9. Apr 10, 2013 #8
    Alright. No problem.

    Useful Equations


    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

    how confident are you in this answer...
    I am running out of "tries" in my online homework.
    (no offense meant; I wouldn't have had any idea of other strategies w/o your help)
     
  10. Apr 10, 2013 #9
    I'm pretty darn confident this time. Conservation of energy never fails. The reason we can't use cons of momentum is because you'll need to know the torque applied by the fly which requires us to know the force the fly applied on the bar. And in order to figure out the force we would need to know the amount of time the fly applied the force which we don't have.

    The alternative is to wait a little while and see if anyone corrects me or has a better idea.
     
  11. Apr 10, 2013 #10
    Well... unfortunately that answer was incorrect also.... I am not sure where to go with this, but I did repost this question as a new thread.
     
  12. Apr 10, 2013 #11
    I apologize, I'm obviously missing something that's leading me down the wrong path. Let me know how to properly do it when you figure it out.
     
  13. Apr 10, 2013 #12

    mukundpa

    User Avatar
    Homework Helper

    I think you were correct for the the angular momentum of the bug about the axis of rotation after jump as mvL

    The angular angular momentum of the rod after jump will be Iω = ML2/3*ω

    Now apply law of conservation of angular momentum.
     
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