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Conservation of Angular Momentum Question

  1. Apr 20, 2015 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    A bird m_b=0.470kg, is flying horizontally at v_b=2.50 m/s, not paying much attention, when it suddenly flies into a stationary vertical rod, hitting it d=25cm below the top. The rod is uniform, L=0.740m long, has a mass of m_r =1.60kg and is hinged at its base. This time the bird goes splat on the rod, and is stuck to it at the point of impact. What is the angular velocity w_f of the bird+rod just after the collision?

    What we have:
    Mass of bird: 0.470 kg
    Mass of rod = 1.60 kg
    Initial Velocity of bird: 2.50 m/s
    Length of rod = 0.740m
    Distance of impact from top of rod = 0.25m
    I_rod = 1/3M*d^2


    2. Relevant equations
    Conservation of Angular Momentum L_0=L_1
    L = mvrsin(theta)
    L= Iw
    v=w*r


    3. The attempt at a solution
    First I calculated the Angular momentum of the bird prior to collision
    L=mrv
    L=0.470kg*(0.740m-0.25m)*2.50m/s

    Sin here is 90 degrees and thus is simply 1.
    L=0.57575

    *NOTE: Did I use the right measurement for r?

    Next, I found the post-collision angular momentum to be:
    L_f=(m_b+m_r)*v_1*r + I_rod*w_f
    L_f=(m_b+m_r)*d^2*w_f+(1/3)*m_r*d^2*w_f

    Simple algebraic manipulation gets us:
    w_f=0.57575/((0.470+1.6)(0.490^2)+(1/3)(1.60)(0.490^2))

    for an answer of 0.921 rads/sec

    My questions/concerns:
    1) Is this the right method to approach this problem? Do I have any need to use conservation of mechanical energy to help solve this problem? All I am looking for is w_f.
    2) Did I use the correct value for d throughout this problem? Namely (0.740-0.25).
    3) Is it okay to be using the moment of inertia about the end of the rod for this problem? (1/3*M*d^2)

    Thank you.
     
  2. jcsd
  3. Apr 20, 2015 #2

    ehild

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    Yes, r=L-0.25.

    The term m_r must not be there. And what is d?
    The moment of inertia of the rod+bird sytem is I (bird) + I(rod). The bird can be considered as a point mass, I_b=m_b r2, and you know the moment of inertia of the rod. The whole system rotates with angular speed w_f


    1) Yes, the method is correct, but you have some errors.
    2)The mechanical energy is not conserved as the bird and rod move together after the collision.
    Well, what are d and r? You seem to mix them.
    3) yes, you can use the moments of inertia about the end of the rod, but you need to use different lengths for bird and rod.
     
  4. Apr 20, 2015 #3

    RJLiberator

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    Interesting.

    Wouldn't m_r be needed here as it is now part of the total system? I'm not quire sure why it would only be m_b needed here?
    I am using d=r is this wrong? I would think d = (0.740-0.25)?

    Hm, I see. So I would have to change my equation to:
    L_1 = (m_b+m_r)*d^2*w_f+(I_rod+I_bird)w_f where I_rod = (1/3)(m_r)(d^2) and I_bird = m_b*d^2

    This makes sense. =)
     
  5. Apr 20, 2015 #4

    ehild

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    The angular momentums add. That of the bird is m_b v_b(final) r_b where r_b = d.
    That of the rod is I w, r = 0.74 m, I =1/3 M_r r^2

    No, it is wrong.
     
  6. Apr 20, 2015 #5

    RJLiberator

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    Okay. So that DOES make sense. The angular momentum of the bird and rod are adding. So the mass of rod is unneeded in the part regarding the bird, and the birds r = 0.49 while the rods d = 0.740.

    So now the corrected equation is
    L_1 = m_b*v*r+(I_rod)w_f
    which simplifies to
    L_1 = m_b*r^2*w_f+(1/3)m_r*d^2*w_f
    where r=0.49 and d=0.740

    Or, do I need it to be (I_rod+I_bird) ?
    If so, why do we need both for this, but not both masses (bird+rod) previously?

    Thank you again.
     
  7. Apr 20, 2015 #6

    ehild

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    It will work now.
    You can factor out w_f, but not the masses: they are multiplied by different length-squares.
     
  8. Apr 20, 2015 #7

    RJLiberator

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    So I do NOT need to use I_b+I_r ?

    If this is the case, then I get:

    w_f= Angular momentum initial/((m_b*r^2)+(I_r))

    w_f=0.57575/((0.47+0.49^2)+((1/3)(1.6)(0.74^2)))
    w_f=1.422 rads/sec
     
  9. Apr 20, 2015 #8

    ehild

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    It is correct!! :cool:
    But you have used I=I_b+I_r!I think that + is just typo.
     
  10. Apr 20, 2015 #9

    RJLiberator

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    Awesome. I appreciate your help, but have one last clarifying question:

    Earlier you stated:
    But here in the end, I did not use this. I only used I_rod.

    What happened here?
     
  11. Apr 20, 2015 #10

    ehild

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    w_f=0.57575/((0.47*0.49^2)+((1/3)(1.6)(0.74^2)))
    I write it out into a readable form.

    ## \omega_f = \frac{0.57575}{0.47\cdot .49^2 + \frac{1}{3}\cdot 1.6 \cdot 0.74^2}##
    The denominator is the sum of the moments of inertia: that of the bird and that of the rod.
     
  12. Apr 20, 2015 #11

    RJLiberator

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    Oi! That makes sense. I can see it clearly now.

    I appreciate your time and help. You have indeed helped me on my path of learning!

    Thank you.
     
  13. Apr 20, 2015 #12

    ehild

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    You are welcome. :oldsmile:
     
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