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Angular velocity after collision?

  1. Aug 30, 2015 #1
    1. The problem statement, all variables and given/known data
    problems_MIT_rayyan_Physics801_Figures_Spinning_rod.png

    A small circular block of mass M traveling with a speed v on a frictionless table collides and sticks to the end of a thin rod of with length D and mass M. The picture shows a top down view of the block and rod on the frictionless table. What is the rod's angular velocity after the collision? Express your answer in some combination of M, v, and D.

    2. Relevant equations
    conservation of angular momentum
    I_cm = M*D^2/12
    I_point mass = Mr^2

    3. The attempt at a solution
    So I guess the part that I'm struggling with is where to put the reference point. I tried the center of mass first:

    L_i = M*v*D/2 + I*omega
    and since omega_i = 0 its just
    L_i = M*v*D/2

    L_f = omega(I_cm + I_point mass) = omega(M*D^2/12+M*D^2/4)

    and using conservation of angular momentum:
    M*v*D/2 = omega(M*D^2/12+M*D^2/4)

    M*v*D/2 = omega*M*D^2/3

    omega = 3*v/(2*D)

    I tried the same thing with the upper end of the rod and got 3*v/(4*D) so I'm obviously doing something wrong. Do I have to include the translation of the center of mass? If I do I'm not quite sure how.
     
  2. jcsd
  3. Aug 30, 2015 #2

    TSny

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    OK, just to be clear, it appears that you are choosing your origin to be at the initial position of the center of mass of the rod, not the center of mass of the whole system.

    OK
    No, the equation above isn't correct.
    Yes. Just after the collision, the center of the rod will have a horizontal "recoil" velocity Vc,f. The velocity of the point mass just after the collision will have a contribution from Vc,f and a contribution from ω.

    Both Vc,f and ω are unknowns. So, you're going to need another equation. Can you think of another conservation principle that is applicable?

    Note: there is a tool bar for formatting. The Σ icon will bring up some mathematical symbols.
     
  4. Aug 30, 2015 #3
    Conservation of energy right? Or can you use conservation of linear momentum to find the velocity of the center of mass?
     
  5. Aug 30, 2015 #4

    TSny

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    Is the collision elastic or inelastic?
     
  6. Aug 30, 2015 #5
    inelastic, so I should use conservation of linear momentum?
     
  7. Aug 30, 2015 #6
    in which case v_f = v/2 but how do i incorporate that into the angular momentum equation?
     
  8. Aug 30, 2015 #7

    TSny

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    It's not that easy to get the final velocity of the center of the rod, Vc,f. You'll need to set up the linear momentum equation carefully.
     
  9. Aug 30, 2015 #8
    p_i = M*v right?
    and p_f = M*v'+M*v' = 2*M*v'
    What am i doing wrong?
     
  10. Aug 30, 2015 #9

    haruspex

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    In general, a reference point for conservation of angular momentum should be either a fixed point or the mass centre of the system. You appear to have taken the mass centre of the rod, but not as a fixed point. Take it it as fixed at the initial position of the rod's mass centre. Then, as TSny says, you get a post-collision contribution from the linear motion of the disc. You can treat that either as purely linear motion, or as the linear motion of the rod's centre plus a contribution from the rotation of the rod.

    Although it's easy to find the post-collision velocity of the system's mass centre, it is, as you found, awkward to use. I would introduce an unknown for the post-collision velocity of the rod's mass centre. The velocity of the disc can easily be expressed in terms of that and the rotation rate. The two conservation equations allow you to solve for those two unknowns.
    I believe this is the path TSny laid out.
     
  11. Aug 30, 2015 #10
    I'm not sure I completely understand, if I take the center of mass of the system 3*D/4 as the reference point will the linear conservation be less awkward to use?
     
  12. Aug 30, 2015 #11

    TSny

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    Yes, I was considering the origin as a fixed point that coincides with the initial location of the center of the rod.
     
  13. Aug 30, 2015 #12

    haruspex

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    No, I'm saying that I regard it generally as simpler not to bother working out where the mass centre of the system is. I just assign a variable to the velocity of each component. But it's a matter of taste.
     
  14. Aug 30, 2015 #13
    so how do I find the final velocity?
     
  15. Aug 30, 2015 #14

    TSny

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    Try to express the final linear momentum of the system in terms of M, Vc,f, ω, and L.
     
  16. Aug 30, 2015 #15
    So would it be correct to say that L_f = m*D^2/3*ω+2*m*D/4*v_f
     
  17. Aug 30, 2015 #16

    haruspex

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    It depends... what's vf here?
     
  18. Aug 30, 2015 #17
    the final velocity of the center of mass of the system
     
  19. Aug 30, 2015 #18

    haruspex

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    Then the equation is wrong. ##\omega \frac D2## is the velocity of the disc relative to what?
     
  20. Aug 30, 2015 #19
    the center of the rod right?
     
  21. Aug 30, 2015 #20
    m*v_cf+m*ω*d/2?
     
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