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Can someone check my answer to a rotational problem?

  1. Jun 6, 2015 #1
    1. The problem statement, all variables and given/known data
    A stick of length L and mass M is in free space and not rotating. A point mass m has an initial velocity v heading in a trajectory perpendicular to the stick. The mass collides and adheres to the stick a distance b from the center of the stick. Find the resulting motion of the two together in terms of their center of mass velocity and final angular velocity.

    2. Relevant equations
    L=IW
    L=rxp
    3. The attempt at a solution

    I set the origin at the center of rotation (in the middle of the stick)
    LET M STAND FOR THE SMALL MASS AND S STAND FOR THE STICK.
    this is a momentum problem so i said [itex] L_i=L_f [/itex] at the time of contact.
    [itex] L_{mi} + L_{si} = L_{mf} + L_{sf} [/itex]
    initial momentm of the stick is 0.
    [itex] mv_ib = I_{stick}w+I_{mass}w [/itex]
    [itex] m_pv_ib=\frac{m_sL^2}{12}w_f + mb^2w [/itex]
    solve for [itex] w_f [/itex]
    [itex] w_f = \frac{m_pv_ib}{\frac{mL^2}{12}+m_pb^2} [/itex]

    i know i didnt solve for center of mass velocity but would we just do a regular momentum problem? (not rotational) becasue there are no external forces and the center of mass wouldnt really be affected by the rotations right?
     
  2. jcsd
  3. Jun 6, 2015 #2

    haruspex

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    It is important to take a non-accelerating reference point. The mass centre of the stick gets accelerated by the impact. You can use the position in space of the mass centre before impact as your reference point. If you do that, the mass m has a post-impact moment which is more than just ##mb\omega##.
     
  4. Jun 6, 2015 #3
    ok so the origin would be the center of the stick and stay at that position (so it doesnt move along with the stick right)? would the post impact of the mass be: r x P + mbw?
     
  5. Jun 6, 2015 #4

    haruspex

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    It depends how you are defining r and P.
     
  6. Jun 6, 2015 #5
    well, P=mv right? and m would be the velocity of the mass (which is the same as the velocity of the stick) and we can take [itex] rsin(\theta) [/itex] (where r is the distance from the origin to the mass )as b (because rsin(theta) is always b). would that be correct?
     
  7. Jun 6, 2015 #6

    haruspex

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    If P is the momentum of the point mass in an inertial frame, you don't need to add ##mb^2\omega## as well. ##b\omega## is the velocity of the mass relative to the rod's centre. Alternatively, if you were to define v as the velocity after impact of the rod's centre then you would use ##mbv+mb^2\omega##
     
  8. Jun 6, 2015 #7
    wait do you mean the total momentum of the stick would be: [itex] mbv + mb^2\omega [/itex] ? isnt that what I said when I said you simply add mvb?
     
  9. Jun 6, 2015 #8

    haruspex

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    Are you mixing up m with M?
    Your post #5 is somewhat confusing because you wrote
    I assume you meant v, not m, and the velocity of the mass is not the same as the velocity of the stick. If v is the velocity of the stick's mass centre then the velocity of the mass is ##v+b\omega##. Its angular momentum about the reference point is mb times that. The total angular momentum about the reference point is ##ML^2\omega/12+mbv+mb^2\omega##.
     
  10. Jun 6, 2015 #9
    ugh yes i messed up. in post 7, i mean the total angular momentum of the mass.... not the stick. that would be correct right?
     
  11. Jun 6, 2015 #10
    also, you would just set that equal to the inital momentum, which I said was [itex] m_pv_ib [/itex] and solve for w, and that would be the correct answer right?
     
  12. Jun 6, 2015 #11

    haruspex

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    If you are defining v as the post impact velocity of the stick's centre, yes.
     
  13. Jun 6, 2015 #12

    haruspex

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    Yes, but you also need to use linear momentum to find the post-impact velocity of the stick. I fear there is confusion here because you have used v for that unknown, but v is the given initial velocity of the mass.
     
  14. Jun 6, 2015 #13
    yes yes, idk why im making so many stupid mistakes. ok, so to do that would just just use [itex] p_f=p_i [/itex] and this would give you the linear velocity right?
     
  15. Jun 6, 2015 #14

    haruspex

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    Yes.
     
  16. Jun 6, 2015 #15
    ok, so doing that I get : [itex] v_f = \frac{m_pv_i}{m_p+m_s} [/itex] , then its a simple plug into the rotational mometum problem. Also, this would be the velocity of the center of mass correct?
     
  17. Jun 6, 2015 #16
    also, do you mind explain to me why only the momentum of the small mass is different than what I posted initially? How come the angular momentum of the stick has no linear velocity factored into it?
     
  18. Jun 7, 2015 #17

    haruspex

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    Yes, but rather than work out where the mass centre of the whole is, it's simpler to let vf be the velocity of the mass centre of the rod and write an expression for the linear velocity of the mass (just after impact) in terms of that.
     
  19. Jun 7, 2015 #18

    haruspex

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    You are taking moments about the point where the mass centre of the rod was just before impact. The velocity of the mass centre of the rod just after impact is along a line passing through that point, so has no angular momentum about it.
     
  20. Jun 7, 2015 #19
    oh is it because the center of the rod is moving in a straight line whereas the mass is moving in a circle?
    would the equation be: [itex] v_{fp} = \frac{m_pv_i-m_sv_{fs}}{m_p} [/itex].... now i have 2 unknowns in this equation. how would be able to solve that?
     
  21. Jun 8, 2015 #20
    hey haruspex... i had another question on this problem. why are we using [itex] L = I\omega^2 [/itex] for the momentum of the point mass after the collision? in another thread I created, some said that Iw^2 is only used for spinning and L= r x p is used for rotating around another object (which the bullet is doing). why isnt the final angular momentum of the mass r x p instead of Iw^2?
     
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