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Angular momentum of a thin uniform rod

  1. Apr 18, 2008 #1
    1. The problem statement, all variables and given/known data
    A thin uniform rod of length 0.810 m and mass M rotates horizontally at angular speed 21.0 rad/s about an axis through its center. A particle of mass M/3.00 attached to one end is ejected from the rod and travels along a path that is perpendicular to the rod at the instant of ejection. If the particle's speed vp is 6.00 m/s greater than the speed of the end of the rod just after ejection, what is the value of vp?



    2. Relevant equations

    Li = Lf
    L = IW
    1/12 * m * r^2 = I for a rod rotating about its center
    mvr* sin(angle between r and v) = I for particle

    3. The attempt at a solution

    (1/12) * M * (.81^2) * 21rad/s = [1/12 * M * (.81^2)]W + [1/3*M * r * v]

    so the problem is I have one equation with two unknowns, and no other equations....
    (I tried to use that v = Wr, but I think that only works for rolling things and the answer didn't work out...)
    Any help would be great! Thanks
     
  2. jcsd
  3. Apr 18, 2008 #2
    The question is not clear. When the rod is rotating at 21 rad/s, where is the particle? on the rod? also rotating with the rod? if so, then vp = 21.0*(0.810/2) m/s
     
  4. Apr 18, 2008 #3
    The particle was on the rod, then when the rod was spinning, it flew off, and the question now is, what tangential speed did it fly off with?
     
  5. Apr 18, 2008 #4

    alphysicist

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    Hi A_lilah,

    I think, if I'm reading the question correctly, that the particle did just not fly off. It was ejected, so something (perhaps a small spring? etc.) pushed if off. That's why the particle ends up faster than the end of the rod.

    Assuming that is correct, you are using the right approach (conservation of angular momentum). However, on the left side of your equation you neglected to include the moment of inertia of the particle--it was spinning with the rod before the ejection.

    As far as the unknowns, you do have a relationship between the v, r, and w. You have not yet used the fact that the particle ends up moving 6m/s faster than the end of the rod.
     
  6. Apr 18, 2008 #5
    So:

    (1/2)(1/3M)(.405^2) + (1/12) * M * (.81^2) * 21rad/s = [1/12 * M * (.81^2)]W + [1/3*M * r * v]

    That adds in the momentum of the particle... but I'm still a little confused about W and v because the particle ends up moving 6m/s faster than the rod, which is a tangential speed, and W is a rotational speed.
     
  7. Apr 18, 2008 #6

    alphysicist

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    What is the tangential speed of the very end of the rod? (How is it related to the angular speed of the rod?)

    Also on the left side of the equation remember that the moment of inertia of a particle is just [itex]m r^2[/itex].
     
  8. Apr 18, 2008 #7
    (1/3M)(.405^2) + (1/12) * M * (.81^2) * 21rad/s = [1/12 * M * (.81^2)]W + [1/3*M * r * v]

    and the tangential speed at the very end of the rod is the speed of the particle (vp) minus 6m/s. unless something is rolling, I don't know any relationships between v and w...
    (for rolling things v = Wr)
     
  9. Apr 18, 2008 #8

    alphysicist

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    If an object is rotating, v = r w is the formula for the tangential speed of that part of the object that is a distance r from the axis.
     
  10. Apr 18, 2008 #9
    (1/3M)(.405^2)(21) + (1/12) * M * (.81^2) * 21rad/s = [1/12 * M * (.81^2)](v/r) + [1/3*M * r * v]
    and M cancels so:

    (1/3)(.405^2)(21) + (1/12)* (.81^2) * 21rad/s = [1/12 * (.81^2)](v/.405) + [1/3* .405 * v]

    =>
    1.148175 + 6.88905 = .054675(v/.405) + .135v

    8.037225 = .135v +.135v

    v = 29.7675m/s

    but this isn't the answer, so I think I may have a math problem...
     
  11. Apr 18, 2008 #10

    alphysicist

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    The two v's on the right side are not the same--one is 6m/s greater than the other.
     
  12. Apr 18, 2008 #11
    1.148175 + 6.88905 = .054675(v/.405) + .135(v+6)

    8.037225 = .135v +.135v + .81

    v = 26.7675

    .... apparently not the right answer either....
    lol~ thank you for being so patient despite my lack of physics know-how!
     
  13. Apr 18, 2008 #12

    alphysicist

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    Is the number in bold correct? I am getting a different answer.

    Also, look at your variable v. With the way you have it in your equation, what does v represent?
     
  14. Apr 18, 2008 #13
    Thank you SO much! :P
     
  15. Apr 18, 2008 #14
    note the angular velocity of the rod after ejection is not v/r, but (v - 6.00)/r
    the final answer is v = 11.5 m/s. Is it right?
     
  16. Apr 18, 2008 #15
    It is!
     
  17. Apr 18, 2008 #16

    alphysicist

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    Glad I could help!
     
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