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Angular momentum problem - am I right?

  1. Feb 14, 2009 #1
    1. The problem statement, all variables and given/known data
    A dumbbell consists of two mass points of equal mass m at opposite ends of a rigid rod of length l and negligible mass. Calculate the angular momentum of this dumbbell as it rotates with constant angular velocity [tex]\omega[/tex] about an axis that makes a constant angle [tex]\theta[/tex] with the dumbbell and passes through its center of mass.


    2. Relevant equations

    In my calculation, I'll use r=l/2

    3. The attempt at a solution I'm not sure if this is correct, but here's what I did:

    L = rXp

    Since there are two masses:
    L=[tex]\Sigma[/tex]Li

    =rXp1 + rXp2 = m[(rXv1)+ (rXv2)]

    =m[(rX(rsin[tex]\theta[/tex][tex]\omega[/tex])+rX(rsin[tex]\theta[/tex][tex]\omega[/tex]]

    =mrsin[tex]\theta[/tex][rX[tex]\omega[/tex] + rX[tex]\omega[/tex]]

    ||L||=mrsin[tex]\theta[/tex][r[tex]\omega[/tex]sin[tex]\theta[/tex]+r[tex]\omega[/tex]sin[tex]\theta[/tex]]

    =2mr2sin2[tex]\theta[/tex][tex]\omega[/tex], r=l/2

    L=(1/2)ml2sin2[tex]\theta[/tex] [tex]\widehat{n}[/tex] ;

    where [tex]\widehat{n}[/tex] is a unit vector perpendicular to both r and [tex]\omega[/tex]
     
    Last edited: Feb 14, 2009
  2. jcsd
  3. Feb 14, 2009 #2
    In cylindrical coordinates, your two particles are located at
    r1 = r sin(theta) er + r cos(theta) k
    r2 = -r sin(theta) er - r cos(theta) k
    the ang velocity vector is
    omega = omega k
    L1 = mr1 x v1 = r1 x (omega x r1) = mr1 x (omega k x (r sin(theta) er + r cos(theta) k))
    = mr1 x r omega (sin(theta) k x er + cos(theta) k x k)
    = mr1 x r omega sin(theta) eth
    = mr^2 omega (sin(theta) er + cos(theta) k) x sin(theta) eth
    = mr^2 omega (sin(theta)^2 (er x eth)+cos(theta) sin(theta) (k x eth))
    = mr^2 omega (sin(theta)^2 k - sin(theta) cos(theta) er)
    It looks to me like this will be the contribution from one mass and the other will be developed similarly. I suggest that the algebra needs to be checked very carefully; I certainly could have made a mistake there, but this is the way I would approach it.
     
  4. Feb 14, 2009 #3
    I have not yet encountered cylindrical coordinates; do you think the way I have done it is correct?
     
  5. Feb 15, 2009 #4

    Doc Al

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    Staff: Mentor

    This is OK. (Realize that r and v are perpendicular.)

    Since r and v are perpendicular, there's no second sinθ factor.
     
  6. Feb 15, 2009 #5
    Doesn't the angular velocity vector point either up or down? Also, the problem doesn't necessarily say that the dumbbell is perpendicular to the axis, so don't I need that sin[tex]\theta[/tex] term?
     
  7. Feb 15, 2009 #6
    Oh... maybe omega does not point up or down. Does it point in the same direction as the tangential velocity? In that case, I would just divide my answer by sinθ, and it should be correct.
     
  8. Feb 16, 2009 #7

    Doc Al

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    Staff: Mentor

    Let's say that the angular velocity vector points along the y axis. The dumbbell's center is at the origin and its axis is at an angle θ from the y axis.

    The tangential velocity is perpendicular to both r and ω, thus the magnitude of r X v is just rv.
    Simply correct your error in your original answer. You have mixed up the angle that the dumbbell makes with ω, which is θ, with the angle that the v makes with r, which is 90°.
     
  9. Feb 17, 2009 #8
    Ok... follow up question.

    Another way to approach this problem is L=Iω
    so I=2m(rsinθ)^2 (since rsinθ is the distance to the axis)
    Then L = 2mωr^2(sinθ)^2 = (1/2)mωl^2(sinθ)^2

    Why does this turn out different?
     
  10. Feb 17, 2009 #9

    Doc Al

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    Staff: Mentor

    Because L=Iω only gives you the component of the angular momentum parallel to the axis of rotation. The angular momentum vector is not parallel to that axis. (Note that the dumbbell is not rotating about an axis of symmetry.)
     
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