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What is the angular momentum of a spinning solid sphere

  1. Dec 27, 2003 #1
    What is the angular momentum of a spinning solid sphere of mass 'm', equitorial velocity 'v', and radius 'r'?

    (The mass of the sphere being uniformly distributed and the spin being along a single axis only)

    I can't figure this out because the common textbook formula for angular momentum (mvr) only applies to cases where the mass is concentrated "at the end of the rope" so to speak. (like swinging around a bucket of water, or a planet going around in it's orbit)

    If anyone out there can help me with this, I'd appreciate it.
     
  2. jcsd
  3. Dec 27, 2003 #2
    first, your textbook is pure crap if it doesn't have something as basic as that or you're a high school student with high school text.

    the formula for angular momentum is L=Iw
    L-angular momentum
    I-moment of inertia
    w-angular velocity (r*v in your case)

    moment of inertia for a uniform sphere is (2/5)m*r^2

    I think you can do the rest
     
  4. Dec 27, 2003 #3
    mistake

    v=rw NOT v*r=w
     
  5. Dec 27, 2003 #4
    And how did you come to this? I get:
    [tex]\int_{0}^{r} d\theta d\phi dr' sin\theta mr' = 2\pi mr^2[/tex]
     
  6. Dec 27, 2003 #5

    Integral

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    The correct expression for the momement of inertia is

    [tex] \int r^2 DM [/tex]

    Are you using the correct Volume element for the sphere?

    The expression given by FulhamFan3 in correct, yours is not.
     
  7. Dec 27, 2003 #6
    I thought it would be, that's why I asked.

    Of course I'm not.
    [tex]\int_{0}^{r} dr' d\theta d\phi mr'^4\sin\theta =\frac{4\pi}{5}mr^5[/tex]

    what am I screwing up now?
     
  8. Dec 27, 2003 #7

    Integral

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    getting closer!

    Perhaps you are using limits of 0-2π for both angular varibles one should only be 0-π
     
  9. Dec 28, 2003 #8
    Nope...

    [tex]m\int_0^{2\pi}d\phi\int_0^\pi d\theta \sin\theta\int_0^rdr' r'^4=m*2\pi*2*\frac{r^5}{5}[/tex]

    i hate it when i get stuck in a mistake like this and can't detach.
     
  10. Dec 28, 2003 #9

    jamesrc

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    You've got two problems:

    1. dm = ρdV as opposed to what you have which looks like dm = mdV. For the sphere, [itex] \rho = \frac{m}{\frac{4}{3}\pi R^2} [/itex].

    2. You're mixing distances. The r in the formula for moment of inertia that Integral gave you is the distance from the axis of rotation. The r in you're formula for dV is the distance from the center of the sphere. If you keep your dummy variables straight, the integral should look more like this:
    [itex] I = \rho \int_0^{2\pi} \int_0^\pi \int_0^R(r\,\sin\theta)^2(r^2\,\sin\theta) dr d\theta d\phi [/itex]

    (you can take the density outside of the integral since it is constant)
     
  11. Dec 29, 2003 #10
    This was the problem. Thanks.
     
  12. Dec 29, 2003 #11
    For more info on this please see
    http://www.geocities.com/physics_world/mech/inertia_tensor.htm

    It is incomplete but may be helpful.

    I have not gotten to the part where I calculate the moment of inertia for a sphere of uniform mass density. I'm assuming that the mass density is uniform. I hope to get to that calculation soon. It's not that hard as I recall.
     
  13. Jan 1, 2004 #12

    Doc Al

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    This is a perfectly good approach, but you can save yourself much work by dividing the sphere into cylindrical shells. If the shells have radius "x" (distance from the axis of rotation), then they have cross-sectional area of 2πx dx with a height of 2√(R2-x2). The integral then becomes:
    [tex] I = \rho \int_0^{R} 4\pi x^3 \sqrt(R^2-x^2) dx[/tex]
    which is easy to evaluate with a simple change of variables.
     
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