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Homework Help: Moment of inertia of two solid spheres, radii each r[0], attached by massless rod.

  1. Aug 18, 2010 #1
    1. The problem statement, all variables and given/known data

    What is: Moment of inertia of two solid spheres, radii each r[0], attached by massless rod which is also of length r[0], so that the distance between the two solid spheres' centres is "3*r[0]"?

    The axis of rotation could either be 1) through both spheres and along the connecting-rod or 2) through the connecting rod so that the two spheres "orbit" about in a radius of 1.5*r[0]. I know how to get either by the parallel and perpendicular axes theorems.


    2. Relevant equations 3. The attempt at a solution

    We stick an axis of rotation through a sphere, and turn, turn, turn:
    [itex]{I_{solid{\rm{ }}sphere,{\rm{ }}radius{\rm{ }}{r_0}}} = \int_{sphere} {{R^2} \cdot dm} = ...{\rm{steps I know}}... = {\textstyle{2 \over 3}}M{r_0}^2[/itex]
    Now: imagine translating: instead of rotation through the sphere’s symmetry axis, let’s rotate through that point on the massless-connecting rod. In general:
    [itex]\begin{array}{c}
    {I_{solid{\rm{ }}sphere,{\rm{ translated by }}{\rho _0}}} = \int_{sphere} {{{(R + {\rho _0})}^2} \cdot dm} \\
    = \int_{sphere} {{R^2} \cdot dm} + \int_{sphere} {{\rho _0}^2 \cdot dm} + \int_{sphere} {2{\rho _0}R \cdot dm} \\
    {I_{solid{\rm{ }}sphere,{\rm{ translated by }}{\rho _0}}} = {\textstyle{2 \over 3}}M{r_0}^2 + M{\rho _0}^2 + 2{\rho _0}\int_{sphere} {R \cdot dm} \\
    \end{array}[/itex]
    In specifics: our problem calls for:
    [itex]{\rho _0} = 1.5{r_0}[/itex]
    My question:
    [itex]\int_{sphere} {R \cdot dm} = ?[/itex]

    I have a feeling I’m going down a blind-alley, though. This is from Giancoli’s book, and I think there’s a shorter way to do the calculation…
     
  2. jcsd
  3. Aug 18, 2010 #2

    vela

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    Re: Moment of inertia of two solid spheres, radii each r[0], attached by massless rod

    That integral would be 0. The first moment is the location of the center of mass.

    You made a mistake in setting up your initial integral. The distance from the axis of rotation to the mass dm isn't simply (ρ0+R); it's [itex]|\vec{\rho}_0+\vec{r}|[/itex], where [itex]\vec{\rho}_0[/itex] is the vector from the axis of rotation to the center of the sphere and [itex]\vec{r}[/itex] is the vector from the center of the sphere to the bit of mass dm.
     
  4. Aug 19, 2010 #3
    Re: Moment of inertia of two solid spheres, radii each r[0], attached by massless rod

    But the location of the centre of mass would have units of metres, not kilogram*metres as that integral has?
     
  5. Aug 19, 2010 #4

    vela

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    Re: Moment of inertia of two solid spheres, radii each r[0], attached by massless rod

    Right. I should have said it's proportional to the location of the center of mass.
     
  6. Aug 19, 2010 #5
    Re: Moment of inertia of two solid spheres, radii each r[0], attached by massless rod

    Oh! I see it now. What you're describing would be the numerator of the mass-centre formula:

    [tex]{x_{CM}} = \frac{{\int {x \cdot \rho (x) \cdot dx} }}{{\int {\rho (x) \cdot dx} }}[/tex]

    ....right? If so, I definitely see why:

    [tex]2 \cdot {\rho _0} \cdot \int {r \cdot dm} = 0 = {\rm{very yes!}}[/tex]

    ...because we have chosen the origin of our coordinate system to be the centre-of-mass of the rotating body (I think without me realizing it), and so the centre-of-mass would have a zero that would kill it before the "missing":

    [tex]\int {\rho \cdot dm} [/tex]

    ...would effect its dividing power on the already-zero-by-coordinate system denominator.

    Okay, you prolly got a little more about how my brain works than you wanted to, but I think you answered my question. (Or maybe, worse yet, I'm wrong and think I'm right). Thanks vela : )
     
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