Moment of Inertia of 2-Spheres Connected by Rod

[bjnartowt]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

We stick an axis of rotation through a sphere, and turn, turn, turn:
${I_{solid{\rm{ }}sphere,{\rm{ }}radius{\rm{ }}{r_0}}} = \int_{sphere} {{R^2} \cdot dm} = ...{\rm{steps I know}}... = {\textstyle{2 \over 3}}M{r_0}^2$
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:
$\begin{array}{c}<br /> {I_{solid{\rm{ }}sphere,{\rm{ translated by }}{\rho _0}}} = \int_{sphere} {{{(R + {\rho _0})}^2} \cdot dm} \\ <br /> = \int_{sphere} {{R^2} \cdot dm} + \int_{sphere} {{\rho _0}^2 \cdot dm} + \int_{sphere} {2{\rho _0}R \cdot dm} \\ <br /> {I_{solid{\rm{ }}sphere,{\rm{ translated by }}{\rho _0}}} = {\textstyle{2 \over 3}}M{r_0}^2 + M{\rho _0}^2 + 2{\rho _0}\int_{sphere} {R \cdot dm} \\ <br /> \end{array}$
In specifics: our problem calls for:
${\rho _0} = 1.5{r_0}$
My question:
$\int_{sphere} {R \cdot dm} = ?$

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…

 

[vela]

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 $|\vec{\rho}_0+\vec{r}|$, where $\vec{\rho}_0$ is the vector from the axis of rotation to the center of the sphere and $\vec{r}$ is the vector from the center of the sphere to the bit of mass dm.

 

[bjnartowt]

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

But the location of the centre of mass would have units of metres, not kilogram*metres as that integral has?

 

[vela]

Right. I should have said it's proportional to the location of the center of mass.

 

[bjnartowt]

Right. I should have said it's proportional to the location of the center of mass.

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

$${x_{CM}} = \frac{{\int {x \cdot \rho (x) \cdot dx} }}{{\int {\rho (x) \cdot dx} }}$$

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

$$2 \cdot {\rho _0} \cdot \int {r \cdot dm} = 0 = {\rm{very yes!}}$$

...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":

$$\int {\rho \cdot dm}$$

...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 : )