(adsbygoogle = window.adsbygoogle || []).push({}); 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 equations3. 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…

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

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