I am asked to compute the moment of inertia about the $z$-axis of a rugby ball in terms of its total mass.

A rugby ball surface is given by the ellipsoid:

$$\frac{x^2}{4} + \frac{y^2}{4} + \frac{z^2}{9} = 1$$

The momentum of inertia of the ball about the z-axis is defined as

$ I=\int_V \rho r^2_\perp\space dV = \rho \int_V (x^2+y^2) \space dV$

where $\rho$ is the constant density of the ball and $r_\perp$ is the orthogonal distance from a point to the $z$-axis, $\sqrt{x^2+y^2}$.

I am confused by a few things in this question. Is the moment of inertia the same as the momentum of inertia about a certain axis of a certain object? I also know that the moment of inertia is defined as $I = \int l^2 dM$, where $l$ is the distance of a mass element $dM$ from the axis, am I supposed to somehow relate this equation to the momentum of inertia equation stated above. Very lost here, all help is greatly appreciated!

  • $\begingroup$ Moment of inertia is the resistance to angular acceleration (around a specified axis). It is to angular momentum what mass is to linear momentum. I'm not sure what momentum of inertia is; given that it's defined with the letter $I$, it might just be another name for moment of inertia. $\endgroup$ – Brian Tung Aug 19 '15 at 3:28
  • $\begingroup$ Two different names for the same thing. The more common name is moment of inertia, however. As for $I = \int l^2 dM$, that's exactly the expression you have above that. Note that $l^2 = x^2 + y^2$ for the axis you have selected and $\rho\,dV = dM$. What you need to do is compute that integral over the volume of the ellipsoid. $\endgroup$ – wltrup Aug 19 '15 at 3:52
  • $\begingroup$ are you trying to calculate the moment of inertia of a uniform solid ellipsoid or of a thin uniform surface of the form of an ellipsoid? $\endgroup$ – David Quinn Aug 19 '15 at 8:17
  • $\begingroup$ I think its a uniform solid ellipsoid @DavidQuinn $\endgroup$ – mnmakrets Aug 19 '15 at 15:21
  • $\begingroup$ Oh okay, to compute the integral over the volume, by defining the $x$ and $y$ bounds, will the integral become: $\rho\int_{0}^{2} \int_{0}^{2} (x^2+y^2) \space dx dy $ since the equation for the rugby ball is as stated above? @wltrup $\endgroup$ – mnmakrets Aug 19 '15 at 16:04

The volume has a circular cross section perpendicular to the $z$ axis, so you can consider it to be the volume of revolution formed by rotating the ellipse $\frac{y^2}{4}+\frac{z^2}{9}=1$ about the $z$ axis.

Considering disc-like elements of thickness $\delta z$ and radius $y$, each element has volume $\pi y^2\delta z$, and hence moment of inertia $\frac 12\pi \rho y^4\delta z$. This is quoting the standard formula for the MI of a disc of radius $r$ and mass $m$ about an axis through the centre perpendicular to the plane of the disc, namely $\frac 12 mr^2$.

Therefore you need to evaluate the integral $$\frac 12\int_{-3}^{3}\pi\rho y^4dz,$$ where $\rho$ is the mass per unit volume, and $m=\rho V$.

There is a standard formula for the volume of an ellipsoid, or you can evaluate $$\int_{-3}^{3}\pi\rho y^2dz$$


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