0
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $$\int_{-3}^{3}\pi\rho y^2dz$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.