# How to calculate the inertia tensor of a spherical cap?

In this question, an attempt is made at calculating the diagonal elements of the inertia tensor of a homogeneous spherical cap, where the $$z$$-axis is the symmetry axis. The mass moment of inertia about the $$z$$-axis is expressed by:

$$M_{zz}=\rho\int_0^{2\pi} \int_{R-h}^R \int_{0}^{\sqrt{R^2-z^2}} r^3 dr dz d\theta$$

where $$\rho$$ is the density, given by:

$$\rho = \frac{m}{\frac{1}{3} π h^2 (3 R - h)}$$

$$R$$ is the radius of the sphere, and $$h$$ and $$m$$ are the height and mass of the cap. The triple integral equation solves as:

$$M_{zz}=\frac{mh}{10(3R - h)}(3h^2 - 15hR + 20R^2)$$

which is correct. However, the following expression is given for the mass moment of inertia about the $$x$$-axis or $$y$$-axis (in the accepted answer to this question):

$$M_{xx}=M_{yy}=\rho\int_0^{2\pi} \int_{R-h}^R \int_{0}^{\sqrt{R^2-z^2}} r(r^2 \cos^2\theta+z^2) dr dz d\theta$$

which solves as (see WolframAlpha):

$$M_{xx}=M_{yy}=\frac{m}{20(3R - h)}(-9h^3 + 45h^2R - 80hR^2 + 60R^3)$$

This is seemingly not correct. If we plug in $$R=5$$, $$h=2$$ and $$m=4.28\times10^{5}$$, we get:

$$I=\begin{bmatrix}7.1246&0&0\\0&7.1246&0\\0&0&2.3836\end{bmatrix} \times10^{6}$$

According to CATIA's "Measure Inertia"-function, this should be:

$$I=\begin{bmatrix}1.2896&0&0\\0&1.2896&0\\0&0&2.3836\end{bmatrix} \times10^{6}$$

For reference (don't mind the definition of the axes here): The question is:

What is the correct expression for the mass moment of inertia about the principal $$x$$-axis/$$y$$-axis (red/green lines in reference figure)?

The geometric centroid (the origin of the axes system) for a spherical cap is given by:

$$z=\frac{3(2R-h)^2}{4(3R-h)}$$

Edit:

Thanks to probably_someone, the answer can be derived as:

$$M_{xx}=M_{yy}=\frac{mh}{80(h-3R)^3}(-9h^3 + 72h^2R - 220hR^2 + 240R^3)$$

The moment of inertia of an object is only defined relative to a particular choice of axes of rotation. You are using a different choice of $$x$$ and $$y$$ axes than the original question.

In the original question, the $$x$$ and $$y$$ axes passed through the center of the whole sphere, meaning they passed through a point on the $$z$$-axis that is below the bottom of the cap. CATIA is calculating the moments of inertia for $$x$$ and $$y$$ axes that pass through the center of mass of the cap. Fortunately, there is an easy way to convert between these two choices: the parallel axis theorem. The theorem states:

Suppose a body of mass $$m$$ has moment of inertia $$I_0$$ about an axis passing through its center of mass. Then the moment of inertia $$I$$ about another axis, parallel to the first and displaced a distance $$d$$ from the center of mass, is given by: $$I=I_0+md^2$$

Applying this to our situation, it's clear that the CATIA calculation gives you $$I_0$$, and we already have $$m$$. Since you defined the center of the whole sphere as the origin, the distance $$d$$ that the axis should be displaced is equal to the $$z$$-coordinate of the center of mass, namely:

$$d=\frac{3(2R-h)^2}{4(3R-h)}$$

So the moment of inertia about $$x$$ and $$y$$ axes passing through the center of the sphere will be:

$$I=I_0+m\frac{9(2R-h)^4}{16(3R-h)^2}$$

Plugging in $$m=4.28\times 10^5$$, $$R=5$$, $$h=2$$, and $$I_0=1.2896\times 10^6$$, we get $$I=7.1246\times 10^6$$, which perfectly agrees with the calculation that you did using the other question.

• Thanks a lot. Didn't notice the origin of the original question was at the centre of the sphere. What is good practice? Should I edit my original question to correct for this mistake? Feb 5 '19 at 18:57
• @woeterb I think it's fine to leave it as is, as the answer specifically references the oversight in the question. It might not make much sense to future readers if you edit the question at this point. Feb 5 '19 at 19:07