Angular dependence of the element of mass of an ellipse

Question

I have an ellipse (a ring, not a disk; its center of mass in $C$) with a constant linear density and mass $m$, with semi-axes $a>b$; $\alpha$ is a dynamical angle describing the orientation of the body in space. $P_1, P_2$ are arbitrary antipodal points on the ellipse, with mass ${\rm d}m$ each; their location is described by the angle $\beta$.

Question: ${\rm d}m$ has to be somehow dependent on $\beta$ (I'll have to sum over all $P_1$, $P_2$ pairs, i.e. integrate over $\beta$), but I'm not sure how. How to relate the mass element ${\rm d}m$ with the angle $\beta$?

Answer 1

The element $dm = \lambda ds$, where $ds$ is the arc length along the ellipse. If the ellipse has equation

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$,

with $a>b$ for sake of definiteness, its parametric representation in terms of $\gamma$, the angle between the major axis and the current point on the ellipse, as seen from the ellipse center, is:

$$x = a \cos\gamma, y = b\sin\gamma$$

from which

$$dm = \lambda \sqrt{a^2 \sin^2\gamma + b^2\cos\gamma} d\gamma$$

From your plot, it seems to me that $\gamma = \alpha + \beta$, from which you should be able to obtain your result. But, I repeat, I am not sure whether $C$, in your diagram, is the ellipse center, and whether the angle $\alpha$ identifies the ellipse major axis. If the answer to both questions is yes, then the above holds.