# Angular speed of an rotating ellipsoid

Consider an ellipsoid described the equation in cartesian coordinates $$(x,y,z)$$ $$\frac{x^2}{a^2} + \frac{y^2}{a^2} + \frac{z^2}{c^2} = 1, \quad \text{where} \quad a < c.$$ This ellipsoid is usually called the prolate spheroid.

Now we introduce rotation around the $$z$$ axis,the angular velocity $$\omega$$ at the point $$x = a,y = a, z= 0$$ is $$\omega = v/a$$. Can we find the formula for angular velocity at any point on the ellipsoid?

What my attempt was: I thought that the angular velocity should be dependent only a the distance from the axis $$z$$, which I denoted $$r = \sqrt{x^2 + y^2}$$. From this i would get $$\omega = \frac{v}{r} = \frac{v}{\sqrt{x^2 +y^2}},$$ where of course $$x$$, $$y$$ are points on the ellipsoid.

But this doesn't seem quite right with my intuition, since the angular velocity as a function of $$r$$ will be the same in a sphere of radius $$a$$. Did I make a mistake or is my intuition wrong?

• Intuition seems wrong. The whole point of the concept of angular velocity of points belonging to a rigid body is that is is the same for all those points. Hint : the points $x,y$ at $z\not=0$ lie on a sphere centered around the $z$-axis. Their $\omega$ is the same as for the points at $z=0$. Their radius $r$ and their linear velocity $v$ is different. Commented Apr 8, 2022 at 17:03