# Angle of rotation of an ellipsoid in a linear shear flow field

I am modeling the motion of an ellipsoid in a linear shear flow field. The ellipsoid is rotating about its shortest semi-principal axis which I have designated the $z$-axis in the body-fixed frame, and so we can consider its motion in the $x$-$y$ plane only. I have derived an expression for $\phi(t)$, the angle of rotation of the ellipsoid in this plane, as a function of time. $\phi$ has a range of $\pi$ and is discontinuous. This makes sense since

$$\phi(t) = \arctan\left( \sqrt{\frac{1+r}{1-r}}\tan \left(\frac{\gamma}{2} \sqrt{1-r^2} t\right)\right).$$

where $0 < r < 1$ is a rational expression involving principal axes and $\gamma$ is the constant fluid shear rate along the $y$ axis.

Physically, the ellipsoid should rotate through the entire $2\pi$ (it is tumbling in the flow). The angle $\phi$ is in fact the Euler angle lying in this plane, which should have a range mod $2\pi$.

Is it possible to identify the angle $-\pi/2$ with $\pi/2$ since the ellipsoid is symmetric? Which is to say, is it possible that my function can actually represent an ellipsoid rotating through a full $2 \pi$, or do I have to assume that this function describes discontinuous motion through $\pi$ rads (and is thus wrong)?

For brevity I have excluded a discussion of the derivation of $\phi$, which I have obtained by following two published results. If it would be useful I can include a derivation and the references.

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