# Circular motion at $Re\ll 1$

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Let's assume a bead of mass $m$ is moving on a circular trajectory in a fluid, and that the typical velocity, the radius of the bead $a$ and the viscoisity of the fluid $\eta$ are such that we are in a regime with $Re\ll1$. What are the driving forces necessary to sustain uniform circular motion with radius $R$ and angular velocity $\omega$?

The derivative of the velocity vector can be written as

$$\frac{d\vec{v}}{dt}=\frac{d\left(\omega R\hat{u}_{\theta}\right)}{dt}=\frac{d\omega}{dt}\hat{u}_{\theta}+\frac{d\hat{u}_{\theta}}{dt}\omega=F_{\theta}\hat{u}_{\theta}(t)+F_{R}\hat{u}_{R}(t)$$

where $\hat{u}_{\theta}$ is the tangent vector at an angle $\theta$.

In order to make the first term of the RHS vanish, I write the force balance tangentially to the circle, in which I have an unknown driving force and the hydrodynamic drag:

$$F_{\theta}=F_{\theta}^{(driving)}+F_{\theta}^{(drag)}=0$$

In this regime I can write

$$F_{\theta}^{(drag)}=-6\pi\eta av=-6\pi\eta a\ \omega R$$

$$\Longrightarrow F_{\theta}^{(driving)}=6\pi\eta aR\ \omega$$

and this gives me the tangential component of the driving force needed.

What about the normal component? In this regime, does $F_{R}=-\omega^{2}R$ make any sense? Does the hydrodynamic drag have a component along $\hat{u}_{R}(t)$?

No. The hydrodynamic force is proportional to the slip velocity ($\pmb{w} := \pmb{u} - \pmb{v}$, where $\pmb{u}$ is the local fluid velocity; that is, the velocity of the background fluid flow as extrapolated to the particle's centre). This means that if the particle is held in a circular orbit it is because there is some additional force acting in the normal direction to avoid it spiraling outward, since $\pmb{w}$ is tangent to the orbit. Once the particle starts its outward spiral, there will be a normal component to $\pmb{w}$, but only then.
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