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 $\hat{u}_{R}(t)$?


One could expect a relationship of the type $F_{R}=-m\omega^{2}R$ to be valid in this context, without any dependence on the hydrodynamic drag. However, this suggests that the inertia of the bead (through its mass) plays a role despite the fact that the inertia of the fluid (via the low $Re$ approximation) doesn't. Is this correct? Also, at low $Re$, torques are linear in angular velocities (is this valid in this case as well? Or only for the intrinsic rotation of an object around an internal axis?)


1 Answer 1


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

Circular motion with drag

Because $\text{Re}$ is very small, Stokes' drag is relevant and a force $\mathbf{F_d}$ needs to be applied in the sense of rotation, to maintain constant tangential velocity $v$:

$$\mathbf{F_d}=6\pi\eta a\mathbf{v}=6\pi\eta a \mathbf{\omega} R$$

But to maintain the centripetal acceleration $\mathbf{a_c}$ a centripetal force $\mathbf{F_c}$ is required:


$\mathbf{F_c}$ is independent from drag and must be supplied separately.

  • $\begingroup$ I am a bit doubtful because usually at low $Re$ inertial effects are neglected, and forces are linear in the velocities, with no accelerations intervening in the dynamics. The relationship you suggest, which is the usual one, seems to contradict this. Also, shouldn't the torque be linear in the angular velocity? $\endgroup$ Commented Apr 10, 2019 at 18:15
  • $\begingroup$ @usumdelphini: Hi, thanks for your comment. No, $F_c \propto \omega^2$, see e.g. here: en.wikipedia.org/wiki/Centripetal_force#Formula $\endgroup$
    – Gert
    Commented Apr 10, 2019 at 18:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.