I am working with a rotary damper (paddle hanging off a shaft attached to the bottom of a rotary table and into a vat of fluid). I know the viscosity of the current Newtonian fluid, I know the flow is laminar, and I know the Q-factor of the damping in this fluid. I need to find a way to get a relation between them to figure out how viscous of a liquid I need to replace the current stuff with to get critical damping.

Is there a simple relationship between Q-Factor and viscosity for a Newtonian fluid based rotary damper in laminar flow?


1 Answer 1


For a damped simple harmonic oscillator, the damping term is linear with the velocity - and the constant of proportionality includes the viscosity of the damping medium.

If the equation of motion is

$$m \ddot{x} + 2 \zeta \omega_0 \dot{x} + \omega^2x = 0 $$

In this equation, $\zeta$ is proportional to viscosity (and contains other terms related to the geometry and surface properties).

The Q is given by

$$Q = \frac{1}{2\zeta}$$

So Q is inversely proportional to viscosity - double the viscosity and Q halves. In an equation, if you have $Q_1$ and you want $Q_2$, then you change the viscosity $\eta$ according to

$$\eta_2 = \eta_1 \frac{Q_1}{Q_2}$$

If you change the regime of the flow (laminar to turbulent, for example) then the above doesn't hold any more. It will also give you trouble if the "memory" of the liquid is long compared to the period of oscillation (in other words, if motion of your system is causing the liquid to start rotating, and it continues to rotate when your disc stops spinning, then you will find the equations will be more complex. I'm not sure I can analyze that - certainly not without some deeper thought.)

  • $\begingroup$ Without a direct equality I can only state that increasing the viscosity should increase the damping (Which I knew from logic anyway) and that logically this should be an inverse proportion for Q. I can't say "if Q = X when eta = Y, then Q = 1/2 when eta = Z", which is what I need to do. I may need added information, but without an equality, even an approximate one, I don't know what that is. Both of your concerns are not issues; the liquid is and stays laminar, and it's own motion has a much faster decay than the motion it is damping. $\endgroup$
    – Elliot
    Jun 27, 2014 at 23:46
  • $\begingroup$ I am not sure I understand the first part of your comment. If flow stays laminar then the answer is "if you need Q to decrease by $\alpha$ then you must increase viscosity by $\alpha$. $\endgroup$
    – Floris
    Jun 28, 2014 at 1:04

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