Force due to drag at low velocities, is equal to some constant times negative velocity


The viscous damping coefficient equals decay constant divided by 2 times mass

$$\gamma = \frac{c_{2}}{2m}$$

So, is $c_{1}$ the same as $c_{2}$?

How is the drag force related to the viscous damping coefficient, what equation is there to relate them? I think the relationship is linear but I'm not certain.

For context, this is for a mass-spring system inside a beaker of water being damped by the friction of the water.


What you have is not actually fully submersed drag equation, or else it would be correct to handle it as damping.

The drag force on a submersed streamlined object is:$$ F_D=\frac{1}{2} \rho\cdot u^2C_DA $$

  • $F_{D} = $ the drag force

  • $ \rho=$ mass density of the fluid

  • $ u= $ the flow velocity

  • $C_D= $ the drag coefficient (function of the Reynold's number)

  • $A=$ is the crosssectional area

  • $\begingroup$ This is not true at low Reynolds numbers, where the drag is proportional to Re and therefore to $v$. Of course you can include that in your formula by making $C_D$ proportional to $1/u$ as $u$ tends to 0, but that isn't very enlightening. $\endgroup$ – alephzero Aug 2 '19 at 0:12

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