# Does an aluminum ball travel at 416.4 meters per second through water?

If you were to measure the viscosity of a fluid you would do the following experiment https://www.wikihow.com/Measure-Viscosity

The formula for dynamic viscosity is the following one:

$n=\frac{2gr^{2}(p_s-p_l)}{9v}=\frac{ms^{-2}m^{2}(kgm^{-3})}{ms^{-1}}=kgm^{-1}s^{-1}= Pa·s$

where n is dynamic viscosity of the liquid in Pa s, g is gravity ($9.81{ms^-}^2$), r the radius of the ball in meters, $p_s$ density of sphere, $p_l$ density of liquid and v the velocity at which the ball travels through the fluid.

This source and many more state that the dynamic viscosity of water is $8.90*{10^-}^4$ Pa s https://www.engineersedge.com/physics/water__density_viscosity_specific_weight_13146.htm

If we plug this value for viscosity and try to find a value for the velocity at witch the sphere travels through the water (in this case). Then we come up with the following.

$8.90*{10^-}^4=\frac{2(9.81)r^{2}(p_s-1000)}{9v}$ (Density of water is 1000kg per meter cube)

Let's say that we drop an aluminum ball of 0.01m radius. Aluminum has a density of 2700kg/$m^3$

$8.90*{10^-}^4=\frac{2(9.81)(0.01)^{2}(2700-1000)}{9v}$

The final value for $v$ is 416.4 ${ms^-}^1$

Is the sphere really traveling through water at that high speed. Is the formula wrong? Or is it my math or what? I'm so confused.

• I didn't check on your math. But a ball creates a 3-D wake behind it as it travels through the fluid. Pressure drag plays a role and so does viscous drag. There will be separation behind the ball if it's speed is too large. I don't know how a formula would include these effects to obtain a value for viscosity. – jjack Dec 10 '17 at 10:48
• @jjack So would you say that this formula to calculate viscosity is not right at all (not accurate)? – Matthew Dec 10 '17 at 10:51
• Sorry, I can't give you a good answer. You'd have to wait for a fluid dynamics guy to answer it. – jjack Dec 10 '17 at 10:55
• read this Wikipedia article on viscosity: en.m.wikipedia.org/wiki/Viscosity, be aware that there seems to be difference between dynamic and kinematic viscosity. There's also a short section on measuring kinematic viscosity. – jjack Dec 10 '17 at 11:00
• @jjack the one I am talking about is dynamic viscosity, check the units they are Pa s as dynamic, while kinematic is m^2/s. Thanks for the article but I already read it so many times :P – Matthew Dec 10 '17 at 11:11

However Stokes' law is not applicable in your example the test for this being to evaluate a parameter called Reynolds' number $R_{\rm e} = \dfrac {\rho_{\rm fluid}R_{\rm shere}v_{\rm terminal}}{\eta_{\rm fluid}}$ which in your example is approximately four million far in excess of the upper limit for Stokes' law to be applicable which is about ten.