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I have a physics exercise tomorrow and it's about a ball falling through a viscous (olive oil, ricin oil) liquid in a tube.

I know this is not needed there but I would like to ask how to compute the length from the surface at which the speed of liquid becomes constant? The balls have around 1 to 4 mm in diameter and it's made of glass.

From the Reynold's number condition(?) I know to use the bigger ball in ricin and smaller in olive oil. (I guess you can ignore the latter, just please how to compute the length.)

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  • $\begingroup$ What's the criteria for "almost constant"? $\endgroup$
    – JMac
    Mar 8, 2017 at 22:02
  • $\begingroup$ Well it is needed but only approximately (sorry for the misunderstanding) $\endgroup$ Mar 8, 2017 at 22:03
  • $\begingroup$ Well I can only tell you the Czech name for it...... $\endgroup$ Mar 8, 2017 at 22:04
  • $\begingroup$ But how close is "approximately"? Do you need to be withing 10% of terminal velocity, 1% of terminal velocity, 0.00001% of terminal velocity? We can't really start to determine the length if we don't know when we can stop. $\endgroup$
    – JMac
    Mar 8, 2017 at 22:05
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    $\begingroup$ What the above comments were trying to say is that the velocity does not ever become exactly equal to the terminal velocity: the difference of the two decreases to zero exponentially with time. So it does not make sense to speak of a certain distance from the surface unless you specify your accuracy: you can calculate distance it takes for the velocity to be within 1% of the terminal velocity, within 0.1% of the terminal velocity etc. $\endgroup$ Mar 8, 2017 at 22:16

4 Answers 4

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What you do is solve the equation $$\frac{4}{3}\pi a^3 \rho \frac{dv}{dt}= \frac{4}{3}\pi a^3g(\rho - \sigma)-6\pi \eta a v$$ subject to the initial condition v = 0 at t = 0. Of course, after you solve for the velocity as a function of time, you then integrate the equation $$\frac{dx}{dt}=v(t)$$ to get the distance as a function of time.

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There is an expression on this page below for terminal velocity through oil

Terminal velocity through oils

$6\pi \eta a \upsilon_0 = 4/3 a^3g(\rho - \sigma)$

where $\eta$ = viscosity
α = radius of the ball bearing
$\upsilon_ο$ = terminal velocity
$g$ = gravitational field strength
$\rho$ = density of the bearing material
$\sigma$ = density of the liquid

Thanks to docscience for this caveat regarding the above equation:

Density, viscosity, probably also can vary with temperature. So don't be surprised if experimental results are different on different days. Also there could be a wall effect if the clearance between the ball and tube is tight, on the order of the boundary layer. I'm guessing formula above is for ball far away from boundaries, and not spinning.

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    $\begingroup$ density, viscosity, probably also can vary with temperature. So don't be surprised if experimental results are different on different days. Also could be a wall effect if the clearance between the ball and tube is tight, on the order of the boundary layer. I'm guessing formula above is for ball far away from boundaries, and not spinning. $\endgroup$
    – docscience
    Mar 8, 2017 at 22:50
  • $\begingroup$ @docscience Thank you, I put your comment in the post, in case of it's deletion later. $\endgroup$
    – user146020
    Mar 8, 2017 at 23:07
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You could solve the differential equation given to you by @ChesterMiller.
This will give you an equation of velocity against time which will show you that in theory the terminal velocity is never reached.
However that does not matter because you can only measure the velocity to a limited accuracy so you can estimate the time for the velocity to reach the terminal velocity minus the accuracy of your velocity measurement.


However might it not be easier to find out whether or not the terminal velocity had been reached whilst doing the experiment?
It would also enable you to get a better estimate of the error in your velocity measurement.

To do this have a series of equally spaced markers, say four?, starting a little way from the surface of the liquid and finishing a little way from the bottom of the liquid (there is an end effect because the liquid is not of infinite extent as Stokes assumed) and using a stopwatch with a lap timing facility time the spheres as they pass the markers.
Your observations should show you whether or not the terminal velocity has been reached and also give you an indication of the accuracy of your measurement of velocity.

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  • $\begingroup$ The times at which markers are passed could be plotted against the distance fallen. Fitting an appropriate curve would give an accurate prediction of terminal velocity even if it had not yet been reached. $\endgroup$ Mar 9, 2017 at 17:10
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For light objects falling through viscous media with laminar flow, the drag force is given by Stokes' Law $F=6\pi \eta av$. The equation of motion of a ball falling with speed $v$ is :
$m\frac{dv}{dt}=\frac43\pi\rho a^3 v\frac{dv}{dx}=\frac43\pi a^3(\rho-\sigma)g-6\pi\eta av$.
Here $\rho, \sigma$ are the densities of the ball and the fluid which it displaces, $a$ is radius, and $\eta$ is viscosity.

Using the substitutions $v=uv_0$ and $x=x_0y$ where $v_0=\frac{2(\rho-\sigma)ga^2}{9\eta}$ is the terminal velocity (found by setting $\frac{dv}{dx}=0$) and $x_0=\frac{(\rho-\sigma)g}{\rho v_0^2}$, the equation of motion can be written in dimensionless form as
$u\frac{du}{dy}=1-u$.
This is separable and can be integrated immediately. The initial condition is $y=0, u=0$ so the solution is
$y=\ln(\frac{1}{1-u})-u$.

As $v \to v_0$ then $u \to 1$ so $y \to \infty$. If we require $v$ to reach $0.99v_0$ then $u=0.99$ so
$y=\ln(100)-0.99=3.615$
ie the ball reaches 99% of terminal velocity after a distance of $3.615x_0$.

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