Coursework help - Stokes Law equation + graphs

In summary: I'll give it a go and see if I can get a linear relationship. In summary, Alex's experiments showed that radius is proportional to velocity and that the slope is equal to g/18μ. He tried to find the equation for the line and determined that r2/v equals g/18μ. He also plotted v/r2 vs Δρ/μ and found that the data all lie on a straight line.
  • #1
astro111
4
0

Homework Statement


Hi there.
For my A2 physics coursework I have been doing an experiment into stokes law, in which I dropped ball bearings of various diameters into a tube filled with a liquid, worked out their terminal velocities and then used Stokes Law to calculate the viscosity of the liquid. My results have been consistent and in line with my expectations. I then did another experiment, the reverse in a way, in which I used my calculated values for the viscosity of the liquids to find the diameter of ball bearing before measuring it to ensure I was gaining accurate results.

However now I am at the stage of writing up my experiment report and I am a little unsure of what graphs to plot. I've been reading and it sounds as though I should plot a graph of radius2 over velocity, however I am unsure as to what this would show. For example, would the gradient show viscosity?

Homework Equations


The equations I've been using are μ=(d2g(Pρ-Ps))/18ν, where (Pρ-Ps) is the density of the ball bearing minus the density of the fluid, d is the diameter of the ball bearing, g is equal to 9.81 and v is the terminal velocity of the ball bearing. I believe this equation can also be written as μ=(2/9)*((Pρ-Ps)/v)*gr2, where r is the radius of the ball bearing.

The Attempt at a Solution


I have tried to rearrange the second equation in order to get r2/v, as the graph would be I think? leaving me with 9μ/(2(Pρ-Ps)g) = r2/v, which I think would mean that r2 ∝ 1/v?
However I am confused as to what my graph would show (i.e what the gradient would be equal to etc) and whether I have rearranged the equation correctly.

As a side note, I am unable to ask my teacher presently as I am currently off school ill and ideally I would like to have finished my report when I go back in.
Any help would be much appreciated, thank you!
Alex
 
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  • #2
Hello Alex, welcome to PF :) Hope you get well soon.

It's good practice to plot experimental results by using the y-axis for what you measure and the x-axis for what you vary. I suppose you measured the time it took the balls to fall from one level to a lower one. And you varied the diameter and the material of the balls.

Another good practice is to try to plot results in such a way that a linear relationship is achieved (at least according to the prediction of the theory). Sometimes that means you need to square, to take a square root or a logarithm.

If you look at your expression, you see you can write ##\displaystyle v = {d^2 \; \Delta\rho g \over 18\mu}## which looks like ##v = C\; d^2##, so you expect a straight line if you plot something that is proportional to v as a function of something like d2.

Here, v = distance / time means that 1/time is a good one to plot vertically. The distance ##\Delta y## is the same in all cases (right?), so you divide that out in a later stage (*).

For the horizontal axis, asw we saw, diameter d doesn't yield a linear relationship in the theory, but d2 does.

With three materials you hope to get get three more or less straight lines, each with slope ##\displaystyle\Delta\rho g \over 18\mu \Delta y##. Draw the best possible straight lines to determine best values for these slopes.

As a next step, you can claim you measured three slopes as a function of ##\Delta \rho##, so you can plot these slopes vertically and ##\Delta \rho## horizontally. That line is expected to have a slope ##\displaystyle g \over 18\mu \Delta y##. Again, draw the best possible straight line to determine its values. Finally, calculate ##\mu##.

(*) because distance is the same for all measurements, the error in the slopes doesn't get smaller if you use more different ball sizes to get a better line. (The error in the distance is 'systematic' and not statistical)
 
  • #3
You have the right idea. Consider plotting v/r2 vs Δρ/μ (so that the left hand side is all geometric and kinematic, and the right hand side is all physical properties). The data should all lie on a straight line.

Chet
 
  • #4
Thank you to both of you, you've helped me greatly!
 
  • #5


Hi Alex,

It sounds like you have conducted a thorough and well-designed experiment for your A2 physics coursework. Stokes Law is a fundamental equation in fluid mechanics and your application of it to calculate the viscosity of a fluid is a great way to demonstrate your understanding of the concept.

In terms of plotting graphs, there are a few options you could consider. One option is to plot a graph of terminal velocity (v) against ball bearing radius (r). This graph would show the relationship between these two variables and could help you to determine whether there is a linear relationship between them. If there is, then you could use the equation v=2/9 * (d^2g(Pρ-Ps))/μ to calculate the viscosity of the fluid.

Another option is to plot a graph of terminal velocity (v) against the inverse of ball bearing radius (1/r). This graph would also show the relationship between these two variables and could help you determine if there is a linear relationship. If there is, then you could use the equation v=2/9 * (d^2g(Pρ-Ps))/μ to calculate the viscosity of the fluid, with the added benefit of having a straight line on your graph.

As for your proposed graph of r^2/v, it is not a commonly used graph in this context and may not provide the most informative results. However, if you are interested in exploring this relationship, you could plot a graph of r^2 against 1/v and see if there is a linear relationship. The gradient of this graph would then be equal to 2/9 * (d^2g(Pρ-Ps))/μ.

I hope this helps and good luck with your report! If you have any further questions, feel free to ask your teacher or seek help from a knowledgeable peer. As a scientist, it's important to seek guidance and collaborate with others in order to improve your understanding and results. Keep up the good work!
 

FAQ: Coursework help - Stokes Law equation + graphs

1. How do I calculate the terminal velocity using the Stokes Law equation?

The Stokes Law equation is used to calculate the terminal velocity of a spherical object falling through a fluid. The equation is Vt = (2/9)(g)(ρs - ρf)(r^2)/η, where Vt is the terminal velocity, g is the acceleration due to gravity, ρs is the density of the object, ρf is the density of the fluid, r is the radius of the object, and η is the viscosity of the fluid. Plug in the values for each variable and solve for Vt to calculate the terminal velocity.

2. What is the relationship between terminal velocity and the radius of the object?

According to the Stokes Law equation, the terminal velocity is directly proportional to the square of the radius of the object. This means that as the radius of the object increases, the terminal velocity also increases. This relationship is important to understand when analyzing the behavior of objects falling through a fluid.

3. How does the density of the fluid affect the terminal velocity?

The density of the fluid has a direct impact on the terminal velocity of an object falling through it. The greater the difference between the density of the object and the density of the fluid, the greater the terminal velocity will be. This is because the larger the difference in density, the greater the buoyant force acting on the object, and the faster it will fall through the fluid.

4. What is the significance of the viscosity in the Stokes Law equation?

The viscosity of the fluid is a measure of its resistance to flow. In the Stokes Law equation, a higher viscosity means that the fluid is more resistant to the motion of the falling object, resulting in a lower terminal velocity. This is why objects fall faster through less viscous fluids, such as air, compared to more viscous fluids, such as honey.

5. Can the Stokes Law equation be used for objects other than spheres?

The Stokes Law equation is specifically designed for spherical objects falling through a fluid. It can be used for other shapes, but only if the object has a similar cross-sectional area as a sphere. For objects with irregular shapes, other equations or methods may need to be used to calculate the terminal velocity.

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