Terminal Velocity from a graph

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Homework Help Overview

The discussion revolves around determining terminal velocity from a Distance-Time graph related to a coursework experiment involving the viscosity of Golden syrup and the motion of ball bearings dropped through it.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss whether to use a curve or a line of best fit for plotting data points and how to interpret terminal velocity from the graph. Questions arise about the relationship between the graph's shape and the concept of terminal velocity, as well as the effects of viscosity on the motion of the ball bearings.

Discussion Status

There are various interpretations of how to approach the graphing and analysis. Some participants suggest using a line of best fit, while others consider the implications of turbulent flow and the need for a theoretical curve fitting. Guidance has been offered regarding the relationship between velocity and the graph's characteristics, but no consensus has been reached.

Contextual Notes

Participants mention constraints such as the requirement to plot points by hand and the focus on calculating viscosity through gradient analysis. There is also a discussion about the assumptions related to the behavior of the ball bearings in the syrup and the impact of viscosity on their motion.

tazmanmaniac
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Im doing a coursework based on the viscosity of Godlen syrup at the momement. I have got my results and have plotted a Distance Time graph for each of my 4 different ball bearings that were dropped through syrup.

Now my first question is do i have to join the points up with a curve and then do a line of best fit or do i just draw a line of best fit, or do i just draw the curve?

Secondly i need to work out terminal velocity from this graph, does that mean the maximum velocity the ball bearing has reached during my graph? and if so then i will just need to use the tangent to work out the maximum gradient point right?

thanks for the help :smile:
 
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One would think that the ball bearing would reach terminal velocity almost immediately after entering the syrup, unless it was at a high temperature, in which case it will reach terminal velocity after traveling a small distance in the syrup. So you can assume that it traveled between the lines at the actual terminal velocity which will thus be equal to the distance between the lines divided by the time to cover this distance.
 
Just do a line of best fit, ask your teacher for specifics.
The terminal velocity is the point where the velocity is constant i.e. not accelerating.
 
ok, what you said is true it reached terminal velocity straight away but the speed started decreasing slowly which i have put down to turbulent flow. So should i just use my gradient of my line of best fit to work out terminal velocity then because its mostly straight due to only a slight decrease in speed.
 
You should fit your points with a type of curve that the theory predicts and let the computer find the parameters which best fit your curve.

For example, if the theory predicts something like v(t)=v_f+(v_0-v_f)\exp(-t/\tau) then fit the data with A+B\exp(Ct) type dependence and read off the parameters.
 
woah sorry i think that's too advanced for my level atm, we're not even using computers for it lol, such amateurs :D we have to plot the points by hand and our overall aim is to find the viscosity of syrup which is done by calculation and also the gradient of V(t) / radius^2 graph. but to do that i obviously need the temrinal velocity which needs to eb found out by hand, so i ahve two choices i can either work out the gradient of the line of ebst fit on ym distance time graph or i can work out the greatest gradient which is right at the start of the graph.
 
Oh and what is the actual physics behind why its reached terminal velocity so soon? is it because the syrups viscosity is so high?
 
The 'drag force' of a sphere in a viscous fluid (i.e. a liquid) is given by:
F_{drag} = -6\pi\alpha\mu v
where \alpha is the radius of the sphere, \mu is the viscosity and v is the velocity of the sphere. As you are dropping the same ball bearing from the same height (or should be :smile:) you can negate the velocity and sphere radius, so the only varible remaining is viscosity. (You were right, but I thought I'd include the equation so you can reference it in you coursework).

Well, you know that you have reach terminal velocity when you curve becomes a straight line, i.e. when \frac{ds}{dt} = k. Where k is constant and your terminal velocity.

Hope this helps :smile:
 
Leg-end, cheers son.
 
  • #10
I assume that you measured the time that it took the bb to drop between two horizontal lines. Did you decrease the distance between the lines and remeasure (or repositioned the two lines at adifferent height). All one can get from such measurements is a sort of average speed between the lines. I guess that you did investigate this and found that the average speed decreased as it went down?
 

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