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Graph of Tension force vs. the square of the speed 
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#1
Apr2713, 03:59 PM

P: 4

Working on a physics lab report regarding uniform circular motion, one of the questions asks to graph the Tension Force vs. the square of the speed. Another part of the same question asks what the slope of the graph represents, which is where I'm stuck.
1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 


#2
Apr2713, 04:13 PM

P: 169

Your graph, or a table of values would definitely qualify as known data, so you should post those. If you did any theory work before this, you probably also have at least one equation you think is related (IE: one with an acceleration or force term, and a V^2 term). We want to help you work through the problem rather than just giving the answer.



#3
Apr2713, 04:19 PM

P: 4

As far as equations, I don't even remotely know which would be useful: I've got one for calculating the speed (v = 2pi/T) and one for one for Tension force (M*g) 


#4
Apr2713, 04:26 PM

P: 169

Graph of Tension force vs. the square of the speed



#5
Apr2713, 04:41 PM

P: 4

I only included the data used to make the graph because I thought that was all that was relevant as far as my question was concerned. The experiment focused on timing how long various masses (starting from 50g and going to 135g) would take to make 20 revolutions. So yes I did vary something besides the velocity, the mass was also varied.



#6
Apr2713, 05:07 PM

P: 169

You should find that the slope has something to do with mass



#7
Apr2713, 05:14 PM

P: 4




#8
Apr2713, 05:44 PM

P: 169

The equations you're looking for is: A_{ctp} = r * v ^{2} and so F_{ctp} = m * r * v^{2} Now, I assume r is constant in your setup, so the mass is what changes the ratio of F to V^{2} Because of that, the slope of any given line segment should show the relative mass between the two data points, if that makes any sense. 


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