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Graph of Tension force vs. the square of the speed

by rstat1
Tags: force, graph, speed, square, tension
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rstat1
#1
Apr27-13, 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
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FireStorm000
#2
Apr27-13, 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.
rstat1
#3
Apr27-13, 04:19 PM
P: 4
Quote Quote by FireStorm000 View Post
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.
My problem was determining what the slope represented. My graph and associated data is in the attached excel file.

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)
Attached Files
File Type: xlsx Book1.xlsx (14.0 KB, 1 views)

FireStorm000
#4
Apr27-13, 04:26 PM
P: 169
Graph of Tension force vs. the square of the speed

My problem was determining what the slope represented. If you still want my graph/data then in its in the attached excel file.
Alright, I took a look at your data and I think I see why you're having trouble. Can you identify the independent and dependent variables in your experiment? It looks like you varied something besides velocity.
rstat1
#5
Apr27-13, 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.
FireStorm000
#6
Apr27-13, 05:07 PM
P: 169
You should find that the slope has something to do with mass
rstat1
#7
Apr27-13, 05:14 PM
P: 4
Quote Quote by FireStorm000 View Post
You should find that the slope has something to do with mass
Well yea I knew that, since the graph is tension force vs speed. I had thought you asked what else I varied.
FireStorm000
#8
Apr27-13, 05:44 PM
P: 169
Quote Quote by rstat1 View Post
Well yea I knew that, since the graph is tension force vs speed. I had thought you asked what else I varied.
Yeah, I was wondering why your graph was all over the place.

The equations you're looking for is:
Actp = r * v 2
and so
Fctp = m * r * v2

Now, I assume r is constant in your setup, so the mass is what changes the ratio of F to V2
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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