Finding experimental Centripetal force from F=4pi^2rM/T^2

In summary, the conversation discusses an experiment involving a centripetal apparatus and the determination of centripetal force using a slope of T^2 vs r^2. The slope was found to be 16.5 s2/m, but there is confusion on how to use this to determine the centripetal force. The plot of the relationship does not go through the origin, and there is a desire to find a value for F using the equation F=4pi^2rM/T^2.
  • #1
HexRei

Homework Statement


I have a second problem I need assistance with but I'll put it in its own thread.

Used centripetal apparatus to run trials with varying radii but constant mass and calibration weight (mg). I have already plotted T^2 vs r^2 and got a slope of 16.5 from the best fit. I now have to solve the equation for T^2 and then use it to determine the centripetal force. I know the slope is T^2/r, but I don't know how to use that.

Mass M = .138kg
Calibration weight= .050 kg so measured force is .490.
slope = 16.5

Homework Equations


F=4pi^2rM/T^2

The Attempt at a Solution


T^2=(4pi^2rM)/F I've tried rearranging it many ways but I can't figure out how to use this to determine centripetal force from the slope.

T^2=(5.92M)/F
T^2*F= 5.92M
(T^2*.490)/5.92=M

.490/5.92=M/T^2
2.92*T^2=M

New try.

T^2=4pi^2rM/F
T^2F= 4pi^2rM
(T^2F)/r=4pi^2M

Do I need to get the slope onto one side of the equation?
 
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  • #2
Hi HexRei, :welcome:
HexRei said:
Used centripetal apparatus
It would help if you described ( or illustrated) this a little more: how is the circular trajectory brought about, what is T, what is r, etc.

PS formulas become a little more legible if you use the superscript button

From your ##T^2 = {4\pi^2rM\over F}## I would expect a plot of ##T^2## versus ##r## to give a straight line with a slope ##4\pi^2 M\over F## (very important: check this has the right dimension of ##s^2/m##)

So I wonder what your plot with a slope of 16.5 s2/m2 looks like ...
 
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  • #3
Thanks, I'll start using the proper notation.

The apparatus was spun by hand (not particularly precise unfortunately) and we did not measure T. The r varied from .15 to .19 during the five trials. This is all the elements we were measuring or attempting to derive.

My plot for that was a nearly perfect 45 degree angle on positive y and x. Points used to derive slope:

y2=1.74 y1=1.08
x2=.19 x1=.15
 
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  • #4
HexRei said:
we did not measure T
Well, then how can you make a plot of ##T^2## versus ##r## :rolleyes: ?

You also want to force yourself to use dimensions: you make a plot with points (correct me if I am wrong)
r1 = 0.15 m, ##\quad## T21 = 1.08 s2
r2 = 0.19 m, ##\quad## T22 = 1.74 s2
and (I hope) a few more points. Because otherwise: how can you conclude that the relationship is indeed a linear one ?

can you post the plot ?
 
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  • #5
BvU said:
Well, then how can you make a plot of ##T^2## versus ##r## :rolleyes: ?

You also want to force yourself to use dimensions: you make a plot with points (correct me if I am wrong)
r1 = 0.15 m, ##\quad## T21 = 1.08 s2
r2 = 0.19 m, ##\quad## T22 = 1.74 s2
and (I hope) a few more points. Because otherwise: how can you conclude that the relationship is indeed a linear one ?

can you post the plot ?

Doh sorry, I thought you meant tension force. I'll try to remember to include my dimensions.

We did measure T for each trial but they were just used to get a T2 for each point pairing on the plot.

Yeah, we used five sets of points, but I used those two for the slope. It was a nearly perfectly linear positive y and x, I'm pretty confident in the slope. My paper graph looks terrible in a photo but here's the plot in excel:
hBcX0m

hBcX0m
hBcX0m
hBcX0m
 

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  • #6
Good. Along the axis I see (dimensionless) numbers. One way to make them valid is to add a legend: ##T^2/s^2## for the vertical axis and ##r/m## for the horizontal. This is a bit puritan: ##\ T^2\ \ (s^2)## and ##\ r\ \ (m) ## is often used also.

The plot looks good and shows a linear relationship. So not the
HexRei said:
I have already plotted T^2 vs r^2 and got a slope of 16.5 from the best fit
in post # 1 but T2 vs r -- as suggested by the equation. And the slope is 16.5 s2/m. Or is it ?

From your two points I get ##{\Delta T^2\over \Delta r} = {0.66\over 0.04} = 15.5\ ## s2/m

There is one big problem remaining, though: how do you explain that the plot of the relationship clearly does not go through the origin as the formula says :wideeyed: ?
 
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  • #7
BvU said:
Good. Along the axis I see (dimensionless) numbers. One way to make them valid is to add a legend: ##T^2/s^2## for the vertical axis and ##r/m## for the horizontal. This is a bit puritan: ##\ T^2\ \ (s^2)## and ##\ r\ \ (m) ## is often used also.

The plot looks good and shows a linear relationship. So not the
in post # 1 but T2 vs r -- as suggested by the equation. And the slope is 16.5 s2/m. Or is it ?

From your two points I get ##{\Delta T^2\over \Delta r} = {0.66\over 0.04} = 15.5\ ## s2/m

There is one big problem remaining, though: how do you explain that the plot of the relationship clearly does not go through the origin as the formula says :wideeyed: ?
That was a typo, and yeah, I just whipped up that graph as an example, normally I'd add units. That slope is probably more accurate. I don't really know why it doesn't go through the origin :/ That's just the data we got from the experiment...

At this point I'm sort of desperate to derive some value for F from the equation though. I don't understand how to use it or even get started past rearranging for T2=. I've tried subbing in values, moving terms around, rewriting F as mg... just don't get it.
 
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  • #8
Wouldn't it be time to merge the two threads ? @haruspex ?

From this slope you get ##{4\pi^2 M\over F} = 16.5 ## s2/m so, unless you have M, all you can find is M/F ...

I've got to run, so I leave you in the competent hands of the entrails inspector
 
  • #9
I do have M! Thanks, I'll give this a whirl. Any tip is helpful.
 
  • #10
BvU said:
Wouldn't it be time to merge the two threads ?
Post #1 in this thread described the other thread as a separate problem, but I'm not sure what the distinction is. @HexRei ?
 
  • #11
haruspex said:
Post #1 in this thread described the other thread as a separate problem, but I'm not sure what the distinction is. @HexRei ?

I think it's safe to merge them. Sorry for the latest response.
 

Related to Finding experimental Centripetal force from F=4pi^2rM/T^2

What is centripetal force?

Centripetal force is the force that acts on an object moving in a circular path, directing it towards the center of the circle.

How can I find the experimental centripetal force?

To find the experimental centripetal force, you need to measure the radius of the circular path (r), the mass of the object (M), and the time it takes for one complete revolution (T). Then, use the formula F = 4π²rM/T² to calculate the centripetal force.

What is the significance of using F=4pi^2rM/T^2 in this experiment?

F=4pi^2rM/T^2 is the formula for calculating the centripetal force in a circular motion. It takes into account the mass and speed of the object, as well as the radius of the circular path. This formula is derived from Newton's Second Law of Motion, which states that the net force acting on an object is equal to its mass times its acceleration.

What factors can affect the accuracy of the experimental centripetal force?

Some factors that can affect the accuracy of the experimental centripetal force include human error in measuring the radius and time, air resistance, and friction. If these factors are not accounted for, the calculated centripetal force may not accurately reflect the actual force acting on the object.

How can the experimental centripetal force be used in real-world applications?

The concept of centripetal force is important in various real-world applications such as amusement park rides, car racing, and space travel. Understanding and accurately calculating centripetal force allows engineers and scientists to design and optimize these systems for safety and efficiency.

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