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

[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?

 

[BvU]

Hi HexRei,

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 ...

 

[HexRei]

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

 

[BvU]

we did not measure T

Well, then how can you make a plot of $T^2$ versus $r$ ?

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 ?

 

[HexRei]

Well, then how can you make a plot of $T^2$ versus $r$ ?

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:

 

[BvU]

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

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 ?

 

[HexRei]

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 ?

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.

 

[BvU]

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

 

[HexRei]

I do have M! Thanks, I'll give this a whirl. Any tip is helpful.

 

[haruspex]

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 ?

 

[HexRei]

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.