B Calculating g with a Conical Pendulum

[ProphetMikey]

TL;DR Summary

Using the conical pendulum to find a value for g, plugging individual measurements into an equation gives accurate results, but graphing the same data and using the gradient gives double the results.

In analysing the conical pendulum, it can be shown that the period is given by T=2pi.sqrt(L.cos(phi)/g) and that therefore, g = 4.pi^2.L.(cos(phi)/T^2).
L = pendulum length, phi is measured at the top of the pendulum (at the point of suspension).

Graphing cos(phi) vs T^2 should produce linearised data with a gradient that can be used to calculate a value for g from measurements done at several periods and angles (a variable speed motor is used to maintain uniform circular motion).

I've had my students conduct this experiment for the last few years, obtaining fairly good results (typically g~11-13 m/s^2) but this year has been very confusing. I think I am just overlooking something simple, but can't figure it out. Maybe I just need more sleep.

Here is the raw data collected by the students: (period was timed over 10 revolutions)
T (seconds) phi (degrees)
1.554 29.5
1.536 36
1.452 49
L = 0.82 m

When I put these into the equation (left as an exercise for you so you don't just copy my mistakes), I get values of g of 11.7, 11.1 and 10.0 m/s^2, but when I graph them and use the gradient to find a value of g, I get g = 21.8 m/s^2.

I've tried this several times and cannot see what I have done wrong. Why doesn't the gradient give the average value of the manual calculations?
Any help would be greatly appreciated!

 

[kuruman]

TL;DR Summary: Using the conical pendulum to find a value for g, plugging individual measurements into an equation gives accurate results, but graphing the same data and using the gradient gives double the results.

In analysing the conical pendulum, it can be shown that the period is given by T=2pi.sqrt(L.cos(phi)/g) and that therefore, g = 4.pi^2.L.(cos(phi)/T^2).
L = pendulum length, phi is measured at the top of the pendulum (at the point of suspension).
Graphing cos(phi) vs T^2 should produce linearised data with a gradient that can be used to calculate a value for g from measurements done at several periods and angles (a variable speed motor is used to maintain uniform circular motion).
I've had my students conduct this experiment for the last few years, obtaining fairly good results (typically g~11-13 m/s^2) but this year has been very confusing. I think I am just overlooking something simple, but can't figure it out. Maybe I just need more sleep.
Here is the raw data collected by the students: (period was timed over 10 revolutions)
T (seconds) phi (degrees)
1.554 29.5
1.536 36
1.452 49
L = 0.82 m
When I put these into the equation (left as an exercise for you so you don't just copy my mistakes), I get values of g of 11.7, 11.1 and 10.0 m/s^2, but when I graph them and use the gradient to find a value of g, I get g = 21.8 m/s^2.
I've tried this several times and cannot see what I have done wrong. Why doesn't the gradient give the average value of the manual calculations?
Any help would be greatly appreciated!

Well, I did the exercise that you assigned and calculated the period on my spreadsheet using L = 0.82 m and g =9.81 m/s² in $$T=2\pi\sqrt{\frac{L \cos\phi}{g}}.$$ A screenshot of the results is shown below. It looks like I did not repeat your mistakes. Did you use a spreadsheet?

 

[kuruman]

I did some plots and I verified that the slope or gradient gives a value for g over 20 m/s². I have a remark and a question.
Remark: Don't expect much accuracy when you have only 3 data points.
Question: What angle did your students report as $\theta$? Are you sure that it is the half-angle of the cone and not the full angle?

Assuming that the students reported the full angle, I divided them by 2 and then plotted $T^2$ vs $4\pi^2 L \cos\phi.$ The acceleration of gravity should be the inverse of the gradient.

The figure below shows the results, such as they are. I did two linear fits
(a) One with non-zero intercept, which is unphysical but puts the line closer to the points (blue line). It predicts $g=0.1704^{-1}=5.9~\rm{m}/\rm{s}^2$.
(b) One with forced zero intercept, which is physical (blue line) and also illustrates why only three points are not enough to determine the gradient. It predicts $g=0.0752^{-1}=13~\rm{m}/\rm{s}^2$.

What's physical must take precedence over what is not even when the physical fit is lousy. However, the truth seems to be somewhere in between. If you average the two numbers with no motivation whatsoever, you get $g=9.6~\rm{m}/\rm{s}^2$.

 

[sophiecentaur]

@ProphetMikey I was just wondering how circular / symmetrical the orbits were. Was there any check of that? A mirror, flat on the table could be easy to arrange - or a lamp suspended over the pendulum. If there were, say a few degrees difference over the two axes, you'd have a possible few percent in uncertainty of results that could bring the 9.6 to more like 9.8 which would be pretty impressive.
What was the launch method used?