Physics / Statistics question related to Atwood Machine and calculus of g

[Jorl17]

Hello everyone.

I am a 12th grade Portuguese student and have a theoretical question to pose, which is probably more related to maths & statistics.

In our class, we made an experiment with an Atwood Machine and deduced the formula for the acceleration of the system (ignoring the string's mass):

$$a = \frac{m_1 - m_2}{m_1 + m_2}g$$

Additionally, we found the easy-to-determine formula for a, from the uniformly accelerated movement (sorry if that doesn't sound correct, as it's what we call it in Portuguese):

$$a = \frac{2x}{t^2}$$

Now, we kept x constant and performed the experiment multiple times. We made the masses vary, but we made sure that $$m_1-m_2$$ was constant.

So, to sum up, g, $$m_1-m_2$$ and x were constant through the experiment.

From these expressions we quickly discovered g for the five times we did this experiment. We always got errors below 5%, with most of them falling below the 1% line. It was always around 9.7, 9.5 or 10.2. The experiment was also coherent, since when we made the masses bigger, the system's acceleration was lower and the time it took to hit the ground was longer -- it was perfect.

We could've found a mean / average value from g by the mean of all our individual gs, but what we did was the following:

Since the expression of a is $$a = \frac{m_1 - m_2}{m_1 + m_2}g$$ and g and $$m_1-m_2$$ were constant, we "wrote" this expression like this:

$$a = k\frac{1}{m_1 + m_2}$$, with $$k = (m_1 - m_2)g$$

So, drawing a graphic (linear regression of the points of the experiment) of a($$\frac{1}{m_1 + m_2}$$) (a in the y axis, as is clear), the slope of that line is k.

Now, since $$k = (m_1 - m_2)g$$, comes that $$g = \frac{k}{(m_1 - m_2)}$$, so we found our "best" g like that, from the slope of the linear regression of the plotting of all a and $$\frac{1}{m_1 + m_2}$$ values, which, I repeat were coherent!

However, here's the problem: this g came up with an awful value, or at least terribly unexpected. It was 11.8! How can it be that this g was 11.8 -- given that it was built out of coherent data that individually was perfect?

We're divided in two groups and our group got that 11.8. The other group, however, got ~18, even though their values were individually close to g (I did not check the coherence, though, but the formula seems to imply coherence is needed).

So, why doesn't this produce the desired results? What is the real difference between this and a plain mean of our g values?

This is not homework, it is an issue that became clear during our class and all of us and the teacher have tried to figure it out, to no avail...

Thanks,

João Ricardo Lourenço / Jorl17

 

[AndyW]

Hello João,

Thank you for sharing your experiment and your question. It's great to see young minds exploring and questioning scientific concepts.

From what I understand, your experiment was focused on determining the acceleration of a system using an Atwood machine, and you found that the acceleration was affected by the masses of the objects involved. You also found that when you plotted the values of acceleration (a) against the inverse of the sum of the masses (1/(m1+m2)), the slope of the line (k) was related to the gravitational acceleration (g) by the formula g = k/(m1-m2). However, you found that this method gave you a different value for g (11.8) compared to using the mean of your individual g values (9.7, 9.5, 10.2).

Firstly, I would like to clarify that your method of finding g using the slope of the line is a valid method, and it is known as linear regression. This method is commonly used in scientific research to determine the relationship between two variables. So, there is nothing wrong with your approach.

Now, let's try to understand why your method gave you a different value for g compared to using the mean of your individual g values. The main reason for this difference is the presence of outliers in your data. Outliers are data points that are significantly different from the rest of the data. In your case, the outlier could be the value of 18 that your other group obtained. When using the mean method, the outlier has a significant impact on the final result, pulling it towards the higher value of 18. However, when using linear regression, the slope of the line is less affected by outliers, as it takes into account the overall trend of the data rather than individual values.

To better understand this, I would suggest plotting a graph of your individual g values and the mean value against the inverse of the sum of the masses. You will see that the mean value is pulled towards the outlier, while the individual values are spread out.

In conclusion, your method of finding g using linear regression is valid, but it is sensitive to outliers. To get a more accurate value for g, it would be best to investigate the outlier and determine if it is a valid data point or not. If it is not a valid data point, it should be removed from the analysis.

I hope this helps to answer your question. Keep up the good work