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I am currently working on a physics project to experimentally confirm the relationship between mass and moment of inertia. The experiment is setup as depicted:

diagram

In this experiment, $r1$ and $r2$ are equal and constant. Attached to the pulley is a 0.1kg mass. As the mass falls the rod with the attached masses will rotate.

In this experiment, I record the time for 5 complete oscillations (ie. $θ = 10pi$) whilst changing the value of M1 and M2. (Note $M1=M2$)

I use this to calculate the angular acceleration ($α=2θ/t^2$) where $θ=10pi$ and $t$ is the average time for 5 complete rotations. I then use the equation $τ=Fr$ to calculate the torque where $F=mg=pullley Mass*9.8$ and $r=radius Of The Central Axis.$

Finally, I then use $I=τ/α$ to calculate the moment of inertia. Note this moment of inertia is for the masses as well as the rotating apparatus.

I then plot the calculated moment of inertia (units: $kgm^2$) against the total added mass (units: kg) (ie. M1+M2) and calculate the gradient/slope in $m^2$. This graph is linear with a non-zero intercept.

Is it possible with this data to confirm the directly proportional relationship between mass and moment of inertia? What should the square root of the gradient (units: $m$) correspond to?

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  • $\begingroup$ Can you describe your set-up more precisely? What are the moving parts? What is oscillating? Are $r1$ and $r2$ equal? Are they equal to $r$ or is $r$ the height of the apparatus? What is the role of $m$? It looks like it's hanging from something that turns. What about the thick vertical line? Does it turn when $m$ moves? $\endgroup$ – Steven Mathey Mar 16 '18 at 13:56
  • $\begingroup$ Couple of comments on the writing. First, presumably by "oscillations" you mean full rotations of the device. Second when you say you "calculate the gradient", I assume you mean that you are finding the slops of a line obtained from graphing the data. But no one can give you any guidance in interpreting the slope if they don't know what data is graphed and how. $\endgroup$ – dmckee --- ex-moderator kitten Mar 16 '18 at 14:21
  • $\begingroup$ "I then use the equation τ=Fr to calculate the torque where F=mg" This is not exactly true, but might be a good approximation as long as $\alpha r$ is very small compared to $g$. Given that your plot yields a good line you are probably OK. $\endgroup$ – dmckee --- ex-moderator kitten Mar 16 '18 at 19:15
  • $\begingroup$ @dmckee Sorry, can you explain further why this would be an approximation. I don't quite understand why. $\endgroup$ – jdoe Mar 16 '18 at 19:16
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    $\begingroup$ The tension in the rope only equals the force of gravity on the hanging weight if the system is in equilibrium. But that weight is accelerating, so the tension is not exactly equal to the $mg$. $\endgroup$ – dmckee --- ex-moderator kitten Mar 16 '18 at 19:21
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This kind of linear graphical analysis is a bit old-school, but it is something that can be done with a pencil and paper and the skills you learn from it can be easily extended to the world of computerized fitting of more complicated functions.


To understand the meaning of your fit parameters, re-cast the theory of your experiment in such a way that it can be read as a slope-intercept line in the plot parameters, and read the interpretation off of the correspondence.

Here the working theory that you are using comes from \begin{align} \tau &= I \alpha\\ \tau &= \left( I_\text{theory} - I_0\right) \alpha \\ \frac{\tau}{\alpha} &= I_\text{theory} - I_0 \\ \color{blue}{I_\text{expr}} &\equiv \color{grey}{2R^2}\color{red}{M} + I_0 \;. \end{align} Then you plot $\color{blue}{I_\text{expr}}$ as the dependent variable (i.e. on the vertical axis} against $\color{red}{M}$ as the independent variable (i.e. horizontal axis).

In analytic geometry the slope-intercept line is $$ \color{blue}{y} = \color{grey}{m} \color{red}{x} + b \;,$$ and we see that on your graph

  • $\color{blue}{I_\text{expr}}$ is playing the role of $\color{blue}{y}$
  • $\color{red}{M}$ is playing the role of $\color{red}{x}$
  • $\color{grey}{2R^2}$ is playing the role of the slope $\color{grey}{m}$

The last bullet point tells you exactly how to interpret the slope of your line.


Questions for the student:

  1. How do you interpret the non-zero intercept?
  2. Why did I write the moment of the inertia in two parts?
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  • $\begingroup$ If $slope=2R^2$ does this prove that mass is directly proportional to inertia? $\endgroup$ – jdoe Mar 17 '18 at 12:49
  • $\begingroup$ Don't deleted and re-post comments in order to badger another user. I didn't answer you the first three times because you should be using your judgement on how to interpret the results. The only thing I'll say is that words like "prove" and "proof" don't show up much in the discussion of experimental results. Data can be "in agreement" with theory. Data can "support" the accepted analysis. $\endgroup$ – dmckee --- ex-moderator kitten Mar 17 '18 at 23:07

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