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I just started physics at uni, and our first labb consists of finding a formula to describe the time it takes for a rolling cylinder to roll down a plane. They can be both hollow and solid. So the current expression I have arrived at is $$ t= c \frac{\sqrt{s}}{\sqrt{g\theta}}\big{(}1+n\big{(}\frac{r_1}{r_2}\big{)}^e\big{)} $$ where t = the time it takes for the cyliner to role down the plane, c is a constant, s is the length of the plane, $\theta$ is the angle of the plane, n is a constant, $r_1$ is the inner radius and $r_2$ is the outer radius and e is the exponent I am currently seeking to find. I also don't know what n is but I think I can find that out if I know what e is. With the other variables I used linear regression of the logarithm of the values to find the exponent for them. But when I tried this with the radius I received a very strange answer. I have found out that apparently it is not as easy to find e, since the radiuses are not a factor like the other ones. How should I go about finding e? I was not able to measure the outer and inner radius separately, and I don't have any data where $r_1 = r_2$.

Edit: I had a thought: If all other variables are constant, and let the entire expression $c \frac{\sqrt{s}}{\sqrt{g\theta}}$ be called $k$, and only $\frac{r_1}{r_2}$ changes, could you still do a regression to find e by doing a linear regression of this expression: $$\mathrm{log}(t-k) = e \cdot \mathrm{log}\big{(}\frac{r_1}{r_2}\big{)} + \mathrm{log}(kn)$$ The result I got from doing this was $e$ is approxamatly equal to 2.

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  • $\begingroup$ Do you have cylinders in different sizes? $\endgroup$
    – R.W. Bird
    Oct 25, 2021 at 15:46
  • $\begingroup$ Am I correct that the point of this lab is to analyze your data, and not to derive this expression from the moment of inertia for a hollow cylinder? Consider adding tags experimental-physics or homework-and-exercises as appropriate. $\endgroup$
    – rob ♦
    Oct 25, 2021 at 16:00
  • $\begingroup$ @R.W.Bird Yes I do $\endgroup$
    – Katerina
    Oct 25, 2021 at 16:39
  • $\begingroup$ @Eli well that's a relief. But the thing is the lab assistent doesn't seem to think this is ok... So is it mathematically correct or was I just lucky? $\endgroup$
    – Katerina
    Oct 25, 2021 at 17:21

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I'm answering on the assumption that you're finding this formula empirically, as an introduction to experimental technique, because students doing their first lab don't usually have the tools to find this expression starting from the moment of inertia for a hollow cylinder. The derivation is cute, but that's not really why we're here.

So I'm guessing that you found the $\sqrt s$ dependence by varying only the path length and doing a regression on your results, and the $1/\sqrt\theta$ dependence by varying only the angle (but, within the bounds of the small-angle approximation) and doing a regression on those results. I'm not sure how you have guessed the form of the term in parentheses; perhaps you had a hint from your lab manual.

However, looking at the complete expression: you're not yet to the point where $c$ is a constant. I suggest you take some data with solid cylinders, who have $r_1/r_2 = 0$, and rule out (or account for) an overall dependence on $r_2$.

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    $\begingroup$ Yes, this is a lab that focuses on experimental problemsolving. The way I got the expression within the parenthesis was through thinking that it cant simply be a factor of r1/r2 since the time would be 0 if r1=0, which is not true. So you world have to have some constant (m + r1/r2). And if you move things around a bit to minimize the number of constants. And this is what I ended up with. So you're saying c isn't a constant? And about the solid cylinders, unless these is a change in the course it takes, it seems to go at the same speed regardless of anything else. I $\endgroup$
    – Katerina
    Oct 25, 2021 at 16:46
  • $\begingroup$ Oh, I see. Excellent reasoning. Do re-check your data for solid cylinders. An intuition check: would a rolling pencil have the same travel time down the slope as the solid wheel of a steamroller? $\endgroup$
    – rob ♦
    Oct 25, 2021 at 17:04
  • $\begingroup$ Well, my intuition says no. But surely it doesn't depend on the mass, because when I did the dimension analysis at the start of this whole thing, i thought mass had something do do with it, but when i included mass in the analysis I got a result that said that 1=0 which is highly unlikely. So does it have to do with the outer radius (or in this case just the radius)? This is a lot harder than I thought it was going to be... $\endgroup$
    – Katerina
    Oct 25, 2021 at 17:17
  • $\begingroup$ An argument against mass dependence of fall time, going back to Galileo. Suppose you had three identical cylinders, rolling down your slope end-to-end, like a pencil that’s been cut into three pieces. If the three bits are identical, they should all have the same roll time. Now put a tiny bit of sticky tape between two of the cylinders. That tiny tape shouldn’t do anything, because those two cylinders were perfectly happy moving together on their own. But the tape means you now have two objects with very different masses, instead of three objects with identical masses. $\endgroup$
    – rob ♦
    Oct 25, 2021 at 21:46

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