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We are supposed to graph velocity squared against period. My instructor specifically said:

  1. The gradient of this graph is acceleration and

  2. We should observe a trend of decreasing orbital velocities at an increasing rate.

My problem is that, from my knowledge, this graph should not be curved. How can an acceleration graph be curved?

I think it should be straight because acceleration is a rate of change, and you cannot have a rate increase at a rate. Further, the centripetal force formula does not suggest the graphical relationship we obtained.

My instructor said it has something to do with the inverse square law, and something being proportional to something and something obeys the inverse square law, hence this curve. I cannot confirm any of my knowledge because I don't have access to the sheet he gave to the class which, I think, showed a derivation of some sort.

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    $\begingroup$ Hi silenceislife, and welcome to Physics Stack Exchange! I do have to say, this question might or might not be appropriate in its current form, but besides that, it's not clear because it's just so long. You've included a lot of pictures of results and a long description of all the details of the experiment, most of which is probably not necessary. Could you edit the question to trim it down to the specific thing you want to ask about? As far as I can tell, it seems like what you really want to ask is why the graph is curved, when you think it shouldn't be curved. (cont.) $\endgroup$ – David Z Dec 13 '14 at 14:45
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    $\begingroup$ If this helps, I can tell you that this is a horrible setup to teach motion under the force of gravity. Now, I know that doesn't help. There are a number of problems with this, even if it's supposed to be a model for centripetal forces. I am not surprised that it doesn't generate good data. Having said that, what's your reasoning that the resulting curve should be straight? $\endgroup$ – CuriousOne Dec 13 '14 at 14:46
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    $\begingroup$ (cont. from 2 comments up) So you can get rid of the results tables, get rid of most of the details of the experiment (e.g. we don't care about the actual masses, we don't care about the specific numerical results, etc.) and just leave a high-level description of how the experiment worked, and focus your question on asking why the graph is curved and telling us why you think it shouldn't be. (If you're not sure what to cut out, I could make an edit to get you started.) $\endgroup$ – David Z Dec 13 '14 at 14:46
  • $\begingroup$ Ok fixed it I guess and I will probably trim it down further tomorrow. Also added my reasoning for it being expected straight. $\endgroup$ – silenceislife Dec 13 '14 at 14:49
  • $\begingroup$ @DavidZ Feel free to cut out whatever you think is unnecessary. $\endgroup$ – silenceislife Dec 13 '14 at 14:56
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I don't know how this is supposed to simulate planetary motion, but we can still look at what data you would expect. The force on the stopper is $\frac {mv^2}r$, which if there is no friction (good luck!) equals the gravitational force on the masses at the bottom. We would therefore expect that for given mass on the bottom, $r \propto v^2$, which doesn't look like your data at all. Were you changing the mass on the bottom as you changed $r$? That would make a better simulation, as the force of gravity is stronger when the satellite is close to the planet.

As the gravitational force is an inverse square law, for your simulation $M$, the mass at the bottom should be proportional to $\frac 1{r^2}$. In that case we would have $\frac {v^2}r \propto \frac1{r^2}$, or $rv^2$ is a constant. Your data is not too far from this.

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  • $\begingroup$ r is proportional to v^2 if the weight is fixed and either r or v is varied, but don't the different points on the graph represent different weights? If so, then as I said in another comment, it may be that they were adding weight while the stopper was still spinning to see how r and v shifted in response, in which case you would have to use the fact that angular momentum should be constant, i.e. even as r and v change we should expect a fixed value of L=mvr. $\endgroup$ – Hypnosifl Dec 13 '14 at 17:43
  • $\begingroup$ Come to think of it, I guess if you assume angular momentum is conserved you don't even need to consider F=mv^2/r to get the basic shape of the graph--it should just be r = L/mv, and F=mv^2/r would only be needed to find the actual value of L given a known radius and weight. $\endgroup$ – Hypnosifl Dec 13 '14 at 18:21
  • $\begingroup$ Yes, the mass at the bottom was changed each time we used a different radius. Thanks for this answer, I'll read into it more later. $\endgroup$ – silenceislife Dec 13 '14 at 22:27

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