# Why is the relationship between velocity and radius curved, in circular motion?

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.

• 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.) Dec 13 '14 at 14:45
• 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? Dec 13 '14 at 14:46
• (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.) Dec 13 '14 at 14:46
• In this experiment the central force is Mg where M is the mass of weights, and the condition for circular orbit is that F = (m v^2)/r, where m is the mass of the stopper, and ideally if weight is added while the stopper is still 'orbiting' as opposed to stopping and restarting it each time, the angular momentum should be constant, without knowing the exact moment of inertia of the stopper it should be approximately mvr. So using these equations it should be possible to calculate how v would vary with r ideally when M is changed. Dec 13 '14 at 15:29
• @silenceislife - nice question and interesting experiment. I am a bit confused because you talk about plotting velocity squared against period, but your graph shows radius against orbital velocity - can you check that? Normally we plot y against x where y is on the vertical axis and x on the horizontal - so I was expecting your graph to show velocity squared on the vertical axis and period on the horizontal axis. -- also i am wondering if it should be velocity squared against radius in your graph ??? (p.s. welcome to Physics SE)
– tom
Dec 13 '14 at 15:50

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.