How does angular acceleration change with revolutions? So, for a section of my EPQ (A-Level, Extended Project Qualification), I am trying to analyse a hypothetical circular accelerator using the angular motion equations for constant angular acceleration.
The issue is, I am trying to look at how the required acceleration to reach a fixed linear velocity ($v$, where $\omega r=v$) changes when the number of revolutions ($R$) the "particle" makes changes, but I seem to be finding an odd relationship between these values. The radius and final velocity are fixed.
So far, I have used the equation $\omega = 2\pi n$ to calculate the revolutions per second ($n$) of a particle travelling at (final) velocity $v$, around a circle radius $r$ (using $\omega r = v$). I rearranged the above equation to get $\frac{v}{r} = 2\pi n$, which then I again rearranged to find n -> $n = \frac{v}{2\pi r}$. By using the constant angular acceleration equation $\omega = \omega _{0} + \alpha t$, I formed an equation for $t$, using $n$ and $R$ (revolution to complete), thinking that $t = \frac{R}{n}$ (because it would $\frac{1}{2}$ seconds to complete 5 revolutions ($R=5$) at a rate of 10 rev/s ($n=10$). Therfore, combining these equations and rearranging, I found that $\alpha = \frac{v^2}{2\pi r^2} \times \frac{1}{R}$ (R has been taken out for clarity and $\omega_{0} = 0$, i.e. it started from rest).
The issue I have here is, when changing R, the resulting $\alpha$ value results in a non-linear relationship (like $y=\frac{1}{x}$). Surely by changing R, you are just changing the distance travelled (like the $s$ in SUVAT), so the resulting required acceleration should change linearly, given that the velocity and radius stay fixed.
I would be very grateful if anyone could point out where I have gone wrong, or explain to me why this is the case.
 A: I think you are confusing linear and angular acceleration (a and $\alpha$). 
Firstly, lets call the number of revolutions n (which I would say is the more conventional choice).
If I understand you correctly, you want to know what angular acceleration will accelerate a particle from $v_{0}$ to $v_{1}$ in n revolutions of a circle of radius r.
You are right that increasing n (the total number of revolutions) increases the displacement. The distance travelled, $S = 2 \pi r n$. If the radius of the circle is constant, you correctly identified that reaching a particular linear velocity is equivalent to reaching a particular angular velocity as $\omega = \frac{v}{r}$. Additionally, $\alpha = \frac{a}{r}$. Given that this is the case, you can see that all SUVATS have direct angular equivalents.
$$v_{1}^{2} = v_{0}^2 + 2aS $$
has the following angular equivalent:
$$\omega_{1}^{2} = \omega_{0}^{2} + 2 \alpha \theta $$
where $\theta = 2\pi n$.
So,
$$\alpha = \frac{\omega_{1}^{2} - \omega_{0}^{2}}{4\pi n} = \frac{v_{1}^{2} - v_{0}^{2}}{4\pi r^{2}n}$$
To get to linear acceleration:
$$a = \alpha r = \frac{v_{1}^{2} - v_{0}^{2}}{4\pi rn}$$
This makes sense. If you double the number of revolutions (n), you half the acceleration as you have doubled the distance travelled (as per the linear case). If you double the radius, you double the path length ($2\pi r n$) and half the required acceleration as per the above expression for a.
