# Why is the angle linearly related to time in uniform circular motion?

We say that a particle undergoes uniform circular motion if it travels a circular path at constant speed. If we assume that the center of curvature is at the origin, then in polar coordinates ($r$,$\theta$), we have that $r$ is constant and only the angle $\theta$ changes.

Most of the sources I've looked at simply declare or assume that the angle must vary linearly with time: $\theta=\omega t$ for a constant $\omega$ (angular velocity). Why is this? I see that there is an analogy with motion in a straight line insofar as the motion varies only in 1 coordinate, but is there a more mathematical justification?

• Take the time derivative of your angular expression and it yield angular speed is constant. This is the requirement for uniform circular motion. Commented Feb 23, 2016 at 4:54
• In a sense I'm asking why constant speed implies constant angular speed.
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Commented Feb 23, 2016 at 5:02
• Because $s=r\theta$ gives the length of an arc of the circle. Commented Feb 23, 2016 at 10:44

Definitions are not right nor wrong, they are either good or bad. Many people take that obeying $r(t) = R$ and $\theta(t) = \omega t$ is the definition of uniform circular motion.

Now your question here is a little different. You suggest to define "uniform circular motion as a motion in a circular path with constant speed" and you want to conclude that $\theta(t) = \omega t$.

Based on your definition the trajectory must be of the form $\mathbf{r}(t) = (R\cos(\theta(t)),R\sin(\theta(t)))$ for some $R\in \mathbb{R}$. What's the velocity? Well it's just the time derivative of $\mathbf{r}$:

$$\mathbf{v}(t)=\mathbf{r}'(t)=(-R\sin(\theta(t))\theta'(t), R\cos(\theta(t))\theta'(t)).$$

The speed, thus, is the norm of the velocity. In that case the speed is

$$|\mathbf{v}(t)|^2=(-R\sin(\theta(t))\theta'(t))^2+(R\cos(\theta(t))\theta'(t))^2=\theta'(t)^2,$$

from this we conclude that $|\mathbf{v}(t)|=\theta'(t)$. Now based on your definition of uniform circular motion, $|\mathbf{v}(t)|$ is constant. Call this constant $\omega\in \mathbb{R}$, thus we have $|\mathbf{v}(t)|=\omega$ and from this you conclude that $\theta'(t)=\omega$ and hence $\theta(t)=\omega t$.

• I think that you have dropped an $R$ in your answer? Commented Feb 23, 2016 at 5:34

In a sense I'm asking why constant speed implies constant angular speed.

Ask a mathematician to define a radian. They'll tell you that a radian is that angle for which the arc-length along the unit circle equals one. And that the arc-length along a two radian angle is two and so on.

This is not an assumption, it is actually the definition of an angle.

## Text I wrote when I misunderstood the question.

Typically this subject is first taught to students who are not yet ready to deal with the problem of showing that a particular motion is or is not uniform.

And it doesn't matter, because despite the emphasis we put on the uniformity, you can always consider a period of motion short enough that the speed doesn't change much and use that speed to find the instantaneous centripetal acceleration.

So the only real thing that matters is that the motion is along an arc that reasonably approximates a segment of a circle.

You'll come back to rotations in a short time and learn about tangential acceleration and who to combine it with centripetal acceleration to find the total acceleration vector of a object in non-uniform circular motion and then all will be well with the world again.

Most of the sources I've looked at simply declare or assume that the angle must vary linearly with time: $θ=ωt$ for a constant $ω$ (angular velocity). Why is this?

This is simply the definition of uniform circular motion. Motion is "circular" if $r$ is constant. It is "uniform" if the circular motion has a constant speed.

In a sense I'm asking why constant speed implies constant angular speed

Having a constant $r$ gives us that the position is constrained only to $\theta(t)$

$$\vec{x} = \left(R(t),\theta (t)\right)\rightarrow\left(r,\theta (t)\right)$$

and constant speed gives us that $\frac{d}{dt}r\cdot\theta(t)=0$, so $\frac{d}{dt}\theta(t)=0$, so $\theta=\omega t$. If $\theta(t)$ was anything but a constant times $t$ then $\frac{d\theta}{dt}\ne0$, so it wouldn't be uniform circular motion.