Solution Verfication (and question about): Path equation of a point on a rolling circle

Question

I was given the task to come up with an equation $p(t)$ describing the path of a point on a circle rolling along the x-Axis (with $t$ describing time). The center of the circle has a constant velocity of $(v,0)$. I came up with the following: \begin{equation} p_x(t)=r\cdot\cos\left(\frac{3\pi}{2}-vt\right)+vt =-r\sin(vt)+vt \end{equation} \begin{equation} p_y(t)=r\cdot\sin\left(\frac{3\pi}{2}-vt\right)+r=-r\cos(vt)+r =r(1-\cos(vt)) \end{equation} I was wondering if this is the correct solution. Comparing my solution with the equation of a cycloid I noticed that (in the x-component) instead of adding "$vt$" I should add "$r\cdot vt$" but I can't seem to figure out why.

Answer 1

If the linear speed of the centre is $v$, and the circle rolls without slipping, then the angular speed $\omega$ of rotation is given by $$v=r\omega$$.

So the angle turned at time $t$ is $\omega t$, while the horizontal displacement of the centre is still $vt$.

Therefore in your trigonometric expressions, you should replace $vt$ with $\omega t$ or $\frac{vt}{r}$, but keep the additional $vt$ part of the $x$ coordinate.

So the final equations are $$x=vt-r\sin\left(\frac{vt}{r}\right)$$ $$y=r\left(1-\cos\left(\frac{vt}{r}\right)\right)$$