Problem on parameterizing a cylinder rolling up a hill

Consider a cylinder of radius $r$ rolling up a hill of incline $\theta$ at constant speed $v$. Initially the point of contact is $(0,0)$. Find the trajectory of the point initially contacting the hill.
• Solution

Recall that
The importation types of motion in this problem are translation of the cylinder up the hill and rotation about the center of the cylinder.
We will parameterize by $t$ and assume that the cylinder's contact point moves uphill at speed $v$.
We write that $\bfx(t) = \bfx_\text{c}(t) + \bfx_\text{r}(t)$, where $\bfx(t)$ is the position of the designated point at time $t$, $\bfx_c(t)$ is the position of the center of the cylinder, and $\bfx_r$ is the position of the designated point relative to the center of the cylinder.

Finding $\bfx_\text{c}(t)$

Now we find the position of the center of the cylinder at time $t$.
At any point in time, the center of the cylinder is a distance $r$ away from the plane in a direction perpendicular to the plane.
Because this vector has unit length, we observe that the center of the cylinder is always a vector $\langle -r \sin \theta, r \cos \theta\rangle$ away from the contact point with the plane.
After time $t$, the cylinder has moved uphill a length $vt$. The contact point after time $t$ has position $vt \langle \cos \theta, \sin \theta\rangle$.
Hence $$\bfx_c(t) = \langle vt \cos \theta - r \sin \theta, vt \sin \theta + r \cos \theta \rangle.$$

Finding $\bfx_\text{r}(t)$

Relative to the center of the cylinder, the designated points rotates around it at constant angular speed.
The designated point always has distance $r$ from the center of the cylinder. To find the point, we need its angle relative to the center as a function of time.
By the no slip condition, the distance the cylinder travels uphill equals the arclength swept around the small cylinder.
After time $t$, the cylinder moves a distance $vt$ uphill. This corresponds to an angle of $vt/r$ along the cylinder.
At time $t=0$, the angle of the designated point is $\theta - \pi/2$.
After time $t$, the angle of the designated point is $\theta - \pi/2 - vt/r$. Note the minus sign because the rotation is clockwise.

Combining

We conclude that \begin{align}
\bfx(t) = \langle vt \cos \theta - r \sin \theta + r \cos( \theta - \pi/2 - vt/r), \\ vt \sin \theta + r \cos \theta + r \sin( \theta - \pi/2 - vt/r ) \rangle
\end{align}