009C Sample Final 3, Problem 9

A wheel of radius 1 rolls along a straight line, say the  $x$ -axis. A point  $P$ is located halfway between the center of the wheel and the rim. As the wheel rolls,  $P$ traces a curve. Find parametric equations for the curve.

Foundations:
Many concepts in physics involve the notion of a relative frame. For example, if I'm in a box dropped from an airplane, I won't be moving relative to the box. However, I'm still heading towards the ground with acceleration $10\,{\textrm {m/sec}}^{2}.$ Say it drops for 5 seconds, so the box is going $50\,{\textrm {m/sec}}$ when it hits the ground. Even if I jump with all my might and pull off something like $5\,{\textrm {m/sec}}$ of upward velocity, I'll still feel the impact of hitting the ground at

$50-5\,{\textrm {m/sec}}=45\,{\textrm {m/sec}}.$ Essentially, equations of motion can often be broken into parts, and then added up.

Solution:

Step 1:
If a wheel of radius one is resting at the origin, its axis will be at the point  $(1,0).$ For this solution, we will assume the point  $P$ is below the axle, although the problem does not state the position of  $P$ .  When the wheel rotates clockwise, it will move to the right. Since the length of the arc defined by an angle  $\theta$ on a circle of radius  $R$ is   $L=R\cdot \theta =1\cdot \theta =\theta ,$ the wheel will roll forward the length of the arc, which is just  $\theta .$ Moreover, the axle's  $x$ position will change in the same manner, while the height of the axle will always be fixed at  $y=1$ .  This means we can describe the position of the axle as a function of  $\theta ,$ or

$a(\theta )\,=\,(\theta ,1).$ Step 2:
Since the wheel is rotating, we also know that our point  $P$ will rotate around the axle. As described in the problem, it is halfway between the rim and the center/axle, so it is  $1/2$ unit away from the axle, and will rotate clockwise. Using our trig relations (while looking at the image), we find that the position of  $P$ relative to the axle can be described as
$p(\theta )\,=\,\left(-{\frac {1}{2}}\sin \theta ,-{\frac {1}{2}}\cos \theta \right).$ Notice that when  $\theta =0,$ the point would be at the position

$p(0)\,=\,\left(-{\frac {1}{2}}\sin 0,-{\frac {1}{2}}\cos 0\right)\,=\,\left(0,-{\frac {1}{2}}\right),$ which is half a unit directly below the axle. This is shown as a gray "ghost" dot in the image, while the black triangle and circle represent the situation at  $\theta =\pi /4.$ Step 3:
We therefore have a frame (the axle) that is moving, and a point  $P$ that is moving relative to the frame. To get the movement relative to the stationary "world", we simply add them up to find
$P(\theta )\,=\,a(\theta )+p(\theta )\,=\,\left(\theta -{\frac {1}{2}}\sin \theta ,1-{\frac {1}{2}}\cos \theta \right).$ $P(\theta )\,=\,\left(\theta -{\frac {1}{2}}\sin \theta ,1-{\frac {1}{2}}\cos \theta \right).$ 