# 009C Sample Final 3, Problem 9

A wheel of radius 1 rolls along a straight line, say the  ${\displaystyle x}$-axis. A point  ${\displaystyle P}$  is located halfway between the center of the wheel and the rim. As the wheel rolls,  ${\displaystyle P}$  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 ${\displaystyle 10\,{\textrm {m/sec}}^{2}.}$  Say it drops for 5 seconds, so the box is going ${\displaystyle 50\,{\textrm {m/sec}}}$ when it hits the ground. Even if I jump with all my might and pull off something like ${\displaystyle 5\,{\textrm {m/sec}}}$ of upward velocity, I'll still feel the impact of hitting the ground at

${\displaystyle 50-5\,{\textrm {m/sec}}=45\,{\textrm {m/sec}}.}$

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  ${\displaystyle (1,0).}$  For this solution, we will assume the point  ${\displaystyle P}$  is below the axle, although the problem does not state the position of  ${\displaystyle P}$.  When the wheel rotates clockwise, it will move to the right. Since the length of the arc defined by an angle  ${\displaystyle \theta }$  on a circle of radius  ${\displaystyle R}$   is   ${\displaystyle L=R\cdot \theta =1\cdot \theta =\theta ,}$  the wheel will roll forward the length of the arc, which is just  ${\displaystyle \theta .}$

Moreover, the axle's  ${\displaystyle x}$  position will change in the same manner, while the height of the axle will always be fixed at  ${\displaystyle y=1}$.  This means we can describe the position of the axle as a function of  ${\displaystyle \theta ,}$  or

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

Notice that when  ${\displaystyle \theta =0,}$  the point would be at the position

${\displaystyle p(0)\,=\,\left(-{\frac {1}{2}}\sin 0,-{\frac {1}{2}}\cos 0\right)\,=\,\left(0,-{\frac {1}{2}}\right),}$

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  ${\displaystyle \theta =\pi /4.}$

Step 3:
We therefore have a frame (the axle) that is moving, and a point  ${\displaystyle P}$  that is moving relative to the frame. To get the movement relative to the stationary "world", we simply add them up to find
${\displaystyle P(\theta )\,=\,a(\theta )+p(\theta )\,=\,\left(\theta -{\frac {1}{2}}\sin \theta ,1-{\frac {1}{2}}\cos \theta \right).}$
${\displaystyle P(\theta )\,=\,\left(\theta -{\frac {1}{2}}\sin \theta ,1-{\frac {1}{2}}\cos \theta \right).}$