Rolling Axles

November 30, 2023 Shreyas Comments 0 Comment

1. Introduction

While browsing YouTube one day, I came across a video covering a problem from the 2016 JEE Advanced physics exam, an engineering entrance exam in India. The problem had several parts connecting to the following physical setup:

In its simplest form, we can consider a circular wheel of radius $r$ connected by its center to one end of a rigid rod of length $l$ , with the other end fixed on the ground. The wheel is able to roll without slipping on the ground, and as a result, the rod will pivot in a circular path about its fixed end. In the problem, the rod/wheel system is spinning about its axis with a certain angular velocity, from which we could deduce how fast the rod is turning about its pivot.

We could begin our approach by noting that since the rod has length $l$ and the wheel has radius $r$ , the bottom portion of the wheel touches the ground at a distance of $\sqrt{l^2 + r^2}$ from the pivot. If we suppose that the wheel has rotated one full time, then it has rolled the length of its circumference, $2\pi r$ . Since it rolls without slipping, the wheel’s contact point with the ground has moved that same distance on the larger circle of radius $\sqrt{l^2 + r^2}$ . In summary, one rotation of the wheel moves it by $2\pi r$ , whereas it would take a distance of $2\pi \sqrt{l^2 + r^2}$ for the rod to pivot back to its original location. Thus, it should take $\frac{\sqrt{l^2 + r^2}}{r}$ wheel rotations to get one full rod revolution. This result is also the one implied by the exam’s answer sheet.

However, the longer I thought about the situation, the more I felt that something was amiss.

2. My Proposal

The main confounding factor in this situation is the angle between the plane of the wheel and the ground. If the wheel were perpendicular to the ground, then I would more easily believe the previous result, but the pivot setup ensures that it always makes a fixed acute angle of $\arctan\left(\frac{l}{r}\right)$ with the ground.

Why is this relevant? Forgetting the rod, consider the extreme case where this angle approaches $180^\circ$ , meaning the wheel becomes horizontal with the plane, and rolls on the outside of a “hoop” lying flat on the ground. Now, we have a case of the so-called Coin Rotation Paradox: if you roll a coin around another fixed coin of the same size, you actually need two rotations to get back to the starting point, not one! I believe that the same principle behind this paradox should show itself in our wheel-axle problem as well.

3. Formalizing Rotation

What does it actually mean for the wheel to “roll without slipping”? I chose to formalize it as a combination of “spinning”, i.e. rotation in a fixed location, and “sliding,” i.e. gliding over the surface $\textit{while maintaining the same contact point}$ . I convinced myself of this by marking an initial point $X$ where the wheel contacts the ground, and rolling the wheel until $X$ returns to the ground. This net motion can be broken into a sequence of two steps: first, the wheel $\textit{spins}$ on its axis one time until $X$ returns to the ground. Then, it $\textit{slides}$ on top of the point $X$ until reaching its final position, whose distance from the initial position equals the wheel’s circumference. The final result is exactly as if it rolled without slipping.

Knowing this, how does it affect our wheel-rod system? Suppose we apply this decomposition principle: mark a point $X$ on the wheel, and roll it until $X$ returns to the ground. How many rotations has the wheel completed? First, we spin it once on its axis, returning $X$ to the ground and producing one rotation. Then, we slide it a distance of $2\pi r$ across the large circle of radius $\sqrt{l^2 + r^2}$ . Here is where the contradiction occurs: due to the nontrivial angle between the wheel and ground, I believe this sliding phase also contributes a nonzero rotation to the wheel.

4. Sliding

Rigid bodies can move in many ways throughout three dimensions, and determining how much they have “rotated” about their own axis can be confusing, so I envisioned a tool to do so. Consider a rigid planar wheel that is free to move throughout space, and pick two reference points $A$ and $B$ on the wheel. Further, pick a unit vector $\vec{n}$ normal to the wheel, i.e. parallel to the wheel’s axis to standardize orientation. Break the wheel’s motion into infinitesimal steps, and keep track of $A$ and $B$ as they move. At each step, suppose the wheel is contained in plane $P$ , the point $A$ displaces by $d\vec{A}$ and $B$ displaces by $d\vec{B}$ .

Firstly, observe that any motion perpendicular to $P$ (parallel to $\vec{n}$ ) cannot affect rotation about the wheel’s axis; only motion parallel to $P$ can do so. Thus, we denote the projections of $d\vec{A}$ and $d\vec{B}$ on $P$ as $d\vec{A}_P$ and $d\vec{B}_P$ , and we will only work with these from now on.

Now, imagine observing the motion from the perspective of point $A$ ; point $B$ will appear to move with displacement $d\vec{B}_P - d\vec{A}_P$ . Because the wheel is rigid, the distance $AB$ cannot change, and so this vector $d\vec{B}_P - d\vec{A}_P$ must be perpendicular to $\vec{AB}$ . Thus, it contributes an observed rotation of $d\theta = \frac{|d\vec{B}_P - d\vec{A}_P|}{AB}$ , where the magnitude in the numerator is positive or negative depending on the relative orientation of $d\vec{B}_P - d\vec{A}_P$ , $\vec{AB}$ , and $\vec{n}$ (i.e., positive when moving “counterclockwise” around $\vec{n}$ and negative when moving “clockwise”). This can be formulated more rigorously (and perhaps more easily) with cross products, but I wanted to try and provide a more physically-intuitive interpretation. To obtain the total rotation, this quantity would be integrated over a finite time.

Now, let’s apply a similar procedure to our main problem. Let our reference points $A$ and $B$ respectively be the wheel’s center, $O$ , and the contact point with the ground, $X$ . By our definition of sliding, $X$ will always remain in contact with the ground. As the wheel slides around the circular rim of radius $\sqrt{l^2 + r^2}$ , convince yourself that the instantaneous motion of both $O$ and $X$ is always contained in the wheel’s plane, so no projections are required. At this point, we could analyze each infinitesimal rotation element and integrate over the entire pivot revolution, but the symmetry of the situation allows for an easier approach.

Observe the above figure: the center $O$ travels around the orange circle, while the edge point $X$ travels around the purple circle. Here, we have arrived at the key: due to the tilt of the wheel, the purple circle must be $\textit{larger}$ than the orange circle: thus, from the perspective of $O$ , the extra distance that $X$ travels on the purple circle manifests as a net rotation.

In one total revolution, $X$ covers a distance of $2\pi\sqrt{l^2+r^2}$ . By elementary geometry, the radius of the orange circle is $\frac{l^2}{\sqrt{l^2+r^2}}$ , so the center $O$ travels a length of $2\pi\frac{l^2}{\sqrt{l^2+r^2}}$ . Therefore, the excess distance $O$ perceives $X$ to travel is:

$\begin{equation*} 2\pi\left(\sqrt{l^2 + r^2} - \frac{l^2}{\sqrt{l^2 + r^2}}\right) = 2\pi\left(\frac{l^2 + r^2 - l^2}{\sqrt{l^2 + r^2}}\right) = \frac{2\pi r^2}{\sqrt{l^2 + r^2}} \end{equation*}$

Because $OR = r$ , this extra distance corresponds to a rotation of angle

$\begin{equation*} \frac{1}{r}*\frac{2\pi r^2}{\sqrt{l^2 + r^2}} = \frac{2\pi r}{\sqrt{l^2 + r^2}} \end{equation*}$

over the course of a revolution. When viewed from the rod’s pivot point on the ground, this will produce a rotation opposite to the rod’s revolution: if the rod is revolving counterclockwise, the induced rotation from sliding will be clockwise.

5. Putting it Together

Now, we have everything we need to answer the question of how many rotations the wheel makes over the course of one revolution. Suppose the rod revolves counterclockwise when viewed from its pivot on the ground. We break the rolling into spinning and sliding: since the since the wheel traverses an arc length of $2\pi\sqrt{l^2 + r^2}$ , the spinning effect contributes a counterclockwise rotation of $\frac{2\pi\sqrt{l^2 + r^2}}{r}$ radians. As we saw in the previous section, the sliding effect contributes a $\textit{clockwise}$ rotation of $\frac{2\pi r}{\sqrt{l^2 + r^2}}$ radians. Thus, one full revolution contributes a counterclockwise rotation of

$\begin{equation*} 2\pi\left(\frac{\sqrt{l^2 + r^2}}{r} - \frac{r}{\sqrt{l^2 + r^2}}\right) \end{equation*}$

radians. Since $2\pi$ radians corresponds to one rotation, my final conclusion is that one revolution of the rod corresponds to

$\begin{equation*} \frac{\sqrt{l^2 + r^2}}{r} - \frac{r}{\sqrt{l^2 + r^2}} = \frac{l^2}{r\sqrt{l^2 + r^2}} \end{equation*}$

rotations. The original JEE problem, gave the relation $l = \sqrt{24}r$ , in which case our formula yields $\frac{5}{1} - \frac{1}{5} = \frac{24}{5}$ rotations per revolution. Interestingly, the first answer choice on the test (which was marked as correct) seems to imply that there are 5 rotations per revolution, which is what we would expect without considering the additional sliding effect.

Let me know in the comments who you think is correct!

Shreyas' blog – Ignited Minds

I love math and science. This is a blog about things I think of, or work on, in my spare time.