By Juan Reyero, Jaime Fernández and Utpal Sarkar.

What is the maximum slope that a unicycle can climb? Trying to answer this question leads to some elegant results, and sheds some light on the dynamics of unicycle riding.

We shall assume the rotational inertia of the wheel, pedals and legs to be negligible. A further simplifying assumption is that the rider is putting all his weight into the pedal throughout its trajectory as it is going down, and that this is the only impulsion force. Which implies that the rider is not pulling up from the saddle, nor accelerating the body upwards with respect to the unicycle, nor making a horizontal force in the pedals. We know this assumption to be too stringent for expert unicyclists, who will typically be able to do all these things, but we hold it to be a reasonable approximation to what a unicyclist of normal skill would do when pressed to climb a slope at the limit of the unicycle’s possiblities.

We’ll concern ourselves with the equation of motion in the \(x\) axis, as depicted in the figure. The only forces in the \(x\) direction are \(\vec{m\ddot{x}}\), the component of \(\vec{mg}\) in the \(x\) direction, and \(\vec{T}\). Therefore

\[ m g \sin{\alpha} + m\ddot{x} = T. \]

There are two torques applied to the zero-inertia wheel: that of the traction force, \(T R\), and the one done by the force on the pedal, \(m g r \|\sin{\theta}\|\):

\[ m g r \|\sin{\theta}\| = T R. \]

Taking into account that \(R \ddot{\theta} = \ddot{x}\) and combining the above equations we get the equation of motion along the \(x\) axis,

\begin{equation*} \ddot{\theta} - g\frac{r}{R^2}\|\sin{\theta}\| + \frac{g}{R} \sin{\alpha} = 0. \tag{1} \end{equation*}Imposing \(\ddot{\theta} = 0\) we get

\[ \sin{\alpha} = \frac{r}{R} \|\sin{\theta}\| \]

This is the limit slope at which the unicycle would be in equilibrium, neither going up or down, for a given position of the cranks. For the position of maximum torque, \(\theta=\pi/2\), the slope is maximum and

\[ \sin{\alpha} = \frac{r}{R}. \]

If we multiply (1) by \(\dot{\theta}\) we can integrate, to get:

\begin{equation*} \frac{1}{2}\left(\dot{\theta}^2 - \dot{\theta}_0^2\right) + g\frac{r}{R^2}(\cos{\theta} - \cos{\theta_0}) + \frac{g}{R} \sin{\alpha}(\theta - \theta_0) = 0 \tag{2} \end{equation*}If at the end of a cycle the angular velocity of the cranks is the same as it was at the beginning, the motion is periodic and sustained over time. Thus the conditions are \(\theta_0=0\) and \(\dot{\theta}(\pi)=\dot{\theta_0}\). Introducing these values in (2) we get:

\[ g\frac{r}{R^2}(\cos{\pi} - \cos{0}) + \frac{g \pi}{R} \sin{\alpha} = 0, \]

from which

\[ \sin{\alpha} = \frac{2}{\pi} \frac{r}{R}. \]

This defines the maximum slope that can be climbed.

When climbing a maximum slope we can replace \(\sin{\alpha}\) by the expression above in the dynamic solution. If \(\theta_0=0\) the equation becomes, multiplying by two:

\[ \dot{\theta}^2 - \dot{\theta}_0^2 + 2\frac{g r}{R^2}(\cos{\theta} - 1) + 4\frac{g r}{\pi R^2} \theta = 0 \]

The condition for this to have a solution is that \(\forall \theta, \, \dot{\theta}^2 > 0\) and, therefore,

\[\dot{\theta}_0^2 - \left( 2\frac{g r}{R^2}(\cos{\theta} - 1) + 4\frac{g r}{\pi R^2} \theta\right) > 0. \]

This condition will define the minimum \(\dot{\theta}_0\). As it has to be true for all \(\theta\), the most limiting condition is when the term in parenthesis is maximum, which will happen when

\[\left( 2\frac{g r}{R^2}(\cos{\theta} - 1) + 4 \frac{g r}{\pi R^2} \theta\right)' = 0, \]

\[-2\frac{g r}{R^2}\sin{\theta} + 4 \frac{g r}{\pi R^2} = 0\]

and therefore

\[ \sin{\theta} = \frac{2}{\pi}. \]

This gives \(\theta = 0.69\) radians (39.54 degrees) and, substituting on the equation for minimum \(\dot{\theta}_0\), we get

\[ \dot{\theta}_0^2 > 4.126\frac{r}{R^2}. \]

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