Better not stop pedaling

If we have our unicycle on flat ground, it exerts a force $mg$ on the ground below. When the plane is tilted at an angle $\theta$ , the normal force become $F_N = \mu mg\cos\theta$ . Similarly, the force along the plane is given by $F_\parallel = mg\sin\theta$ .

In order for the unicycle to remain at a given height, it must exert a force equal in magnitude to $F_\parallel$ up the plane. The force due to the torque $\tau$ is given by $F_\tau R$ so that $F_\tau = \tau / R$ .

Equating this with the force down the plane, we have $\begin{aligned} F_\tau &= \tau/R \\ &= mg\sin\theta. \end{aligned}$

Thus we find $\boxed{\tau = mgR\sin\theta}.$

Note, this is the magnitude of the force up the plane, but the force is provided by the normal force. Thus, it must be the case that $\begin{aligned} F_\tau &\leq F_N \\ mg\sin\theta &\leq \mu mg\cos\theta \\ \tan\theta &\leq \mu. \end{aligned}$ So that $\theta_\textrm{max} = \arctan \mu$ . Above this level of incline, the unicycle could not proceed up the incline. For $\mu=1$ , this corresponds to $\theta = \SI{45}{\degree}$ .

if we draw a straight line along the axle of the unicycle then its perpendicular distance from the point where the wheel touches the incline is rsin15 and along that direction the force is mg, so torque is t=mgr sin15