A Unicycle on a Slope
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.
Assumptions
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.
Equation of motion
We'll concern ourselves with the equation of motion in the
axis, as depicted in the figure. The only forces in the
direction are
, the component of
in the
direction, and
. Therefore
There are two torques applied to the zero-inertia wheel: that
of the traction force,
, and the one done by the force
on the pedal,
:
Taking into account that
and
combining the above equations we get the equation of motion
along the
axis,
The static solution
Imposing
we get
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,
, the slope is maximum and
The dynamic solution
If we multiply (1) by
we can integrate,
to get:
Sustainability condition
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
and
. Introducing these values
in (2) we get:
from which
This defines the maximum slope that can be climbed.
Periodic solution
When climbing a maximum slope we can replace
by the expression above in the dynamic solution. If
the equation becomes, multiplying by two:
Minimum initial velocity
The condition for this to have a solution is
that
and, therefore,
This condition will define the minimum
. As
it has to be true for all
, the most limiting
condition is when the term in parenthesis is maximum, which
will happen when
and therefore
This gives
radians (39.54 degrees) and,
substituting on the equation for minimum
, we
get
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