Re: Bicycle physics question

On 6/19/2024 6:39 PM, bp@www.zefox.net wrote:

Jeff Liebermann <jeffl@cruzio.com> wrote:

On Tue, 18 Jun 2024 00:16:01 -0000 (UTC), <bp@www.zefox.net> wrote:
>

While out for a motorcycle ride this morning a question
applicable to both bicycles and motorcycles came to mind:
>
When a bike/cycle is leaned into a turn, its center of gravity
is lowered.

>
Gravity doesn't move.  However, your center of mass does move and is
lowered.
<https://en.wikipedia.org/wiki/Center_of_mass>
>

That would seem to remove some potential energy.

>
True, but it's a tiny amount of energy.
>
Potential_energy = mass * gravity * height
or
joules = kg * 9.8 meters/sec^2 * meters
>
Notice that it's the same change in potential energy whether you're
moving of standing still.  You could be riding furiously or at a
traffic light, and the change in potential energy would be the same.
Your forward motion is also not involved in the potential energy
calculation, because it is perpendicular to force vector (gravity).
>
If you were to lean the bicycle over 1/2 meter and you and your
bicycle weigh 80 kg (176 lbs), the change in potential energy would
be:
Potential_Energy(change) = 80 * 9.8 * 0.5 = 392 joules or 392
watt-seconds
>
<https://www.omnicalculator.com/physics/potential-energy>
I like calculators that allow me to mix metric and imperialist units.
>

To undo the lean, the wheels have to be steered back under
the CG, which requires pedal effort on the bicycle and extra
throttle on the motorcycle.

>
Correct.  Assuming 100% efficiency (most of which is lost in
compressing the tires), in the above example, you will need to supply
392 joules of energy to return to an upright position.  Note that the
energy is supplied only in the upright direction (perpendicular to the
ground) and does not involve anything in the forward direction.
>
There are some interesting comments in this discussion:
<https://www.bikeforums.net/advocacy-safety/288303-what-makes-bike-turn.html>
>

But, leaning a bike/motorcycle doesn't seem to make it go
perceptibly faster, so if it takes work to stand it back up,
where did the energy of leaning over go?

>
It didn't go anywhere.  It's all POTENTIAL energy, not kinetic energy.
You can use potential energy to do work.  Only kinetic energy can do work.

        ^
      can't <-typo?
Potential energy can certainly do work, think of a trebuchet. Potential
energy is lost in leaning. Tom thinks it's going into tire friction,
we all seem to agree the amount is smallish compared to the KE of
the bike and dissipation caused by air drag making it hard to detect..

I haven't thought deeply about this, but I suspect there's no practical change in potential energy due to the lean. But I'll admit my thinking on this is both somewhat fuzzy and esoteric. Here goes:
The leaning of the bike+rider occurs only when the bike is turning - that is, undergoing a lateral acceleration. The amount of lean is precisely what's necessary to balance the vector sum of gravity (or its acceleration) and the lateral acceleration.
One of the concepts that kick-started Einstein toward relativity was the fact that gravity (or its acceleration) and linear acceleration are indistinguishable. Specifically, he realized that no measurement done inside an isolated elevator can tell whether it's gravitational force or upward acceleration that causes a passenger to stay in contact with the floor.
So it's logical to treat as identical sorts of vectors both the upward force on a cyclist (fighting gravity) and the lateral force on a turning cyclist (pushing him into a curve). It's the resultant of those two forces against which potential energy is determined. And again, the cyclist's angle is always what's necessary to exactly balance that resultant.
So in that reference frame, there is no reduction in potential energy. The bike+rider's CG is always the same distance above the tire. No change in PE, no resulting change in KE.
Again, a bit fuzzy. I haven't bothered to work out all the details. But for now, that's my bet.
--
- Frank Krygowski