Re: [Phys-l] Turning effects of a moving bike

[Steve Highland <shighlan@uslink.net>]

> I've got a question that's puzzling me regarding the classic question of why
> a cyclist needs to lean inwards towards the centre of the circular motion.
>
> Usually we say that the rider leans inwards so that there is rotational
> equilibrium about the CG of the rider right? The frictional force F produces
> a moment about the centre of gravity G such that the normal contact force
> produces a counterbalance moment. However, while taking the moments about G
> solves the issue of rotational equilibrium, if we take the point of rotation
> to be anywhere else, (say the point of contact of the rider and the road),
> the turning effects no longer appear to be balancing. There'll only be one
> moment due to the weight of the bike and biker. Where's the other force
> ensuring that the biker does not fall to the ground? Or am I doing something
> wrong?
>
> Can anyone help?

Years ago when I was teaching a mechanics course using a book by Hibbeler I
learned that you need to include an "r x ma" term in the torque equation if
you're not taking torques about the center of mass.

If you want to choose the contact point as your origin for computing torques
then the net torque doesn't add up to zero -- it adds up to the 'torque' of
the ma vector acting through the center of mass point.

I just worked the leaning rider problem using both the CM as the origin and
the contact point as the origin and the answers came out the same.


This issue has had me thinking for many years about whether the condition
for rotational equilibrium is a consequence of Newton's Laws or is something
imposed in addition to them -- what do the rest of you think?

Steve Highland