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

["Bob Sciamanda" <trebor@winbeam.com>]

Hi Jora,
The point of contact between tire and road is NOT an inertial origin. 
Torques relative to this accelerating origin are not equal to the time rate 
of change of system angular momentum.  There is an added term in the 
Torque/Angular momentum relation which restores equilibrium relative to this 
accelerating origin.
A reference of this general situation is the mechanics book by Becker 
(equatio 8-24 in the 1954 edition).

Bob Sciamanda
Physics, Edinboro Univ of PA (Em)
trebor@winbeam.com
http://www.winbeam.com/~trebor/
----- Original Message ----- 
From: "Lim Hwee Ke" <autumnrainz@gmail.com>
To: <phys-l@carnot.physics.buffalo.edu>
Sent: Tuesday, August 19, 2008 3:00 AM
Subject: [Phys-l] Turning effects of a moving bike


> Hi.
>
> 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?
>
> Cheers
> Jora
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