Re: [Phys-l] circular motion FBD

[Paul Nord <Paul.Nord@valpo.edu>]

On Oct 31, 2011, at 6:49 PM, John Denker wrote:

> > So the
> > forces/directions on the rider are different than those on the tire?
>
> They can't be much different.  The essential physics of the turn is 
> the same in all cases:  There is a force vector perpendicular to 
> the  velocity vector, and that changes the direction of the momentum
> vector without changing its magnitude.

The direction of the forces should be pretty much the same.  Magnitude of the forces will be very different.

The force on the tires needs to be large enough to accelerate* the the mass of the bike and the rider around a certain radius.  The force on the seat needs to only accelerate the mass of the rider.  Note that by saying "accelerate" here I don't mean changing the speed, rather it is a change in the direction of the motion.

It might be needlessly complicated to deal with the bike/rider system as anything but a point mass.  But if you must ask what the forces are on the rider, imagine him floating freely down a straight road.  How much upward force is required to keep him hovering over the ground?  Next ask: When going around a corner how much force is required to change his direction of motion around a particular radius?  F=(mv^2)/r

Paul