Re: [Phys-l] circular motion FBD

[John Denker <jsd@av8n.com>]

On 10/31/2011 04:24 PM, Anthony Lapinski wrote:
> When a car turns in a horizontal circle, there are three planar forces (on
> the tires). The upward normal force balances the downward weight (mg). 

True.

> The unbalanced force is static friction, which points to the center of the
> circle. 

True, in the terrestrial frame
(but not in the frame comoving with the driver).

> There also is static friction which moves the car forward.

I wouldn't have said that.  The car is going to move forward on
its own, in accordance with the first law of motion.  No forward
force is required, except maybe to overcome drag .... and even
then I wouldn't say that this "moves the car forward".

> What about a motorcycle turning in a circle? I am a bit confused with the
> FBD on the tires. Does the normal force still point upward since the road
> is still horizontal? 

You can define "normal" to mean anything you want ... but most
folks define it to mean normal to the road, in which case this
is exactly the same as the 4-wheel situation.

You could equally well call this the vertical component.

> The driver leans inward toward the curve, so he/she
> presses more into the seat. So the normal force on the rider is angled
> inward (and greater than mg). 

That's not the "normal" force as defined above.  I would call that
the total force in the transverse plane.  It can be resolved into
a lateral component and a vertical component.

> There's also friction of the seat. 

That is small, possibly zero, and irrelevant even if nonzero, since
it is internal to the cycle+rider system.  It's also not different
from the 4-wheel case.

> 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.