Re: [Phys-l] circular motion FBD

["Anthony Lapinski" <Anthony_Lapinski@pds.org>]

Thanks for the help. I was thinking in the terrestrial frame.

Also, doesn't static friction propel cars and bikes (and people walking)
forward? If you stop pedalling or stepping on the gas, the moving vehicle
will eventually stop.


Forum for Physics Educators <phys-l@carnot.physics.buffalo.edu> writes:
> 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.
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