Re: Car acceleration

["John S. Denker" <jsd@MONMOUTH.COM>]

Gary Turner wrote:
> From a FBD, it appears as though the friction between the car and the road
> provides the force to accelerate the car.

Yes.  That seems pretty clear.

> But what is that friction?  It
> is not kinetic because these surfaces are not sliding.

OK.

> It cannot be
> static, because static can do no work - yet the car gains kinetic energy as
> a result of that force.

I wouldn't have said that.

Surely the car gains momentum from the road via
friction.  But I see no evidence that it gained
energy via that channel.  In particular, if I draw
an imaginary boundary enclosing the car, excluding
the road, there is no energy flow across that
boundary.  The gain in KE is associated with a
loss in stored energy in the fuel tank.

(BTW this does _not_ go to show that the alleged rule
that "static friction can do no work" is true in
general.  If this were a cable car, it could attach
itself to the moving cable via static friction and
extract lots and lots of energy.)

> It must be something alse, possibly a quasi-static
> friction.

The difference between static and quasi-static is
immaterial in this situation.

> But where does that come from?  Is it due to the flexing of the
> tire?

No.  There is no reason to believe that flexing
of the tire plays any role in the aforementioned
increase in KE.

> What happens then if you take a rigid tire, with a small footprint and only
> static friction.  Could you accelerate with this?

Sure.

> Another option I looked at - consider the wheel/axle as one system and
> the 'car' as another.  There are three pairs of forces, the friction
> between the tire and the ground, the contact in the axle housing and
> the 'drive - force' from the engine/drive shaft.
> Suppose the whole thing is going to move off to the right, with the wheel
> rotating CW.  The friction will produce a CCW torque on the wheel, the axle
> housing (assume frictionless) will produce no torque, so the drive must
> produce a CW torque.  That will produce a CCW torque on the car - balanced
> presumably by the normal on the rear wheels.

All true.

> If the drive and friction push the wheel to the right and the housing to
> the left, it is still possible to produce a force on the car to the right
> as well.  This force will act over a distance - and will do work.  The
> force on the wheel will act over a distance and will do work.

Yes.

> Now, is that sufficient?  Is the car going to move?  I have relied on
> static friction only here and the end result is that the force is between
> the engine and tires.

Sounds good to me.

> [The analogy of a skater pushing off a wall is not completely valid - that
> force is exerted over a finite time and hence a finite distance.]

It seems to me that the analogy is essentially perfect
in this case.  In particular, consider the following three
cases:
 -- skater pushing off the wall
 -- pushing a punt along using a pole, then resetting
    the pole and pushing again
 -- pushing with one piece of tire, then the next, then
    the next, then the next, as it rolls along.

The repetitive but piecewise-static action of the punter
serves as sort of a "missing link" between the once-only
push of the skater and the continuous pushing of the tire.
To make the relationship even more clear, imagine replacing
the tire with nothing but a lot of radial spokes.  Each
spoke makes static contact for a time, like a punt-pole,
then the next spoke takes over.  In the limit of innumerably
many spokes, this becomes roughly equivalent to a solid tire.