Chronology Current Month Current Thread Current Date
[Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]

Re: L2-"Negotiating" a curve.



I'll have a stab at Ludwik's L2 question. I'm feeling kinda L2ish at the
moment.

At 11:00 03/11/99 -0500, Ludwik Kowalski wrote:
We often say that the centripetal force acting on a
turning car (turning left or right on a flat surface) is the
force to static friction. Here are questions about this.

I am pushing a cart which has four parallel wheels. The
applied force (from behind) is just big enough to keep
the speed constant. The child driver steers the front
wheels to the left (front wheels are no longer parallel
to the back wheels) and the cart starts turning left.

We say that the centripetal force is frictional. Must the
external pushing force be increased to make sure the
cart moves along a circle with the same constant speed
or not? How does the "extra force needed", if any,
depend on the radius of the path curvature?


I guess the front wheels will have a static friction force acting on them,
perpendicular to their rolling direction. The unbalanced component must be
strictly centripetal for uniform circular motion of the trolley, so I
deduce that the tangential part of this static friction force, acting
backwards, must be balanced by the pusher, for constant speed. This
suggests that, yes, you have to push in order to turn at constant speed.
But what about energy? The KE of the trolley is unchanged and there's no
sliding, yet you are doing work pushing in the tangential direction.
Hmmm.... maybe turning is not so simple. Can it be that you do something
other than just pushing tangentially? For a car the situation seems
different depending on whether the vehicle has front-wheel drive or rear
wheel drive.

If all four wheels were locked in the parallel orientation
then I would have to "push harder to compensate for the
sliding friction".

I'm not sure what picture you're trying to convey here.

So much about the cart for which I am
an external combustion engine. A pedaling cart has an
internal engine, the child. Does the power of that engine
(work per unit time) must increase when the curve is
negotiated?

Now a real car. To simplify we may ignore the engine
and assume that the initial speed (when the turning starts)
was already present. Suppose somebody was pushing the
car up to that initial moment. A four wheeled car is not a
particle. So how should I explain the centripetal force
acting on its center of mass when the front wheels are
not parallel to the back wheels?


I think this is simpler: the centripetal force does not act on the centre
of mass. It's the unbalanced component of the friction acting on the
wheels. This tends to cause the vehicle to roll, and there will be a
corresponding opposite moment due to the normal reaction on the outer
wheels being greater than that on the inner wheels. Now if you want to use
the non-inertial reference frame of the vehicle itself and have a
centrifugal force acting on it, then this *will* act on the centre of mass
- for the same reason that the "force of gravity" acts on the centre of
mass, to refer to one of the few recent threads that I've had time to read,
but which may not be strictly L2!

A banked road is conceptually easier to deal with than
the flat road because here the centripetal force is mostly
gravitational.

Well, no. Not if the circle is horizontal. The centripetal force is the
horizontal component of the normal reaction from the banked road. I was
warning my class about this just this morning! It's an fme (frequently made
error).

But the question of the "mechanism" by
which it appears at the center of mass when the front
wheels are no longer parallel to the back wheels must
still be answered, somehow. Help will be appreciated.
I am in Chapter 6 of Serway, pages 155 and 156.
Ludwik Kowalski


ciao, Mark.


Mark Sylvester
United World College of the Adriatic
34013 Duino TS
Italy.
msylvest@spin.it
tel: +39 040 3739 255