Re: L2-"Negotiating" a curve.

[Ludwik Kowalski <KowalskiL@MAIL.MONTCLAIR.EDU>]

I still do not know how to justify (explain) the net
centripetal force in terms of four external forces acting
on my simplified model. It is now a tricycle without
pedals. I draw it in the form of a large equilateral triangle,
the view from above. Two wheels at the base (back side
where the pushing force is applied) and one wheel at the
top (where the steering column is).

The axis of the back wheels is parallel to x and the planes
of all three wheels are parallel to y. If you want to participate
please make a drawing and use the same notation. The
platform (in the shape of an arrow head) moves in the y
direction, the speed is constant because the sum of four
external forces is zero. The forces are:

Fp (my push along the y axis),
Fr (rolling friction on the back right wheel),
Fl (rolling friction on the back left wheel) and
Ff (rolling friction on the front wheel).

The magnitudes of the three rolling friction forces are
identical; they are equal to Fp/3.

It is easy to make the free body diagram in this case.
One force along y and three forces along the -y. I am
assuming the tricycle is a rigid body (except for tires)
and that the mass of each wheel is negligibly small.

Now the front wheel is turned to the left. Why does
the net force, directed to the left, appear? Why should
there be a torque? I tried to answer this by changing
the direction of Ff but the conclusion was that the
platform should turn to the right. This must be wrong,
it conflicts with what actually happens.

I know what to say to avoid the dilemma, I was
doing this for years. But this time I am not
comfortable. Please help if you can.
Ludwik Kowalski

Mark Sylvester (referring to a four wheel vehicle) wrote:

> 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? ......