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?
If all four wheels were locked in the parallel orientation
then I would have to "push harder to compensate for the
sliding friction". 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?
A banked road is conceptually easier to deal with than
the flat road because here the centripetal force is mostly
gravitational. 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