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Re: L2-"Negotiating" a curve.



Arlyn DeBruyckere wrote in part:

On the banked curve the vertical component of N is NOT equal to mg. If it
implies that on the diagram then the diagram is not to scale or it suggests
something that is not correct. The vertical component of N on a banked curve
will be less than -mg because "some of mg" is used as the centripetal force.

This statement is exactly why it is dangerous to allow beginning
students to think in terms of "centripetal force" rather than
centripetal acceleration. NSL is a statement of cause and effect. The
left hand side, the "net force," is the cause part, and the right
hand side, the "ma," is the effect part. In the case of circular
motion, that part of the acceleration that is directed centripetally,
is entirely due to forces being applied by outside agents--strings,
gravity, wing lift, magnetic fields, etc., etc. There is no single
force that can properly be named "the centripetal force," since that
is the resultant of all the other forces. There is no "centripet"
that exerts a centripetal force. Such a force should never appear on
a free-body diagram, since it is (at least part of) the net force
which causes the right hand side of NSL to be what it is. Allowing
students to think about a centripetal force can get them in all sorts
of problem-solving trouble, because they try to include it with all
the other forces and then cannot understand why they get crazy
results. I have seen experienced HS physics teachers get in trouble
with this.

We all know that if we are willing to use an accelerated reference
frame (rotating), then there appears a centrifugal force that can
properly be incorporated in the force diagram, and advanced problem
solvers use this technique to simplify the solution of certain types
of problems all the time. But for beginning students they should be
strictly limited to talking about centripetal acceleration, and the
forces that give rise to that acceleration, none of which is properly
called a centripetal force. In this sense, I think that Hewitt did
indeed "blow it" on his example.

In other words, just like mg causes a horizontal force on an inclined plane
(=mgsin(angle) for the force *along the plane*, part of this is
horizontal {too
many angles and forces to describe without sketches}) it will also cause a
horizontal force in the circular motion. Remember centripetal force in this
case *IS* the net force. The net force causes uniform circular motion and is
directed to the center of the circle (if we are using the "standard" earth as
the frame of reference).

As noted above, the comparison between mg and the net force making an
object go in a circle is not valid. Gravity has a readily
identifiable origin that is outside of the context of the system
under investigation. Centripetal force does not, and always ends up
being the resultant of other forces.

In the case of a ball being twirled in a vertically oriented circle,
mg provides part of the force that makes it go in a circle, part of
the time, but is trying to make it do other things during the rest of
the circle. But if one looks at the (vector) sum of mg and the
tension in the string, and equates that sum to the "ma" on the right
hand side, then there will be a component of "a" that will have the
form "v^2/r", and can be called the centripetal acceleration. Its
magnitude will vary depending on the point the ball is in its circle,
being a maximum and the bottom and a minimum at the top. If you add
to these two forces a centripetal force, then try to get a sensible
answer. It can't be done.

Since centripetal forces are always made up of forces applied by
other sources, it makes no sense to include them at all. Just use the
actual forces and let the centripetal part be associated with
acceleration. Pedagogically, this makes so much sense that I cannot
understand why anyone does it any other way.

Hugh


Hugh Haskell
<mailto://hhaskell@mindspring.com>

Let's face it. People use a Mac because they want to, Windows because they
have to..
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