I think I am in agreement with John Mallinckrodt and Leigh concerning
the centripetal force and free-body diagrams. Just to be sure, let me
explicitly state what I do.
I have a handout in which I provide some guidelines I have found
helpful for drawing free-body diagrams. In that handout I say the
following:
* * * *
If the object is following a curved path, there must be a component of
the net force pointing perpendicularly to the velocity vector. This
component is often called the centripetal force. However, the
"centripetal force" is not a force in itself, it is a component of the
other forces that you have already drawn on your free-body diagram.
Therefore, do not draw a centripetal force vector on your free-body
diagram. The trick is to make sure that you get the acceleration
correct. For curved paths you must realize there is an acceleration
component in the direction perpendicular to the velocity. The
magnitude of this acceleration is v^2/r. If you include that
acceleration when you set F(net) = ma, then everything should work out
correctly even though there isn't any force labeled "centripetal force"
anywhere in your drawings. Align one axis of your coordinate system
with the velocity, suppose this is x. Align one of the other axes with
the instantaneous center of curvature of the path, suppose this is y.
a(y) = v^2/r, and F(y)(net) must equal ma(y).
* * * *
Notice, at this point I am staying out of the discussion about
action/reaction pairs, etc. I just wanted to react to the recent
discussion about whether the centripetal force should be drawn on
free-body diagrams. I believe it should not.
Michael D. Edmiston, Ph.D. Phone/voice-mail: 419-358-3270
Professor of Chemistry & Physics FAX: 419-358-3323
Chairman, Science Department E-Mail edmiston@bluffton.edu
Bluffton College
280 West College Avenue
Bluffton, OH 45817