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Re: The old centrifugal force



My question was asked in the context trying to
construct a free-body force diagram for an object
sliding counter-clock-wise along a looping track,
for example in the two o'clock location. The first
arrow, representing the force m*g, and pointing
down has already been drawn. What other
arrows must be drawn and how should they be
explain them to students?

Perhaps I haven't followed this thread carefully enough, but I would
hope the answer would be something like this:

Since the object is in contact only with the track, the only
remaining force to indicate is the force of contact with the track.
Contact forces should already have been explained to the students in
terms of the deformations--compressive or tensile for the normal
component and shear for the frictional component--that *always*
accompany "contact."

1) The diagram MUST SHOW ALL FORCES ACTING
ON THE SLIDING OBJECT. We assume that forces
due to frictional and Coriolis effects are negligible.
Please be specific, I need this for a Monday class.

If the frictional force is negligible then the contact force is
purely normal to the track surface. There is no coriolis force (or
centrifugal force for that matter) in the inertial frame that I
assume you are working in.

2) What is the nature of the force acting on the track?
(Am I still allowed to say "what causes it?"). I know
what Millikan say about this force. In the "A First
Course in Physics," copyright 1906, he and Gale
wrote: Inertia manifesting itself in this tendency
of the parts of rotating systems to move away from
the center of rotation is called centripetal force."

That sounds very peculiar to me. Why would anyone label a "tendency
... to move away from the center of rotation" using the adjective
"centripetal"?

Accepting this I would say something like this:

a) the sliding object exerts a force on a track;
it is the force of inertia directed away from the
center (not necessarily along the radius).

I don't see any need to make it that complicated. The force on the
track is due to contact with the sliding object. It is the Newton's
third law interaction partner of the contact force on the object.
(I'm not completely sure why you care about the force on the track,
but I don't think it offers any particular challenges to
understanding.)

b) That centripetal force ("due" to rotation) does
not act on the sliding object. But the spring-like
reaction to that force must be drawn.

? Now I'm confused. Maybe I'm not sure what "the system" is any
more. When you say "that centripetal force" do you mean the contact
force on the track? Because it is directed away from the interior of
the curve I would not call it centripetal. And because we are not
analyzing the problem in a rotating frame, I would not call it
centrifugal. (Furthermore, even if we were analyzing the problem in
a rotating frame, I would not call it a centrifugal force.
Centrifugal forces are not the result of contact with other objects.
Contact forces are contact forces, inertial frame or not.

c) We draw the second force and give it a name,
such as constrain force, C. The direction of that
force is neither radial nor tangential, for example,
at some angle from the vertical m*g.

? Do you mean the force of contact on the object? If so, then why
isn't it radial? I thought you said friction was negligible. Are
you, perhaps considering a noncircular track? If so, then "radial"
isn't a very meaningful term except insofar as it means
"instantaneously normal" anyway.

d) The net force, R, is the sum of mg and C. The
radial component of R is association (or causing
if you prefer) centripetal acceleration while the
tangential component is "responsible for" the
change in the in the instantaneous speed.

Despite my prior confusion, this sounds pretty reasonable.

e) The mass of the object was given. Knowing
the radius of the loop, and the instantaneous
speed, one can calculate the centripetal force at
the two o'clock location.

I wouldn't say it that way for fear of inducing students to think
that mass times acceleration IS force. Instead I would say that one
can calculate the radial component of the acceleration (from v and r)
and one can multiply it by the mass to get a quantity that, by
Netwon's second law, must be *equal to* the radial component of the
net force.

Likewise, knowing the
tangential acceleration one can calculate the
tangential component of R.

I *really* wouldn't say that. How do you "know" the tangential
acceleration? It seems to me that what you know is the tangential
component of the net force. You can divide that by the mass to
obtain a quantity that, by Newton's second law, must be *equal to*
the tangential acceleration.

--
John Mallinckrodt mailto:ajm@csupomona.edu
Cal Poly Pomona http://www.csupomona.edu/~ajm