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Thanks, Joe. Let me see if I understand. Suppose a conical pendulum w=
ith no friction or air drag. The "centripetal force" on the ball is t=
he horizontal component of the string pulling it inward. The "centrif=
ugal force" on the string is opposite to the centripetal force on the=
ball, and acts to pull outward on the end of the string. Ball and st=
ring (c.o.m.) have different accelerations because each has a differe=
nt mass and different other forces acting on it. Is that what you're =
saying?
I thought earlier posts were implying a centrifugal force on the ball=
, which exists only in a frame of reference rotating with it, resulti=
ng from the ball's inertia.
skip=20
-----Original Message-----
=46rom: Joseph Bellina [mailto:jbellina@SAINTMARYS.EDU]
Sent: Monday, November 17, 2003 11:07
To: PHYS-L@lists.nau.edu
Subject: Re: Centrifugal force
A number of folks have responded to this well. I just want to point =
out
that will due respect, your question implies that you may be confusin=
g
issues having to do with the second law, the forces acting on a singl=
e
object, and the third law, equal but opposite forces acting on differ=
ent
objects.
This is a common problem for students, and shows up often through the
use of the words "equal and opposite" that has a very different meani=
ng
in the context of the 2nd law than it does in the context of the thir=
d.
It is a good question to ask ourselves and our students to be sure we
understand the conceptual difference.
cheers,
joe
On
Mon, 17 Nov 2003, Kilmer, Skip wrote:
I've never really understood Physics teachers' distaste for the phr=as=3D
e, centrifugal force. Doesn't N3 tell us that for every centripetal=f=3D
orce on an object there is an equal centrifugal force on another ob=je=3D
ct?
skip=3D20
Joseph J. Bellina, Jr. 574-284-4662
Associate Professor of Physics
Saint Mary's College
Notre Dame, IN 46556