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Re: The old centrifugal force



I am not going to correct my message again
after reading what JohnM wrote.

On Saturday, Nov 15, 2003, John Mallinckrodt wrote:

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

This typing error was corrected in the message posted
several minutes ago.

? 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.

OK, it should be radial when friction is negligible.
I am not the best person to deal with this problem.

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.

OK, I understand. It is a pedagogical issue worth addressing.

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.

Tangential acceleration can be determined experimentally,
for example, by filming the object at high speed. This would
produce data to find the rate at which speed is changing.
Ludwik Kowalski