Re: The old centrifugal force

[Rick Tarara <rtarara@SAINTMARYS.EDU>]

This is just like the Ferris Wheel which I used extensively to discuss this.
Look for the forces that 'cause' the centripetal acceleration at 12,9,6,and
3--rider facing left and wheel rotating counter-clockwise (also no safety
bar).

To answer your question--look at the 6 position.  Two forces--the earth
pulling down on the person and the chair bottom pushing up.  But the up push
has to be greater than the down pull by Mv^2/R.  The seat is pushing up with
more force than just the person's weight.  Your track is pushing on the ball
with more force than just the normal weight component--otherwise there is no
acceleration.  The ball is trying to move in a straight line but the track
has to apply enough force to keep the direction changing.

BTW:  This really is similar to your car problem.  The car wants to keep
moving in a straight line but the frictional interaction between the tire
and the road keeps it from doing so (sliding).  It is the inertia of the car
that is equivalent to the push you apply with your foot against the floor to
get the floor to push you forward.

Rick

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Richard W. Tarara
Professor of Physics
Saint Mary's College
Notre Dame, Indiana
rtarara@saintmarys.edu
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----- Original Message -----
From: "Ludwik Kowalski" <kowalskil@MAIL.MONTCLAIR.EDU>
To: <PHYS-L@lists.nau.edu>
Sent: Friday, November 14, 2003 6:44 PM
Subject: The old centrifugal force


> YES, I KNOW, THIS TOPIC ALSO HAS BEEN
> DEBATED IN THE PAST; SEVERAL TIMES.
>
> On Thursday, Nov 13, 2003, I wrote:
>
> > The textbook I am using has a provocative
> > conceptual question: "Explain why Earth is not
> > spherical in shape, but bulges at the equator."
> > What answer is expected?
>
> And here is another puzzle. In the same chapter
> a roller coaster problem is presented. Instead
> of analyzing the roller coster let me refer to a small
> object sliding (negligible friction) along the inner
> surface of a vertical loop. Suppose the sliding is
> counter-clock-wise, as in the textbook illustration.
>
> The object is in the 5 o'clock position. I draw the
> m*g as a vertical vector pointing down. I decompose
> it into two components: radial (away from the center)
> and tangential (clock-wise direction). Then I draw
> the normal force vector which has the same length
> as the radial component of m*g but the opposite
> direction. So far all is the same as when r is infinite
> (an object sliding along an inclined plane).
>
> But if I do just that then the net radial force is
> zero. This is not possible. An additional radial,
> force directed toward the center (m*v^2/r) must
> be added to "account for" (or to "explain," or to
> "cause," if you prefer) circular motion. But what is
> the nature of this force? Keep in mind that the
> the normal force, due to the radial component of
> m*g, has already been introduced. Another normal
> force must be present. What kind of constraint
> force is it? It can not be another reaction to the
> radial component of mg.
>
> In the rotating frame of reference I would say
> that the centripetal normal force is the reaction
> force due to the centrifugal force. The centrifugal
> force acts on the circular road and the constraining
> (centripetal) force is nothing else but the reaction
> force from the road. But how can the above
> question be answered without introducing the
> concept of the centrifugal force?
>
> Please share your answer. But before doing
> this draw the free-body diagram for the same
> object at the two o'clock location. Here m*g
> does not naturally decompose into radial and
> tangential components. Show that your way
> of explaining things work for both positions?
> Ludwik Kowalski