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Re: An Integral oF Zero



Tom, N = mv^2/r + mg*cos(Theta), as you wrote. The first term is always
positive, but the second term goes through zero and becomes negative as
Theta advances through pi/2. If the first term is not large enough, N
will go to zero, contact with the loop will be lost - resulting in a
cessation of any frictional force.

Different problem: If the ball rides a hoop which threads through a hole
in the ball, then N can go negative, the above equation is still valid,
a negative N simply meaning that the hoop is pushing out instead of in
(which you defined as the meaning of a positive N). However in
evaluating the frictional force you must use f = mu* |N| , the direction
of f always being given by -v (opposing the velocity).

Hope this is useful

-Bob
--

Bob Sciamanda sciamanda@edinboro.edu
Dept of Physics sciamanda@worldnet.att.net
Edinboro Univ of PA http://www.edinboro.edu/~sciamanda/home.html
Edinboro, PA (814)838-7185

The simplest schoolboy is now familiar with facts for which Archimedes
would have sacrificed his life.
Souvenirs d'enfance et de jeunesse. - Ernest Renan

********************************
Tom McCarthy wrote:

I have a question regarding what seems to be a straight forward integral
yielding null results. Suppose a ball rolls on a vertical loop-de-loop with
friction providing more than just torque (i.e., more than one point of
contact--the realistic case). Also, suppose that you wish to calculate the
velocity necessary to just make it around the loop, so that the frictional
force is zero at the top, where the normal force is zero. The equation for
friction is: f = (mu) N, where N = mv^2/r + mg cos(theta), where theta = 0
at the bottom of the loop. Now, the centripetal term can be written in
terms of potential energy of the balls release point, assuming this is how
the ball is getting its energy to make it around the loop (neglect the
friction the ball has rolling down the ramp while traveling to the loop,
since this is calculated in a straightforward manner.
To get the energy lost due to friction, an integral of the form,
Scos(theta)d(theta), from 0 to pi must be calculated but this gives zero!
If I break it up into two parts and ignore the signs, I will get a net
result which is probably correct. How can this integral be performed
without having to intervene half way through? I mean, the work integral
along the path should produce the correct result without tampering with the
process.
Thanks.

Tom McCarthy