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Re: [Phys-l] Sharing a problem for students



On Dec 24, 2007, at 10:56 AM, Bob Sciamanda wrote:

Ludwik,
It might help to realize that the effective potential U(r) is more
than just a helpful visualization tool for the inertial observer
(helping him to abstract from the angular motion and consider only the
one-dimensional radial motion, It is also the actual view which a
rotating observer would take of the planet. The added term in U(r) is
precisely the potential energy which a rotating observer would ascribe
to the centifugal force which he sees affecting the radial motion of
the planet. The radial motion is all that he sees, since he rotates
along with the angular motion of the planet. The rotating observer
indeed sees one dimensional motion in a potential energy well - no
paradox here.

I would say that there is no paradox for the inertial observer either,
because he realizes that this is not one-dimensional motion. Looking
only at V(r) ignores the important complicating effects of the angular
motion.

I found an appealing analogy for the "effective potential energy" of circular motion. Consider a familiar problem of finding the maximum height for a ball projected vertically with the initial speed v. The answer is in:

K=U

where K is the initial kinetic energy and U is the potential energy at the maximum height. As usual, we choose U=0 at y=0. Thus

ymax=v^2 / (2*g)

A person in an elevator can experimentally confirm this formula, but only when the elevator is either at rest or moves without acceleration. In the presence of acceleration, a, the energy formula above must be modified by adding a potential energy term that is not due to gravity. The maximum height can then be calculated from

K=Ueff,

where Ueff = U - m*a*y = m*(g-a)*y

The m*a is a pseudo force (not due to any agent) and the m*a*y is the corresponding pseudo potential energy. The direction of a pseudo force is opposite to the direction of the frame's acceleration, a. Here are numerical illustrations (for g=9.8 m/s^2 and v=5 m/s, declaring down as positive).

a=0 --> h=1.27 m
a=6 m/s^2 --> h=3.29 m The elevator is gaining speed on the way down or loosing speed on the way up.
a= 9.5 m/s^2 (the elevator is nearly in free fall) --> h=20.8 m (make sure the elevator has a high ceiling)
a=12 m/s^2 (gaining speed faster than in free fall) h= -2.84 m (silly answer, the ball will stay at the ceiling).

a= - 6 m/s^2 (for example when the elevator is gaining speed on the way up) h=0.79 m
a -10 m/s^2--> h=0.63 m
a=1000 m/s^2 --> =0.012 m.

Even more trivial illustration is a box moved along a horizontal surface, with acceleration a. Think about an observer in the box and a frictionless slider, on the floor. The only force acting of the slider (it can be measured with a spring force-meter) is m*a; the direction of this pseudo-force is opposite to the direction of the acceleration. Suppose the slider is at rest on the left side of the box. Its apparent (effective weight, pushing against the left wall, is m*a. Ignoring vertical motion of the slider we can say that its potential energy, U'= m*g*x, increases when x (the distance from the left wall) increases. That is the pseudo potential energy associated with the pseudo force m*a. In other words, the entire potential energy is "pseudo," even a small part of it is not due to an agent, like earth or magnet. Suppose the slider is pushed to the right with an initial velocity v. Find the maximum possible x. The solution can again be found by comparing two energies, kinetic and potential. The only difference is absence of g. I know that all this trivial. But it helps to digest the idea of the effective potential energy (as opposed to real potential energy) for a planet orbiting the sun. Yes, pseudo forces and pseudo potentials are real.
_______________________________________________________
Ludwik Kowalski, a retired physicist
5 Horizon Road, apt. 2702, Fort Lee, NJ, 07024, USA
Also an amateur journalist at http://csam.montclair.edu/~kowalski/cf/