Re: Conservation of angular momentum

On May 28, 10:36 am, "Greg Neill" <gneill...@xxxxxxxxxxxxxxx> wrote:
"Peter" <Poakfi...@xxxxxxx> wrote in message

news:1180358905.501021.260270@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx





I'll describe briefly an actual experiment involving a collision
between a point object (a steel cylinder with round ends) and an
extended object (a steel, equal-arm lever). The cylinder has a mass m
andangularvelocity w, and the lever (a thin rod of length L) has
mass 6m and moment of inertia 6mL^2/12. The cylinder hits
approximately the center of the right arm of the lever and stops on
impact. For simplicity, let us assume this collision is elastic, and
kinetic energy is conserved. Sinceangularmomentumr x p must also be
conserved, the initialangularmomentumof the cylinder and the final
angularmomentumof the lever must be equal, i.e., m(L/4)^2 w =
6mL^2/12 (w/8). Thus, the initial kinetic energy of the cylinder is
(1/2)m(L/4)^2w^2, and the final kinetic energy of the lever is
(1/2)6mL^2/12(w/8)^2. But then, the final kinetic energy of the lever
is only 1/8 of the initial kinetic energy of the cylinder. This cannot
be: we assumed the collision was elastic, and kinetic energy should
have been conserved. Someone might say: the cylinder could not have
hit the center of the right arm of the lever. I say, O. K., where
should have it hit, then? The answer is that no matter where the
cylinder hits, eitherangularmomentumr x p, or kinetic energy is not
conserved in this collision. This outcome reveals a fundamental
problem with present-day understanding of this part of mechanics. This
is what I am talking about.

A cylinder with round ends is not a point object.
You give this cylinder anangularvelocity, but do not
describe its center of rotation. You have not taken into
account theconservationof linearmomentum, nor specified
how the lever is supported (if at all). You can't just
assign kinetic energy to one "outlet" (rotational energy)
without showing that it is the only form that the kinetic
energy can take.- Hide quoted text -

- Show quoted text -

The cylinder acts as a point object; it does not rotate. The reason it
is a steel cylinder and not a steel ball, is that the ball rolls after
impact, and the cylinder, which is guided by a track -so it can be
aimed accurately- does not. The lever is mounted on a vertical axis,
on top of a table. The angular momentum of the cylinder m(L/4)^2w is
with respect to the axis. Since the cylinder stops on impact with the
lever, it has no linear momentum after the collision. In a collision
between steel objects, the coefficient of restitution is very high,
and it can be assumed not much heat is generated in the collision.

Peter


.

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