Re: Conservation of angular momentum

On May 28, 2:24 pm, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
SperM.hotmail.com> wrote:
"Peter" <Poakfi...@xxxxxxx> wrote in messagenews:1180373932.320360.123700@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On May 28, 11:51 am, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
SperM.hotmail.com> wrote:
"Peter" <Poakfi...@xxxxxxx> wrote in messagenews:1180358905.501021.260270@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

[snip]

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.

The fundamental problem is entirely with your past-, present- and,
without any doubt, future-day understanding of this part of mechanics.

Dirk Vdm- Hide quoted text -

- Show quoted text -

I see you did not bother to read and understand what I said.

Yes, I have read it and I have read your other 'contributions'.
I have understood that you have no intention to do something
about your deplorable but entirely self-inflicted condition of
arrogant ignorance. But if it makes you feel less lonely, what's
the problem, right?

Dirk Vdm- Hide quoted text -

- Show quoted text -

If I am so wrong, it should be easy for you to pinpoint my errors. Why
don't you?

Peter

.

Relevant Pages

  • Re: Conservation of angular momentum
    ... translate (so that the total kinetic energy is the sum ... If a cylinder moving in a straight line hits a pivoted lever, ... After the collision, the lever and what it is attached to will ...
    (sci.physics)
  • Re: Conservation of angular momentum
    ... extended object (a steel, equal-arm lever). ... The cylinder has a mass m ... the initial kinetic energy of the cylinder is ... cylinder hits, eitherangularmomentumr x p, or kinetic energy is not ...
    (sci.physics)
  • Re: Conservation of angular momentum
    ... extended object (a steel, equal-arm lever). ... The cylinder has a mass m ... the initial kinetic energy of the cylinder is ... cylinder hits, eitherangularmomentumr x p, or kinetic energy is not ...
    (sci.physics)
  • Re: Conservation of angular momentum
    ... extended object (a steel, equal-arm lever). ... The cylinder has a mass m ... Since angular momentum r x p must also be ... the initial kinetic energy of the cylinder is ...
    (sci.physics)
  • Re: Conservation of angular momentum
    ... extended object (a steel, equal-arm lever). ... The cylinder has a mass m ... the initial kinetic energy of the cylinder is ...
    (sci.physics)