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Old May 5th 08 posted to sci.physics.relativity,sci.physics
El Enrrabadore-mor[_2_]
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Posts: 173
Default Time dilatation in circular motion


"Greg Neill" escreveu na mensagem
m...
"El Enrrabadore-mor" wrote in message


You've missed the fact that during a given amount
of time a torque T = dL/dt have moved the body a
given angle theta (which is a basic coordinate).


No, I didn't miss anything. A torque doesn't have to
be accompanied by an angular movement any more than
a static force does.


You did miss exactly what you're saying now, but
never mind.


You forget the fact that derivatives are of the form:
d/dt f(x(t)) = x df/dt
x - is an angular displacement

Energy = Torque . angular displacement (dot product)


No. Work is torque x displacement.


I understand if you blame the 'dot product' and replace
it by a simple multiplication sign.
Actually, I've the strong feeling that I've messed up something
about the derivative argument, because what I can get
is a power balance, not an energy balance. Hence,
it is a velocity that comes out, not an angular displacement.

For circular motion:
Work (Kg m^2 s^-2) = Torque (Kg m^2 s^-2) *
* angular displacement (rad)
(relative to the center of rotation).



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