On 14 Oct 2003 22:41:51 GMT, a (Dave Baker) wrote:
Subject: Compressor Motor: HP v.s. Amps?
From: Gary Coffman
Date: 14/10/03 23:33 GMT Daylight Time
Message-id:

On 9 Oct 2003 05:44:20 -0700, jim rozen wrote:
But for a pendulum, the displacement
and the force are still orthogonal.

No it isn't. The pendulum swings down
and in, or up and out. The up and down
components are in line with the force.
So work is done in each half of the
swing. It is just that the positive work
while the pendulum is swinging down
is offset by the negative work done
as the pendulum is swinging up. So
net work is zero.

The dot product is still zero.

Depends on over what part of the swing you calculate
it. If you picture the pendulum end points as a and c,
and call the midpoint b. Then from a to b positive work
is done, and from b to c negative work is done. If you
calculate directly from a to c, of course, the result is
zero. That's because you're adding the work done in
the two halves of the swing together.

Wac = Wab + (- Wbc)

Since, ignoring friction, Wab and Wbc have the same
magnitudes, but opposite signs, Wac should always
equal zero.

I'm not convinced this recent line of argument about zero net work is actually
valid in the first place. All a pendulum does is convert potential energy into
kinetic and vice versa. The only work that is done is against the friction in
the pivot and that is a scalar quantity and irreversible.

That's not what the definition of work says, though. The definition contains
no requirement of irreversibility. In fact it is easy to show that net work can
be done in a system which is nondissipative.

Feynman analyzes pendulum motion as I have above precisely to show
why a closed reversible system must exhibit both positive and negative work
in order to be consistent with both the definition of work and the principle
of conservation of energy.

Gary