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<DIV><FONT face="Courier New" size=2>Sarah:</FONT></DIV>
<DIV><FONT face="Courier New" size=2></FONT> </DIV>
<DIV><FONT face="Courier New" size=2>Sure, angular properties are =
similar in
algebraic form to linear ones, but they are in no way interchangeable =
any more
than spinning a flywheel is the same as shooting it out of a =
cannon. To
calculate the energy or momentum of a purely rotating object using =
linear
formulas will give you the wrong answer. Ideal approximations =
are
often used in engineering, but you must first consider their =
eligibility.
</FONT></DIV>
<DIV><FONT face="Courier New" size=2></FONT> </DIV>
<DIV><FONT face="Courier New" size=2>Rotating objects have a moment =
of inertia
that depends on their shape, mass/density distribution and the location =
and
orientation of their axes of rotation. A piano hammer pivoting =
about an
axis will not behave like a point-mass rotating on a string for two very =
important reasons. </FONT></DIV>
<DIV><FONT face="Courier New" size=2></FONT> </DIV>
<DIV><FONT face="Courier New" size=2><U>Minor Reason</U>: The hammer =
shank and
knuckle have mass; mass that is not neglible and that will affect =
its
radius of gyration.</FONT></DIV>
<DIV><FONT face="Courier New" size=2></FONT> </DIV>
<DIV><FONT face="Courier New" size=2><U>Major Reason</U>: The shank =
is also
rigid, so the mass of the hammer head wouldn't even act from its center =
of
gravity <EM>even if the shank had zero mass</EM>. The center of
<EM>percussion</EM> of the whole hammer (which is nowhere close to the =
cg of the
head <EM>or</EM> the whole hammer) would determine its behaviour =
and a
non-trivial (though easily calculable) part of the impact =
would be
absorbed by the pivot itself as a result. </FONT></DIV>
<DIV><FONT face="Courier New" size=2></FONT> </DIV>
<DIV><FONT face="Courier New" size=2>Actually calculating this stuff =
isn't
really that tough ususally. I do it every day. But you have =
to
differentiate rotational from linear effects.</FONT></DIV>
<DIV> </DIV>
<DIV><FONT face="Courier New" size=2>Incidentally, roller-coaster =
calculations
wouldn't be done my way <EM>or</EM> your way. Neglecting friction, =
this
would merely involve changes in potential energy with respect to
elevation. Radii and the direction of travel would
be immaterial. This is one of those freshman physics =
problems...but
I'll bet roller-coaster engineers don't get off quite <EM>that</EM>
easy!</FONT></DIV>
<DIV><FONT face="Courier New" size=2></FONT> </DIV>
<DIV><FONT face="Courier New" size=2>Please don't get =
the impression that
I'm trying to be a smart-ass. I'm really just trying to =
help. This
is fun!</FONT></DIV>
<DIV> </DIV>
<DIV><FONT face=Arial size=2>Don A. Gilmore<BR>Mechanical =
Engineer<BR>Kansas
City</FONT></DIV></DIV>
<BLOCKQUOTE dir=ltr
style="PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; =
BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px">
<DIV style="FONT: 10pt arial">----- Original Message ----- </DIV>
<DIV
style="BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: =
black"><B>From:</B>
<A title=sarah@gendernet.org =
href="mailto:sarah@gendernet.org">Sarah Fox</A>
</DIV>
<DIV style="FONT: 10pt arial"><B>To:</B> <A =
title=pianotech@ptg.org
href="mailto:pianotech@ptg.org">Pianotech</A> </DIV>
<DIV style="FONT: 10pt arial"><B>Sent:</B> Wednesday, December 17, =
2003 11:16
PM</DIV>
<DIV style="FONT: 10pt arial"><B>Subject:</B> Re: Key Inertia</DIV>
<DIV><BR></DIV>
<DIV><FONT face=Arial size=2>Hi Don,</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2><<Thirdly, the dynamic motion =
of the hammer
has been described herein as a linear problem, which it is
not. >></FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>Ah, yes, true. However, the =
same sorts of
principles most of us have been arguing apply equally to linear and =
angular
systems. If linear systems are so hard for everyone to =
understand,
angular systems would make most people's brains bleed! It's =
perhaps more
useful, I think, when talking about transfer of energy from one =
component to
another to another to another, to pretend like we're talking about a =
linear
system, even though it really ain't so. At least then it's =
possible to
see how energy is lost in the system, which was really my only point =
in the
first place -- before I got mired down in this whole =
thing.</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>But since you raise the point, =
k</FONT><FONT
face=Arial size=2>inetic energy for a point mass "orbiting" =
rotating
about a point can also be described as (mv^2)/2. Also, =
velocity is
angular velocity times radius, torque is tangental force times =
radius,
etc. S</FONT><FONT face=Arial size=2>o it's really not so =
different from
a linear system, is it? Granted, angular moments of inertia are =
useful
for describing rotation of complex mass distributions, but for =
describing
something like a key lead, which is almost a point mass, aren't linear =
terms
really close enough? If my physics professors had asked us to =
describe
the kinetic energy of a roller coaster in angular terms, using the =
radii of
curves in the track and "torque" exerted by gravity, I think there =
would
surely have been a mutany! ;-)</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2>Peace,</FONT></DIV>
<DIV><FONT face=Arial size=2>Sarah</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial =
size=2></FONT> </DIV></BLOCKQUOTE></BODY></HTML>