This is a multi-part message in MIME format. ---------------------- multipart/alternative attachment Sarah: 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. =20 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. =20 Minor Reason: The hammer shank and knuckle have mass; mass that is not = neglible and that will affect its radius of gyration. Major Reason: The shank is also rigid, so the mass of the hammer head = wouldn't even act from its center of gravity even if the shank had zero = mass. The center of percussion of the whole hammer (which is nowhere = close to the cg of the head or 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.=20 Actually calculating this stuff isn't really that tough ususally. I do = it every day. But you have to differentiate rotational from linear = effects. Incidentally, roller-coaster calculations wouldn't be done my way or = 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 = that easy! 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! Don A. Gilmore Mechanical Engineer Kansas City ----- Original Message -----=20 From: Sarah Fox=20 To: Pianotech=20 Sent: Wednesday, December 17, 2003 11:16 PM Subject: Re: Key Inertia Hi Don, <<Thirdly, the dynamic motion of the hammer has been described herein = as a linear problem, which it is not. >> 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. But since you raise the point, kinetic 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. So 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! ;-) Peace, Sarah ---------------------- multipart/alternative attachment An HTML attachment was scrubbed... URL: https://www.moypiano.com/ptg/pianotech.php/attachments/14/d5/18/78/attachment.htm ---------------------- multipart/alternative attachment--
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