> From: Jim Crumley <crumley at belka.space.umn.edu> > Alas, I must pick some nits. By all means. > > Audio CD's start at 500 RPM and slow down to 200 RPM. That's 1x. > > The fast drives go CAV at one speed, and are fastest at the end of > > a disc, so let's go with 52x200 or 10,400 RPM. > > Hopefully your numbers are right here - I didn't check. That one I'm very sure of. Then again, it is a reported speed, and is probably rounded for marketing literature. Looking for citations, found this: http://www.powerlabs.org/cdexplode.htm in which a guy mounts a CD onto a Dremel tool for 35000 RPM fun. (With a picture!) Here we go: http://hypertextbook.com/facts/2000/LawrenceFung.shtml > > The speed of the edge of a disc v=wr which is the rotational > > speed times the radius, or 10400RPM * 6cm or 62400cm/min or > > (uhm... times 1min/60s times 1m/100cm) 10.4m/s (!) or > > 62400 cm/min (times 60min/1hr times 1mi/160934cm) = 23.26 mph. > > w in v=wr is angular frequency, which is radians per second. > "Regular" frequency is w/(2 pi). Or v = 2 pi f r. So the actual > speed at the edge is 392000 cm/min = 65.3 m/s = 146 mph. You are correct, of course. Pesky angular units! The second site I referenced confirms your calculation. > > CD's have a mass of about 20g. A 1g chunk at 10.4m/s should > > have 1/2 * m * v * v energy or about 1/20 (need help with unit > > here...) joule? Or the same energy as a kilogram dropped from > > a height of 5.5mm. (U = mgh) > > A 1 gram chunk at 65.3 m/s would have 2.1 joules of energy. > While the disk is spinning, its kinetic energy = 1/2 I w^2. I, > the moment of inertia is 1/2 m r^2. So, > KE_rot = 1/2 (1/2 m r^2) (v/r)^2 = 1/4 m v^2 = 16 J , with > a CD mass of 15 grams. For comparison, this is the same as the > kinetic energy of a golf ball going 59 mph (26 m/s) - which would > be slow for a golf ball. And that's all probably correct too, but as soon as the 1g chunk breaks free, it's just a chunk going in a straight line (for an instant) at 146mph. I used the linear momentum (1/2 m v^2) equal to potential of mgh, but using the new numbers, the drop height would be 217mm or 8.6 inches. Thanks for the correction, Chris _______________________________________________ TCLUG Mailing List - Minneapolis/St. Paul, Minnesota http://www.mn-linux.org tclug-list at mn-linux.org https://mailman.real-time.com/mailman/listinfo/tclug-list