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Rotational Physics

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Wed May 21, 2008 5:49 am PostPost subject: Rotational Physics
Harvey
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Problems:

Using SI (MKS),

1. Graph the MOI of a symmetrical rotating cylinder 258 grams with a radius of 72mm from rest to 1250 RPM's.

2. Graph the angular momentum for #1

3. Graph the Kinetic Energy for #1

Please show the formulas used in the provided math format. Example:

`I=\frac { mr^2}{2}` where I = moment of inertia in kg m, m = Mass in kg, r = Radius in meters. (Scaler)

#1 will graph as a flat line from 0 to 1250 with a value of .000669 kilogram meters squared.

#2 Using `L = I\omega` where L = Angular Momentum, I = MOI = .000669, ω = Angular Velocity = `2\pi\f` we get a linear xy graph where x = RPM and y = L = x 0.000700575161750524

#3 Using `KE_{rotational}=\frac{I\omega^2}{2}` we get a non-linear xy graph where x = RPM and y = `KE_{rotational}` starting with 0 and ending with 5.71 Joules


Last edited by Harvey on Wed May 28, 2008 9:49 am; edited 6 times in total
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Wed May 21, 2008 10:39 pm PostPost subject:
alsetalokin
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Ooh, this is too much like School!!

I thought this was sort of interesting...
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Thu May 22, 2008 2:18 am PostPost subject:
Harvey
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alsetalokin wrote:
Ooh, this is too much like School!!

I thought this was sort of interesting...


...like School... that's a good thing, right? Wink

Yes I have seen those equations before and the first thing I wondered is where are the references to ambient charges? IOW, I suppose he is referring to air at sea level, at some barometric pressure, humidity and temperature. Given some standard for these, he then has to quantify the ambient charge which can be quite different from one area to another. Then, is it air in sunlight or air in darkeness and does the presence of photons matter? (It certainly seems to matter in the case charge buildup for lightning).

So what are we looking at? It seems to be an avalanche of Ions triggered by a small energy quotient but then we need to move 1 cubic meter of air for another charge set or suffer the reionization process at an energy loss relative to `f_{ion}`.

I'm not saying he's not correct here, I'm just saying its a bit convoluted and has some loose ends. Wink

Edit: I also find it interesting that his Testatika Schematic leaves out the capacitors. But his Single Disk description certainly does have a nice photo. Smile
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Fri May 23, 2008 10:35 am PostPost subject:
Harvey
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Next Problem:

Chart the necessary force vectors to accelerate the above MOI from 0 to 1250

In order to be meaningful there must be a time frame. Reversing the rundown time 120 seconds for the WhiPMag we can average the force needed to accelerate to 1250 RPM over that time frame.

Given 120 seconds overall time to reach 1250 RPM from 0 we will find an acceleration of 1.090831 Rad/s. The expected force to be applied tangent at 72mm is 0.010136 Nm throughout the 120 seconds. Thus the chart is a flat line across the RPM.

For Tangent Force we have `F=\frac{\tau}{r}` where F is force in Nm, r is radius in meters.

`\tau = I\alpha` : Torque = Moment of Inertia times Angular Acceleration

`\alpha = frac {\Delta\omega}{\Delta t}` Angular Acceleration = Delta Angular Velocity divided by Delta Time


Last edited by Harvey on Sun Nov 02, 2008 6:20 am; edited 2 times in total
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Tue May 27, 2008 1:15 pm PostPost subject: Stator Calcs
Harvey
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Next Problem:
(Note: You can click on the formulae to enlarge them)

Perform the previous problems for the stators.
For ease of calculations the stator will be considered an evenly distributed mass. Use the values of 10.000000000000000000000000000000000000000000000000000000000000000 grams,
15.875000000000000000000000000000000000000000000000000000000000000mm in diameter,
5000.000000000000000000000000000000000000000000000000000000000000000 RPM.


`I=\frac{mr^2}{2}` where I = moment of inertia in kg m, m = Mass in kg, r = Radius in meters. (Scaler)

`I=\frac{.01*\(frac{.015875}{2})^2}{2}=0.00000031501953125` kg m

`L=I\omega = I2\pi\f` Where `f=\frac{5000}{60}=83.\bar3`
`L=0.00000031501953125 * 2\pi * 83.\bar3 = 1.6494384085205004929415356487277e-4`

`KE_{rotational} = \frac{I\omega^2}{2} = \frac{0.00000031501953125 * (2\pi * 83.\bar3)^2}{2} = 0.086364393112614076151899688225363


Last edited by Harvey on Sun Nov 02, 2008 6:22 am; edited 3 times in total
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Wed May 28, 2008 9:59 am PostPost subject:
Harvey
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Next Problem:

With the forgoing information, how many gravitational units (g's) are involved in flipping the stator from GW to AGW with a rotor RPM of:

800

500

300

150
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Thu May 29, 2008 12:37 am PostPost subject:
alsetalokin
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Oh, so That's what they look like!

I usually have scripts turned off, so I have been seeing the raw formulae as entered, rather than the pretty displays.

But in spite of that, I dislike the artificial appearance of precision caused by the excessive proliferation of meaningless digits masquerading as data.

By which I mean, of course, that you cannot have more significant digits in your result, than you have in your least precise input data. I'll allow five sig digs in the moment calculation, assuming you can measure to the thousandth of a millimeter and weigh to the thousandth of a gram(I can't, with ordinary shop tools; I'd have to wash my hands and go into the metrology room), so call it
3.1502 x 10e-7 kg m
Needless to say, most of the digits in the Kinetic Energy calculation are meaningless, even if the first few may be approximately correct.

But thanks for doing the work, it's just hard to see past all those numerals...
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Thu May 29, 2008 1:32 am PostPost subject:
Harvey
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Quote:

I dislike the artificial appearance of precision


There, I fixed the apperance to be real precision Mr. Green

Edit: Feel free to round up as much as you like Smile
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