Causality in mechanics2.dfw

File: Causality in mechanics2.dfw Date: 9/9/2014

Time: 8:00:33 PM

We illustrate Chapter 9 of "Neo-Newtonian Mechanics ..." by Dr. Jeremy Dunning-Davies and myself by actually doing a special case of the Euler integration described in this chapter entitled: "Causality in Mechanics". A.P. French in his "Newtonian Mechanics" discusses gyroscopic nutation and starts out with a system two second order ordinary differential equations both of which have time as the only independent variable. He goes on to work with and simplify this system, but we instead take this system (with its initial conditions) and Euler integrate it so as to illustrate our causality ideas. The physical situation is that we begin with a spinning gyro supported at both ends (and both ends at the same height); and at t = 0, we un-support one end and observe its behavior now that it is supported at just one end. Our first ODE is (where ¾ is the precession angle measured from the i3 axis): #1:

CaseMode := Sensitive

#2:

InputMode := Word

#3:

I1·÷ D¾ = ——————·² I3

where I1 and I3 are the moments of inertia of the rotor about the axes i1 and i3, where i1 is horizontal and along the gyro shaft (assumed massless) and i3 is the vertical axis. We have that D¾ = 0 at t = t1 = 0, with our notation being that Dz is dz/dt for any variable z. Our second ODE is: #4:

M·g·L·D² = I2·D²·DD² + I3·D¾·DD¾

where ² is the nutation angle measured from the vertical i3 axis and being ¹/2 at t = 0. We note that I2 = I3, and so we can only utilize I3. Also, D¾ = 0 and D² = 0 at t = 0. Since we can only Euler integrate when all our constants are numbers, we arbitrarily set M = 0.5 kg, g = 9.806 (observed) meters per second^2, L = 0.02 meter, ÷ = 50 radians per second, and I2 = I3 = M L^2 + I1/2, and I1 = 0.2 (using the parallel axis theorem). We try to solve this two-system simultaneously for DD¾ and DD²:

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#5:

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H„ I1·÷ † ‚ SOLVE¦¦D¾ = ——————·², M·g·L·D² = I2·D²·DD² + I3·D¾·DD¾¦, [DD¾, DD²]¦ P… I3 ‡ ƒ

#6:

[]

No solution at all since our first ODE is only first order in time t, and so we must time differentiate the first ODE, and replace our first equation by this time derivative to obain the second order ODE: #7:

I1·÷·D² DD¾ = ————————— I3

We note that ÷, the rotor spin angular velocity is constant in time and so D÷ = 0. We attempt to solve our new system: #8:

H„ I1·÷ † ‚ SOLVE¦¦DD¾ = ——————·D², M·g·L·D² = I2·D²·DD² + I3·D¾·DD¾¦, [DD¾, DD²]¦ P… I3 ‡ ƒ „ D²·I1·÷ L·M·g - D¾·I1·÷ † ¦DD¾ = ————————— U DD² = —————————————————¦ … I3 I2 ‡

#9:

Our new two ODE system is then: #10:

D²·I1·÷ DD¾ = ————————— I3

#11:

L·M·g - D¾·I1·÷ DD² = ————————————————— I2

Substituting the values of our constants mentioned above: #12:

#13:

D²·0.2·50 DD¾ = ——————————— 0.1 0.02·0.5·9.806 - D¾·0.2·50 DD² = ———————————————————————————— 2 0.1 + 0.5·0.02

Simplifying: #14:

DD¾ = 100·D²

#15:

DD² = 0.9786427145 - 99.8003992·D¾

We now proceed with our Euler integration of this two ODE system Page: 2


Time: 8:00:33 PM

just above. Our initial condition on D² is D²1 = 0 at t = t1 = 0, where our notation is D²1 = D²(0) [= D² at t = t1 = 0]. Thus we see from our first ODE that at t = t1 = 0: #16:

DD¾1 = 100·D²1 = 0

That is, DD¾ = 0, i.e. it vanishes. Our second ODE yields at t = t1 = 0: #17:

DD²1 = 0.9786427145 - 99.8003992·D¾1 = 0.9786427145

Since D¾1 = 0. Thus DD²1 is just: #18:

0.9786427145

Thus DD²1 FAILS to vanish ... even though DD¾ does vanish!!! We continue with our Euler integration using a time step of Wt = 0.01 second: #19:

D¾2 = Wt·DD¾1 + D¾1

Here, we mean by D¾2 the quantity D¾(t2), where t2 = Wt + t1 = 0.01, and t1 = 0. We know from above that D¾0 = 0. #20:

Thus we have:

D¾2 = 0.01·0 + 0

Simplifying: #21:

D¾2 = 0

Thus D¾2 also vanishes as does D¾1. We now examine ¾2: #22:

¾2 = Wt·D¾1 + ¾1

Substituting: #23:

¾2 = 0.01·0 + 0

Simplifying: #24:

¾2 = 0

Thus ¾2 vanishes as does ¾1. Page: 3


Time: 8:00:33 PM

We now turn to DD²2: #25:

D²2 = DD²1·Wt + D²1

Substituting: #26:

D²2 = 0.9786427145·0.01 + 0

Simplifying: #27:

D²2 = 0.009786427145

Thus D²2 fails to vanish, but D¾2 does vanish. #28:

²2 = Wt·D²1 + ²1

Substituting: #29:

¹ ²2 = 0.01·0 + ——— 2

Simplifying: #30:

¹ ²2 = ——— 2

Thus there is NO change in the nutation angle at t = t2 = 0.01 second ... just as there is no change in the precession angle ¾ as yet either. We now calculate DD¾2 by substituting into our first system ODE: #31:

DD¾2 = 100·D²2

Substituting: #32:

DD¾2 = 100·0.00978642745

Simplifying: #33:

DD¾2 = 0.9786427449

Next we calculate DD²2 in much the same way: #34:

DD²2 = 0.9786427145 - 99.8003992·D¾2

Substituting: #35:

DD²2 = 0.9786427145 - 99.8003992·0

Simplifying:

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#36:

Time: 8:00:33 PM

DD²2 = 0.9786427144

We can now Euler integrate to obtain D¾3 (at t = t3 = 2 Wt = 0.02) ... and also D²3, ¾3 and ²3: #37:

D¾3 = Wt·DD¾2 + D¾2

Substituting: #38:

D¾3 = 0.01·0.9786427449 + 0

Simplifying: #39:

D¾3 = 0.009786427449

Next we calculate ¾3: #40:

¾3 = Wt·D¾2 + ¾2

Substituting: #41:

¾3 = 0.01·0 + 0

Simplifying: #42:

¾3 = 0

Thus, at two time steps out, ¾ (denoted then by ¾3) still vanishes so that precession has not as yet begun!!! We can now calculate D²3 (at t = t3 = 2 Wt): #43:

D²3 = DD²2·Wt + D²2

Substituting: #44:

D²3 = 0.9786427449·0.01 + 0.009786427145

Simplifying: #45:

D²3 = 0.01957285459

Finally, we calculate ²3: #46:

²3 = D²2·Wt + ²2

Substituting: #47:

¹ ²3 = 0.009786427145·0.01 + ——— 2

Simplifying:

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-5 #48:

²3 = 9.786427144·10

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¹ + ——— 2

Therefore, we see that the nutation (from origional nutation angle ¹/2) has begun at the second time step out (with t = t3 = 2 Wt) BUT THE PRECESSION HAS NOT AS YET BEGUN AT t = t3!!! Thus the nutation begins BEFORE the precession in this example of A.P.French's!!! This means that nutation and precession are related as cause to (later on) effect here ... much as French alluded to; however, USING the exact (continuum mathematics) solution, one MUST conclude that nutation and precession ARE simultaneous. Never-the-less, the (usual) continuum mathematical model (yielding a "closed form solution) MUST be deemed to be ONLY an approximation to the physical phenomena, and so cannot be considered to be more real than the physical phenomema ... however useful and convenient it may well be!!!

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