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# Re: [Phys-L] simple Mathematica question (one more typo fixed)

In a mathematical sense there is no problem at all, because y(t)=100 is a perfectly valid solution to the energy-conservation differential equation.

This "unphysical" solution comes from taking Newton's 2nd Law for a single conservative force:
[$m\ddot{y} = -U'(y)$]

and multiplying both sides by the velocity in order to get a "first integral":
[$m\ddot{y}\dot{y} = -U'(y)\dot{y} \Longrightarrow \frac{d}{dt}\left[\frac{1}{2}m\dot{y}^2 + U(y)\right] = 0$]

In doing so, you introduce another possible (spurious) solution to the equation, in which [$\dot{y} = 0$] identically. Multiplying both sides of the N2 equation by zero effectively erases the dynamics.

Cheers,
- Craig

On 12/8/20 12:17 PM, Carl Mungan via Phys-l wrote:

[EXTERNAL SENDER - PROCEED CAUTIOUSLY]

Since people seem to be in a talkative mood today, maybe somebody can help me a with a simple (?) matter. It involves dropping a rock. I’m using Mathematica but you don’t have to be an expert in that to see the issue.

If we use conservation of mechanical energy, then we find the downward speed of the rock (choosing upward to be +y with the ground at y=0) is v_y = - Sqrt[2*g*(y0-y)].

Suppose we choose y0 = 100 m and g = 9.8 m/s/s and ask Mathematica to do a numerical solution from t = 0 to 4 s:

s = NDSolve[{y'[t] == -Sqrt[2 9.8 (100 - y[t])], y == 100}, y, {t, 0, 4}]

Plot[Evaluate[y[t] /. s], {t, 0, 4}]

Can you guess what happens here? It’s a failure. The rock just sits at rest in the air at 100 m.

If I instead start the evaluation at 99 m height, so the rock has already fallen 1 m, then it works just fine:

s = NDSolve[{y'[t] == -Sqrt[2 9.8 (100 - y[t])], y == 99}, y, {t, 0, 4}]

Plot[Evaluate[y[t] /. s], {t, 0, 4}]

But I want to start the rock at 100 m. How do I nudge the rock so it starts falling?

I must be missing something. My actual problem is like this, but more complicated. The above is just a test case for debugging purposes.

Who can tell me what I need to do? -Carl

ps: Converting to acceleration does work but I don’t see why I should have to do that:

s2 = NDSolve[{y''[t] == -9.8, y == 100, y' == 0}, y, {t, 0, 4}]

Plot[Evaluate[y[t] /. s2], {t, 0, 4}]

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Carl E. Mungan, Professor of Physics 410-293-6680 (O) -3729 (F)
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