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# How do I solve for time?

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And I guess I really meant:
Let y' ' = d2theta/dt2
L{sin(k*theta)} = k / (s2 + k2)
L{sin(k*theta)} -> k / (s2 + k2) + c3*s / (s2 + k2)

Are these valid assumptions to make? or does the laplace transform only work on functions of time?

And the final result would be:
J*y(theta) + c1*theta + c2*theta2 - m*g*L*sin(theta) + c3*cos(theta) = 0

I uh, used two different variables named t... all a mote point if I missed something in the original equation.

....
So, are you saying that theta is actually a function, dependant on t?

J*d2theta(t)/dt2 = -m*g*L*sin(theta(t)) ?

quote:

Still, its good to hone up on those old heavy math skills sometimes, .

Absolutely, I feel like I'm leaking.

Magmai Kai Holmlor
- Not For Rent

Edited by - Magmai Kai Holmlor on October 12, 2001 9:17:15 PM

Edited by - Magmai Kai Holmlor on October 12, 2001 9:21:53 PM

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quote:
Original post by Magmai Kai Holmlor
And I guess I really meant:
Let y'' '' = d2theta/dt2
L{sin(k*theta)} = k / (s2 + k2)
L{sin(k*theta)} -> k / (s2 + k2) + c3*s / (s2 + k2)

Are these valid assumptions to make? or does the laplace transform only work on functions of time?

Your first equation, defining y'''', is correct. Its basically the same thing as saying y = theta, or y'''' == theta''''. Its just a change of the variable name from theta to y. Both are functions of time, t.

The Laplace transform transforms a function in the time domain into a function in the frequency domain. Your transforms of L(sin(k*theta)) are incorrect unless you say that theta = t = time. That''s a linear function that says theta grows to infinity with time. And again, for the nonlinear pendulum this assumption is wrong since theta does not grow to infinity, it just oscillates between a maximum and minimum value. (Well, if its a frictionless pendulum and the bob can move 360 degrees around and you apply a large enough initial force or velocity then theta could go to infinity. But it still won''t grow linearly with time since gravity will cause it to accelerate and decelerate as it moves around. (And a pure linearized ODE solution won''t capture this case.)

quote:

And the final result would be:
J*y(theta) + c1*theta + c2*theta2 - m*g*L*sin(theta) + c3*cos(theta) = 0

I didn''t look to closely at this, but the "J*y(theta)" part is wrong. Remember, y and theta are the exact same so you would never have a y(theta). You would have a y(t) in the time domain or y(s) in the frequency domain. The transform of J*y''''(t) is J*s2y(s) plus the two initial condition terms.

quote:

So, are you saying that theta is actually a function, dependant on t?

Yes! That''s exactly it! theta(t) = y(t). Theta is the solution we''re looking for, and it specifies the motion of the pendulum or object over time. Theta is exactly the same as your variable y.

quote:

Absolutely, I feel like I''m leaking.

Ha!

I say we quit while we''re ahead, .

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.