Solving System of Non-Linear Diff. Equations

  • Thread starter izotop
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  • #1
izotop
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Hi,

I'm new to this forum. It looks like an amazing source of information and people here seem to be helpful and knowledgeable.

Briefly, I'm an undergrad student currently studying Mechanical Engineering at Concordia University, Montreal, Canada.

The first equation seems a bit tricky to me. In fact, I've been working for a while on these two equations without being able to find the right way for solving them.

Here's what I would like to do:
1) Linearize the first equation
2) Take the laplace transform of both equations
3) Use Cramer's rule to solve the system of linear equations
4) Take the inverse laplace of the two equations thus obtained

Is it possible to linearize the first equation?
 

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  • #2
tiny-tim
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welcome to pf!

hi izotop! welcome to pf! :smile:

(have a theta: θ :wink:)

can't you just completely solve the second equation for t (it looks like ordinary shm), then chuck all the θ stuff in the first equation over to the RHS?
 
  • #3
JJacquelin
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Regarding the unknown function x(t), the first equation is a linear ODE.
 
  • #4
izotop
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hi izotop! welcome to pf! :smile:

(have a theta: θ :wink:)

can't you just completely solve the second equation for t (it looks like ordinary shm), then chuck all the θ stuff in the first equation over to the RHS?

I first thought of solving the second equation but since my initial conditions are all zero and there is no input function in my second equation, theta is simply going to end up being zero.

I attached picture of the actual physical problem I need to solve.
 

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  • #5
tiny-tim
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hi izotop! :wink:
I first thought of solving the second equation but since my initial conditions are all zero and there is no input function in my second equation, theta is simply going to end up being zero.

(hmm … surprisingly neat handwriting! :smile:)

so it's a pendulum hanging from a block free to roll along a horizontal rail in the plane of the swing …

but the initial conditions aren't all zero … when θ' = 0, θ is some maximum θ0 :wink:
 
  • #6
izotop
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hi izotop! :wink:


(hmm … surprisingly neat handwriting! :smile:)

so it's a pendulum hanging from a block free to roll along a horizontal rail in the plane of the swing …

but the initial conditions aren't all zero … when θ' = 0, θ is some maximum θ0 :wink:


Regarding the hand-writing, if it's not clear on paper, it won't be clear in my mind!

Now. The problem.
You're right, it's a pendulum "m" free to rotate about point O located on Mass "M". The mass "M" is not quite "free" to move since there is some friction acting on it (Friction = -bx'). Other than that, as you can see, there is the inertial force, the tension of the cable and the input force u(t) acting on it.

I know what you mean: when [tex]\theta[/tex]' is maximum, [tex]\theta[/tex]=0. Similarily, when [tex]\theta[/tex]'=0, [tex]\theta[/tex] is maximum.

But at time=0-, I assume [tex]\theta[/tex] and [tex]\theta[/tex]' to be zero since the system is in static equilibrium.

At time=0+, the input function u(t) which I assume to be the unit step function comes into play. But still, at t=0+, [tex]\theta[/tex] and [tex]\theta[/tex]' are zero.

It seems like I can't solve the two equations independently since the motions of the two masses are definitely linked.

Does that make sense?
 
  • #7
izotop
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Oh and I may add that on my drawing, the angle [tex]\theta[/tex] looks like its initial value is NOT zero because I put a certain angle there. I deliberately put an angle to be able to build up my equations.
 
  • #8
tiny-tim
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ah, i get it now! :smile:

(i didn't appreciate the input force u :redface:)

yes initially θ = θ' = 0

i think the problem is that you can't do τ = dL/dt about O because O isn't moving parallel to the centre of mass (there'd be an error term of vO x mvm)
 
  • #9
izotop
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T = dL/dt? I don't get that part. L is the length of the cable joining the two masses. T is the tension in the cable.
 
  • #10
tiny-tim
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no, not T, τ (for torque :wink:) …

you have J = ml2 and Jθ'' = -lmgsinθ …

that's τ = dL/dt with angular momentum L = Jθ' …

it doesn't work for moments (torques) about O.
 

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