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I have these systems:

$x''=\frac{-g(2m_1+m_2)\sin x-m2g\sin(x-2y)-2\sin(x-y)m_2(y'^2L_2+x'^2L_1\cos(x-y))}{L_1(2m_1+m_2-m_2\cos(2x-2y))}$

$y''=\frac{2\sin(x-y)(x'^2L_1(m_1+m_2)+g(m_1+m_2)+g(m_1+m_2)\cos(x+y'^2)L_2m_2\cos(x-y))}{L_2(2m_1+m_2-m_2\cos(2x-2y))}$

Which I want to linearize.

I have $x'=w$, $y'=z$

I get $w'=\frac{-g(2m_1+m_2)\sin x-m2g\sin(x-2y)-2\sin(x-y)m_2(z^2L_2+w^2L_1\cos(x-y))}{L_1(2m_1+m_2-m_2\cos(2x-2y))}$

$z'=\frac{2\sin(x-y)(w^2L_1(m_1+m_2)+g(m_1+m_2)+g(m_1+m_2)\cos(x+z^2L_2m_2\cos(x-y))}{L_2(2m_1+m_2-m_2\cos(2x-2y))}$

And was trying to input them into this matrix $A=\begin{pmatrix}0&1&0&0\\?&?&?&?\\0&0&0&1\\?&?&?&?\\\end{pmatrix}$ with $x=\begin{pmatrix}x\\w\\y\\z\\\end{pmatrix}$

Im not sure how to deal with terms like $-g(2m_1+m_2)\sin x$. I'm use to linearizing systems which have simple coefficients in front of the terms.

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1 Answer 1

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You are linearizing for small movements around $x=y=0$, so that all quadratic and higher-degree contributions in $x,y,x',y'$ can be neglected. Among other things, $\sin x$ gets replaced by $x$, $\cos(x-y)$ by $1$ etc. What remains in linear terms after this pruning is \begin{align} w'=x''=\frac{-g((m_1+m_2) x-m_2y)}{L_1m_1} \\ z'=y''=\frac{2g(m_1+m_2)(x-y)}{L_2m_1} \end{align} Note that I had to interpret some missing closing parentheses.

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  • $\begingroup$ Did you mean $sin(x)=0$? since $x=0$? $\endgroup$ Oct 5, 2020 at 19:46
  • $\begingroup$ I said "around $x=0$", or in other symbols, for $x\approx 0$. $\endgroup$ Oct 5, 2020 at 19:47
  • $\begingroup$ I still don't see how you have a $-g$ term in $x''$. I get that $w'=\frac{m_2(z^2L_2+w^2L_1)}{L_1(2m_1)}$ with $x=y=0$. $\endgroup$ Oct 5, 2020 at 21:29
  • $\begingroup$ Close to zero, but not zero. $z^2$ and $w^2$ are quadratic and thus do not occur in the linearization. $\endgroup$ Oct 5, 2020 at 21:58
  • $\begingroup$ So I set the quadratic terms to $1$? I also don't get why $\sin x$ is replaced by $x$ $\endgroup$ Oct 5, 2020 at 22:10

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