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Suppose a bead mass $m$ that is free to slide in on a wire that is rotating about the $z$ axis through origin $O$ with angular velocity $\omega$ and the wire is made into a parabolic shape such that $z=r^2/2a$ where $z$ is the vertically upward from $O$ and $r$ is the horizontal distance from $O$.

I have got the second order differential equation related to $r$ such that $(a^2+r^2) \ddot r +r\dot r^2 = (a^2 \omega^2-ga)r$.

I have now got to form the linearised equation about $r=0$, which is given as $a^2 \ddot r=(a^2 \omega^2-ga)r$ but I have no idea how to get to this linearised equation. My teacher told me it's something to do with computing $r(t)=p+\epsilon(t)$ where $p$ is the equilibrium position, which is $r=0$ in our case but I do not know how to do so.

My idea: Do I plug $r(t)=p+\epsilon(t)$ into the differential equation and collect order $\epsilon$ terms?

UnsinkableSam
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  • Is $\ddot{x}$ a typo perhaps? Should be $\ddot{r}$ instead? – StarBug Jun 11 '19 at 20:31
  • @StarBug ah yes my apologies! – UnsinkableSam Jun 11 '19 at 20:37
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    To linearize around 0, you put $p=0$, then plug $r=\epsilon$ into the equation, and then remove all nonlinear terms in $\epsilon$. Of course, this is the same as directly removing all nonlinear terms in $r$. In your case, the nonlinear terms are $r^2\ddot{r}$ and $r\dot{r}^2$, which leaves you with the linearized equation you stated. – StarBug Jun 11 '19 at 20:47
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    You "linearize" an equation simply by dropping any non-linear terms! Here the equation is $(a+r^2)r''+ rr'^2= (a^2\omega^2- ga)r$. The "non-linear" terms are $r^2$ and $rr'^2$. Without those, we have $ar''= (a^2\omega^2- ga)r. – user247327 Jun 11 '19 at 20:53

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