Linearize a function about a solution to check for stability

• DanielA
In summary, the homework question is at graduate level, but the area I'm having trouble on I feel is at an undergraduate level. I'm not sure if I don't know how to pull the approximation off with this equation or if I messed up my expansion or what. If I do small angle approximation, I get ##\theta_c \approx 0## ie ##\cos^{-1}(\frac{\omega_0}{\omega}) \approx 1## so ##\theta_c \approx 0##. However, in a fast spin approximation, I got ##\theta_c \approx \frac{\pi}{2} ## or ## \frac{3\pi

DanielA

<Moderator's note: Moved from a technical forum and thus no template.>

Technically the homework question is at graduate level, but the area I'm having trouble on I feel is at an undergraduate level.

In the question we studied a particle rotating on a vertical hoop that is also rotating about the z axis. Our goal was to find where the particle would be stationary and whether the particle was stable at that point or not. After creating my Lagrangian and getting my equation of motion, $$\ddot \theta = \frac{1}{2}\omega^2\sin(2\theta) + \frac{g}{r}\sin(\theta)$$ I end up with 4 solutions. 2 coming from ##\sin(\theta) = 0## and two from ##\frac{g}{a} + \omega^2\cos(\theta) = 0## , where a is the radius of the hoop and ##\omega## is the angular velocity of the hoop about the z axis. The first two are trivial to check for stability. I linearized the equation of motion about those points and examined the result.

However, for my last two solutions, I tried to linearize and I ran into a problem.
Solution: $$\theta_c = \cos^{-1}\frac{-g}{a\omega^2}$$ where I just renamed my solution to "##\theta critical##""
So, I linearize my equation of motion about my solution and I end up with $$\frac{1}{2}\omega^2(\sin(\theta_c) + 2\cos(\theta_c)(\theta-\theta_c)) + \frac{g}{r}(\sin(\theta_c) + \cos(\theta_c)(\theta-\theta_c)$$ I don't know what ##\sin(\theta_c) ## is. My professor somehow did a small angle approximation to get around the issue that I did not understand and that is where my problem lies. I'm not sure if I don't know how to pull the approximation off with this equation or if I messed up my expansion or what.

Just to note, I know all the details of the question besides this is correct as it's a graded assignment and I talked to my professor about the rest.
As another aside, does anyone know where I can learn equation analysis methods (approximations, simplification, turning higher order DE into lower order, etc)? There's a ton of it in CM and I was never very good at it.

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Lynda Jhon
If ##\theta_c <<1## you can use
$$\sin(\theta_c)\approx \theta_c$$ and
$$\cos(\theta)\approx 1$$.

Lynda Jhon
eys_physics said:
If ##\theta_c <<1## you can use
$$\sin(\theta_c)\approx \theta_c$$ and
$$\cos(\theta)\approx 1$$.
Sorry, I wasn't clear, I get what a small angle approximation is and how to use it, but I think he did it with a weird variable substitution. The only requirement for this solution to exist was that the rotation of the hoop was greater than the rotation of the particle around the hoop. He found it can be stable if I remember correctly.

Just to note, I measured my theta from the z axis the hoop is spinning about to the x axis. Most of my class did it from the x-axis.
If I do small angle approximation, I get ##\theta_c \approx 0## ie ##\cos^{-1}(\frac{\omega_0}{\omega}) \approx 1## so ##\theta_c \approx 0##. In that case I got an unstable solution. However, in a fast spin approximation, I got ##\theta_c \approx \frac{\pi}{2} ## or ## \frac{3\pi}{2}## where plugging those in I get the ##\frac{3\pi}{2}## solution to be stable and the other unstable. Maybe I just misremembered him, but it doesn't seem to make sense that one side of the hoop would be unstable and the other stable.