Linearize a function about a solution to check for stability

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
  • #1
DanielA
27
2
<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.
 
Last edited by a moderator:
  • Like
Likes Lynda Jhon
Physics news on Phys.org
  • #2
If ##\theta_c <<1## you can use
$$\sin(\theta_c)\approx \theta_c $$ and
$$\cos(\theta)\approx 1$$.
 
  • Like
Likes Lynda Jhon
  • #3
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.
 

1. What does it mean to linearize a function?

Linearizing a function involves approximating a nonlinear function with a linear function in order to make it easier to analyze and understand.

2. Why is it important to linearize a function about a solution?

Linearizing a function about a solution allows us to better understand the behavior of the function near that particular solution. This is especially useful in determining the stability of the system.

3. How do you linearize a function about a solution?

To linearize a function about a solution, we first find the derivative of the function. Then, we substitute the solution value into the derivative to find the slope of the tangent line at that point. Finally, we use the slope and the solution value to create an equation for the linearized function.

4. What is the purpose of checking for stability when linearizing a function?

Checking for stability allows us to determine whether the system will converge towards the solution or diverge away from it. This is crucial in understanding the long-term behavior of the system.

5. Can linearizing a function guarantee stability?

No, linearizing a function does not guarantee stability. It is simply a tool that helps us analyze the behavior of a function near a particular solution. Other factors, such as initial conditions, must also be considered to determine the stability of a system.

Similar threads

  • Calculus and Beyond Homework Help
Replies
4
Views
152
  • Calculus and Beyond Homework Help
Replies
3
Views
571
  • Calculus and Beyond Homework Help
Replies
6
Views
404
  • Calculus and Beyond Homework Help
Replies
1
Views
301
  • Calculus and Beyond Homework Help
Replies
1
Views
838
  • Calculus and Beyond Homework Help
Replies
1
Views
303
  • Introductory Physics Homework Help
Replies
2
Views
639
  • Calculus and Beyond Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
275
  • Calculus and Beyond Homework Help
Replies
21
Views
853
Back
Top