# Small Angle Approximation to Hoop Oscillator

1. Oct 30, 2016

### FallenApple

1. The problem statement, all variables and given/known data

A point particle of mass m slides without friction within a hoop of radius R and mass M. The hoop is free to roll without slipping along a horizontal surface. What is the frequency of small oscillations of the point mass, when it is close to the bottom of the hoop?

2. Relevant equations
Euler Legrange Equations and taylor approximation for cosine of phi. cosine of phi is approximated as being porportional to phi squared

3. The attempt at a solution
Well, I have the solution in the above. I just don't understand why cos(phi)=1 going from line 3 to line 4. It make sense because the angle is nearly 0, but still, isn't it more proper to do second order taylor approximation?

Also, in line 5, a small angle approximation was used. But why here? Why in the potential energy expression and not the kinetic?

Edit: I think I figured it out. So at the bottom, or very near it, the height which is directly related to distance to the equilibrium point, is very near R. The situation is observed in both the potential and kinetic. But kinetic is tangential, but since the particle is virtually at the bottom, the tangential is in the x, and the displacement forward in the x is very small, almost 0, while the height from the bottom is very small as well but somehow does not approach 0 as fast as the displacement in the x.

Is this correct?

Last edited: Oct 30, 2016
2. Oct 31, 2016

### TSny

You want the differential equations of motion to be accurate to first order in the small quantities $\theta$, $\dot \theta$, $\phi$, $\dot \phi$, $\ddot \theta$, and $\ddot \phi$. Since these equations are obtained by taking derivatives of the Lagrangian with respect to $\theta$, $\dot \theta$, $\phi$, and $\dot \phi$, the Lagrangian needs to be expressed to an accuracy of second order in $\theta$, $\dot \theta$, $\phi$, and $\dot \phi$.

In line 3 you have the term $mR^2 \cos \phi \dot \theta \dot \phi$. Because of the presence of $\dot \theta$ and $\dot \phi$, the $\cos \phi$ factor only needs to be expressed to "zeroth order" in $\phi$ in order for the overall term to be accurate to second order. So, here you can let $\cos \phi \approx 1$.

In line 5 you have $mgR \cos \phi$. For this to be accurate to second order, you need to let $\cos \phi \approx 1 - \phi^2/2$.

3. Oct 31, 2016

### FallenApple

Ah got it. But If I don't change cos to 1 to begin with, then is it nescessary to change the cos to $\cos \phi \approx 1 - \phi^2/2$ ?.

Since the lagragian was originally accurate, I think, then I only need to change on the other side as kind of a balancing act.

4. Oct 31, 2016

### TSny

I'm not sure I understand the question or the statement.

When I'm in doubt about small angle approximations, I usually don't make any approximations in the Lagrangian. I use the exact Lagrangian to derive the differential equations of motion. Then I make the small angle approximation in the differential equations of motion. This requires more writing than making the small angle approximation in the original Lagrangian before deriving the equations of motion. But I often find it's easier to see how to make the correct approximations after deriving the equations of motion.

If you don't make an approximation for $\cos \phi$ in the Lagrangian and then $\cos \phi$ shows up in one of the terms of the equations of motion, then you can simply replace $\cos \phi$ by 1 in the equation of motion since you only need to be accurate to first order in $\phi$ in the equation of motion.

5. Oct 31, 2016

### FallenApple

Oh I guess what I'm asking is that would the equations be correct if I don't make any approximations? I know that its not integrable if I don't.

It makes sense that it is because it's derived from theory. And there is no need for anything of the euler lagrange equations to be of a certain order unless approximations are applied, in which then both sides must be the same order?

6. Oct 31, 2016

### TSny

Yes.
That's right.
Yes.