# Proving stable equilibrium: Rotating circular hoop

• JyJ

## Homework Statement

A circular hoop of radius R rotates with angular frequency ω about a vertical axis coincident with its diameter. A bead of mass m slides frictionlessly under gravity on the hoop. Let θ be the bead’s angular position relative to the vertical (so that θ = 0 corresponds to the bead being at the bottom of the hoop).
Under what conditions can the bead sit at a fixed angle θ_0 not equal to 0 or π?
Show that this situation corresponds to a stable equilibrium, and determine the frequency of small oscillations about this stable equilibrium.

## Homework Equations

In cylindrical coordinates:
$$\vec{v} = \dot{r}\vec{e_r} + r\dot{\theta}\vec{e_\theta} + rsin(\theta)\dot{\phi}\vec{e_\phi}$$

## The Attempt at a Solution

By using Lagrangian methods I found that the equation of motion is:
$$mR^2\ddot{\theta} + mgRsin(\theta)-mR^2(\omega)^2sin(\theta)cos(\theta)=0$$ which agrees with the solution given.
By plugging:
$$\ddot{\theta} = \dot{\theta} = 0$$
For the first question I got the following as a condition on θ_0:
$$\omega = \sqrt{\frac{g}{cos(\theta_0)}}$$
I also then computed the frequency of small deviations from equilibrium by using Taylor expansion with:
$$\theta = \theta_0 + a$$
where a is the small deviation of order h from stable equilibrium. Them by getting rid off terms of order h2 or higher I found the frequency to be:
$$R^2\ddot{a} + gRsin(a+\theta_0) - \frac{gR^2}{cos(\theta_0)}sin(a+\theta_0)cos(a+\theta_0)=0 \\ R^2\ddot{a} + gR(a+\theta_0)-\frac{gR^2}{cos(\theta_0)}(a+\theta_0) = 0$$
which leads to frequency equal to:
$$\frac{\frac{Rg}{cos(\theta_0)}-g}{R}$$
This is how I think the frequency would be found.
Now, the only bit that is left to prove is to show that θ=θ_0 indeed corresponds to the stable equilibrium. I am having difficulties with this part as I am not sure how to show it using the information I have. Typically I would be able to estimate stable equilibrium from the minimum of the potential but in this case my potential is dependent on both y and x and so I am unsure what to do:
$$U = -mgRcos(\theta) = -mgRcos(arctan\frac{y}{x})$$

For the first question I got the following as a condition on θ_0:
$$\omega = \sqrt{\frac{g}{cos(\theta_0)}}$$
This equation is dimensionally incorrect. You probably need a factor of ##R## in the denominator.
I also don't see how you got from the first line to the second.
$$R^2\ddot{a} + gRsin(a+\theta_0) - \frac{gR^2}{cos(\theta_0)}sin(a+\theta_0)cos(a+\theta_0)=0 \\ R^2\ddot{a} + gR(a+\theta_0)-\frac{gR^2}{cos(\theta_0)}(a+\theta_0) = 0$$
For example, ##\sin (a+\theta_0)=\sin a \cos \theta_0+ \cos a \sin \theta_0##. This is exact. For small ##a##, you get (to first order) ##\sin(a+\theta_0)\approx a\cos \theta_0+\sin \theta_0##. It seems that you are assuming that ##\theta_0## is small when it isn't. For the other term you can save yourself some work and write ##\sin (a+\theta_0) \cos (a+\theta_0) = \frac{1}{2}\sin[2(a+\theta_0)]## then expand the sine to first order as before.

This equation is dimensionally incorrect. You probably need a factor of ##R## in the denominator.
I also don't see how you got from the first line to the second.
$$R^2\ddot{a} + gRsin(a+\theta_0) - \frac{gR^2}{cos(\theta_0)}sin(a+\theta_0)cos(a+\theta_0)=0 \\ R^2\ddot{a} + gR(a+\theta_0)-\frac{gR^2}{cos(\theta_0)}(a+\theta_0) = 0$$
For example, ##\sin (a+\theta_0)=\sin a \cos \theta_0+ \cos a \sin \theta_0##. This is exact. For small ##a##, you get (to first order) ##\sin(a+\theta_0)\approx a\cos \theta_0+\sin \theta_0##. It seems that you are assuming that ##\theta_0## is small when it isn't. For the other term you can save yourself some work and write ##\sin (a+\theta_0) \cos (a+\theta_0) = \frac{1}{2}\sin[2(a+\theta_0)]## then expand the sine to first order as before.
Thank you for your reply! Yes, you are correct there is a missing R in the denominator for w_0 and yes, I should not have treated θ_0 as a small quantity - my mistake. This all makes sense now though!
The question also asks to show that θ_0 is a stable equilibrium, how can I can show this mathematically? I rearranged the governing equation with the deduced condition on w_0 and got this:
$$\ddot{\theta} = \frac{g}{R}sin(\theta)(\frac{cos(\theta)}{cos(\theta_0)} -1)$$
and plotted an easier case which is when θ_0 = pi/3 and g/R = 1 (graph here):
$$\ddot{\theta} = 2sin(\theta)cos(\theta) - sin(\theta)$$
from which it doesn't look that pi/3 is an equilibrium point at all, because at this value we get that the whole equation is zero. Am I confusing things up?

I suggest that you go back to the equation of motion
$$mR^2\ddot{\theta} + mgR\sin(\theta)-mR^2(\omega)^2\sin(\theta)\cos(\theta)=0$$
Leave ##\omega## as is; it is a given quantity after all. Rewrite this equation in the vicinity of ##\theta_0## as you started doing
$$R\ddot{a} + gsin(\theta_0+a)-R(\omega)^2\sin(\theta_0+a)\cos(\theta_0+a)=0$$
Use trig identities to expand the trig functions, then expand the trig functions themselves keeping only first order terms in ##a##. For example,
$$\sin(\theta_0+a)=\sin\theta_0 \cos a+\sin a \cos\theta_0 \approx \sin\theta_0 +a \cos\theta_0.$$You should be able to cast this equation into the harmonic oscillator equation in the form ##\ddot{a} + \omega_s^2 a+C=0##, where ##\omega_s## is the frequency of small oscillations about ##\theta_0## and ##C## is a constant related to the equilibrium position.

I suggest that you go back to the equation of motion
$$mR^2\ddot{\theta} + mgR\sin(\theta)-mR^2(\omega)^2\sin(\theta)\cos(\theta)=0$$
Leave ##\omega## as is; it is a given quantity after all. Rewrite this equation in the vicinity of ##\theta_0## as you started doing
$$R\ddot{a} + gsin(\theta_0+a)-R(\omega)^2\sin(\theta_0+a)\cos(\theta_0+a)=0$$
Use trig identities to expand the trig functions, then expand the trig functions themselves keeping only first order terms in ##a##. For example,
$$\sin(\theta_0+a)=\sin\theta_0 \cos a+\sin a \cos\theta_0 \approx \sin\theta_0 +a \cos\theta_0.$$You should be able to cast this equation into the harmonic oscillator equation in the form ##\ddot{a} + \omega_s^2 a+C=0##, where ##\omega_s## is the frequency of small oscillations about ##\theta_0## and ##C## is a constant related to the equilibrium position.
Yes, I did get an equation of a harmonic oscillator and, hence, found the frequency. However, my question is that is this sufficient for showing that ##\theta_0## is indeed stable equilibrium, i.e. that since we have equation of a harmonic oscillator, we conclude that it is stable equilibrium?

Look at the equation you got in ##\ddot{\theta}##. Does it predict a restoring torque for small displacements ##\pm a## from ##\theta_0##?

Look at the equation you got in ##\ddot{\theta}##. Does it predict a restoring torque for small displacements ##\pm a## from ##\theta_0##?
Thank you, I got it!