# Bead on loop of wire in rotating frame

## Homework Statement

Consider a bead sliding without friction on a circular hoop of wire rotating at constant $$\Omega$$, where $$\phi$$ is the angle between the bottom of the hoop and the bead. Find the equation of motion of the bead.
$$\hat{\Omega}=\hat{z}$$

## Homework Equations

$$m\ddot{\vec{r}}=\vec{F}+2m(\dot{\vec{r}} \times \vec{\Omega})+m(\vec{\Omega} \times \vec{r}) \times \vec{\Omega}$$

## The Attempt at a Solution

I started by taking the time derivative (first and second) of $$\hat{r}$$ to get an expression for the above equation in terms of $$\hat{x}$$, $$\hat{y}$$,and $$\hat{z}$$, but after separating the differential equations for each of those directions I have complicated differential equations that I can't solve. For example for the $$\hat{x}$$ direction, I had

$$mr(\ddot{\phi}cos \phi-\dot{\phi} sin\phi)=F_x + m\Omega^2 sin \phi$$
Where $$F_x$$ is the normal force in the $$\hat{x}$$ direction.

Is that the right approach, and if so do you have any idea what I might have done wrong or is more information (steps) required?

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I expect that the equations would be crazy, so I'm not surprised if you got such complicated equation Just a suggestion to simplify the problem: consider the reference frame of the rotating hoop; in this frame, the bead's motion is restricted so it's much easier to describe it. Way easier, because if you use spherical coordinates, in this frame, you only have to consider only one coordinate: $$\phi$$ . Since r is constant, if you want to go back to the rest frame, then all you need is the other coordinate which is determined by the motion of the hoop: $$\theta = \Omega t + \theta_o$$.

In this frame, we have: $$\ddot{\phi}R = \Omega^2Rsin\phi cos\phi - gsin\phi$$
I give up here. I have no idea how to solve it An analysis on small disturbance of the bead around its equilibrium positions might be interesting and noteworthy.

This is a problem ideally solved with the Lagrangian formulation of classical mechanics. Do you know how it works?

If you do, then the solution only takes 3 or 4 lines of work. My fingers are crossed for you.

Yeah, a Lagrangian is really the way to go, but be warned because even the Lagrangian gets a little bit messy. I was working the Lagrangian and saw an explosion of "cot"s and "csc"s.