Bead on loop of wire in rotating frame

[msimmons]

Homework Statement

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

Homework Equations

[tex]m\ddot{\vec{r}}=\vec{F}+2m(\dot{\vec{r}} \times \vec{\Omega})+m(\vec{\Omega} \times \vec{r}) \times \vec{\Omega}[/tex]

The Attempt at a Solution

I started by taking the time derivative (first and second) of [tex]\hat{r}[/tex] to get an expression for the above equation in terms of [tex]\hat{x}[/tex], [tex]\hat{y}[/tex],and [tex]\hat{z}[/tex], 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 [tex]\hat{x}[/tex] direction, I had

[tex]mr(\ddot{\phi}cos \phi-\dot{\phi} sin\phi)=F_x + m\Omega^2 sin \phi[/tex]
Where [tex]F_x[/tex] is the normal force in the [tex]\hat{x}[/tex] 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?

 

[hikaru1221]

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: [tex]\phi[/tex] . 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: [tex]\theta = \Omega t + \theta_o[/tex].

In this frame, we have: [tex]\ddot{\phi}R = \Omega^2Rsin\phi cos\phi - gsin\phi[/tex]
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.

 

[Coto]

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.

 

[Mindscrape]

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.

 

[paranormal]

I would approach this problem by first understanding the physical principles involved. In this case, the bead is sliding without friction, which means that there is no force acting on it in the tangential direction. This means that the only forces acting on the bead are the normal force from the hoop and the centrifugal force due to the rotation of the hoop.

Next, I would consider the forces acting on the bead in the rotating frame of reference. In this frame, the centrifugal force is balanced by the normal force, so the net force on the bead is zero. This means that the bead will move in a straight line at a constant velocity, in the direction of the tangent to the hoop at any given point.

To find the equation of motion, we can use the fact that the velocity of the bead is constant. This means that the acceleration of the bead must be perpendicular to the velocity vector. Using vector notation, we can write this as:

\vec{a}=\dot{\vec{v}}=\dot{\vec{r}} \times \vec{\Omega}

Since the velocity vector is perpendicular to the acceleration vector, we can write the equation of motion as:

\ddot{\vec{r}}=\vec{\Omega} \times (\vec{\Omega} \times \vec{r})

We can then substitute in the given values for \vec{\Omega} and \vec{r} (which is the position vector of the bead in the rotating frame) to get the final equation of motion:

\ddot{\vec{r}}=-\Omega^2 \hat{r}

This equation describes the motion of the bead in the rotating frame. To get the equation of motion in the inertial frame, we can simply rotate the coordinate system to align with the inertial frame, and then substitute in the appropriate values for \Omega and \hat{r}. This will give us the equation of motion in terms of the position vector and its derivatives.

In summary, the key to solving this problem is to understand the physical principles involved and to use vector notation to describe the motion of the bead in the rotating frame. By correctly considering the forces and using the appropriate coordinate system, we can arrive at the final equation of motion.