# Bead on wire tilted at angle theta.

1. Jul 9, 2012

### AbigailM

1. The problem statement, all variables and given/known data
Consider a bead of mass m constrained to slide without friction along a rigid wire that rotates about the vertical at a fixed angle $\theta$ with constant angular velocity $\omega$. Write down the Lagrangian in terms of z as the general coordinate. Find the equation of motion of the bead, and determine whether there are positions of equilibrium. If there are equilibrium positions, are they stable?

2. Relevant equations
$z=rcos\theta$

$U=mgz$

3. The attempt at a solution
$\dot{r}=\frac{\dot{z}}{cos\theta}$

$T=\frac{1}{2}m(\dot{r}^{2}+r^{2}\dot{\theta}^{2})$

$L=\frac{2}{3}m(\frac{\dot{z}^{2}}{cos^{2}\theta}+(z\omega)^{2}/(cos^{2}\theta}))-mgz$

$m\ddot{z}=2z\omega^{2}-mgcos^{2}\theta$

$\ddot{z}=\frac{2z\omega^{2}}{m}-gcos^{2}\theta$

Just wondering if my equation of motion is correct. Thanks for the help

Last edited: Jul 9, 2012
2. Jul 9, 2012

### TSny

Note that $\theta$ is a fixed angle. So, $\dot{θ}$ = 0.

You will need a kinetic energy term that takes into account the azimuthal motion ("swinging around motion") around the axis of rotation at the angular speed ω. So, see if you can construct a kinetic energy term involving ω, z, and $\theta$. Of course, you will still have the kinetic energy term $\frac{1}{2}$m$\dot{r}$2.

3. Jul 9, 2012

### TSny

Note that the distance of the particle from the axis of rotation is rsinθ = ztanθ.

4. Jul 11, 2012

### AbigailM

TSny,

$L=\frac{1}{2}m(\frac{\dot{z}^{2}}{cos^{2}\theta}+(z^{2}\omega^{2})/(cos^{2}\theta))-mgz$

$\ddot{z}=\frac{2z\omega^{2}}{m}-gcos^{2}\theta$

Does this look ok?

Thanks.

5. Jul 11, 2012

### TSny

The kinetic energy associated with the rotation about the z axis would be (1/2)m(ωd)2 where d is the radius of the circular motion about the z-axis. If the bead is a distance r out along the wire, then check that d = r sinθ. d can then be expressed in terms of z by using r = z/cosθ.

So, verify that the second term of the Lagrangian would have a tan2θ in place of the 1/cos2θ.

You should find that the mass m cancels out in the equation of motion for $\ddot{z}$. Also, you have a factor of 2 in the first term on the right hand side of the equation for $\ddot{z}$ that should not be there.

6. Jul 12, 2012

### AbigailM

Awesome that made it perfectly clear and I got the correct answer. Thanks so much for the help!

7. Jul 12, 2012

### AbigailM

Also to find any equilibrium positions I looked at what would make $\ddot{z}=0$.
I got only 1 equilibrium position of $z=\frac{g}{tan^{2}\theta\omega^{2}}$.
I also concluded that this is stable since $\omega$ and $\theta$ are constant.

8. Jul 12, 2012

### TSny

Your equilibrium value for z looks correct to me.

However, I believe it's an unstable equilibrium. One way to see this is to go back to your result for $\ddot{z}$.

$\ddot{z}$ = sin2θ ω2 z - g cos2θ

Imagine placing the bead on the rod at the equilibrium position where $\ddot{z}$ = 0. Then displace the bead a little bit farther out on the rod. This will make z greater than the equilibrium value of z and an inspection of the expression for $\ddot{z}$ shows that $\ddot{z}$ will become positive. So, if you let the bead go from this position, which way would it begin to move?

Similarly, analyze what would happen if you displaced the bead a small amount in the opposite direction from the equilibrium position.

Last edited: Jul 12, 2012