Bead Sliding on Rotating Rod after Motor is Turned Off

[Ichigo449]

Homework Statement

A bead of mass m slides in a frictionless hollow open-ended tube of length L which is held at an angle of β to the vertical and rotated by a motor at an angular velocity ω. The apparatus is in a vertical gravitational field.
a) Find the bead's equations of motion

b) Find the equilibrium position of the bead.

c) What is the minimum angular velocity so that the bead remains at the equilibrium position?

d) Suppose the bead is displaced from rest a distance δ from the equilibrium position and at the exact same time the motor is shut off and the tube can freely spin without friction about its lower endpoint. Find the position of the bead as a function of time while it is still in the tube.

Homework Equations

Lagrange's equations

The Attempt at a Solution

a) Using spherical coordinates to describe the position of the bead only the distance along the rod, r, is a degree of freedom. θ is fixed in the problem to be β, and $\dot θ =ω$. So substituting these constraints into the kinetic energy expressed in spherical coordinates gives:
$L = T - V = \frac{1}{2}m[\dot{r}^{2}+r^{2}ω^{2}sin^{2}β] - mgrcosβ$

From which the equations of motion easily follow as:
$\frac{d}{dt}\frac{\partial L}{\partial \dot{r}} -frac{\partial L}{\partial r} = 0$,

$\ddot{r} -rsin^{2}βω^{2} = -gcosβ$

b) At equilibrium $\ddot{r} = 0$, implying that $r= \frac{gcosβ}{sin^{2}βω^{2}}$.

c) Linearizing the equation of motion gives:
$\ddot{r} -rβ^{2}ω^{2} = -g$, so if ω^{2} > \frac{g}{rβ^{2}}, then the equilibrium position may be stable.
But something seems odd here since normal mode analysis implies that perturbations of this system about equilibrium grow exponentially so stability of the equilibrium position should be impossible.

d) I don't really know where to begin with this part of the problem. The azimuthal angle φ is now a degree of freedom with the motor shut off but I don't know how to relate the two situations.

 

[haruspex]

The Attempt at a Solution

First, I'll just post your attempt with the Latex fixed:
---------------
a) Using spherical coordinates to describe the position of the bead only the distance along the rod, r, is a degree of freedom. θ is fixed in the problem to be $\beta$, and $\dot θ =ω$. So substituting these constraints into the kinetic energy expressed in spherical coordinates gives:
$L = T - V = \frac{1}{2}m[\dot{r}^{2}+r^{2}ω^{2}\sin^{2}(\beta)] - mgr\cos (\beta)$

From which the equations of motion easily follow as:
$\frac{d}{dt}\frac{\partial L}{\partial \dot{r}} -\frac{\partial L}{\partial r} = 0$,

$\ddot{r} -r\sin^{2}(\beta)ω^{2} = -g\cos(\beta)$

b) At equilibrium $\ddot{r} = 0$, implying that $r= \frac{g\cos(\beta)}{\sin^{2}(\beta)ω^{2}}$.

c) Linearizing the equation of motion gives:
$\ddot{r} -r(\beta)^{2}ω^{2} = -g$, so if $ω^{2} > \frac{g}{r(\beta)^{2}}$, then the equilibrium position may be stable.
But something seems odd here since normal mode analysis implies that perturbations of this system about equilibrium grow exponentially so stability of the equilibrium position should be impossible.

 

[haruspex]

θ is fixed in the problem to be β

I think you mean φ is fixed as β.

I did not understand this deduction:

$\ddot{r} -r(\beta)^{2}ω^{2} = -g$, so if $ω^{2} > \frac{g}{r(\beta)^{2}}$, then the equilibrium position may be stable.

To analyse stability, work in terms of the displacement from equilibrium. Obtain the D.E. for that before linearizing. Please post the details of your argument.

 

[Ichigo449]

Working with the displacement from equilibrium, $ r= r_{0} + ε(t)$, the equation of motion becomes:
$\ddot{ε(t)} -gcosβ -ε(t)sin^{2}βω^{2} =-gcosβ$,
which simplifies to,
$\ddot{ε(t)} -ε(t)sin^{2}βω^{2} = 0$.
Linearizing this expression gives:
$\ddot{ε(t)} -ε(t)β^{2}ω^{2} =0$.
In order for stability to occur the solution for ε(t) must be a sinusoidal function of time. Due to this β^{2}ω^{2} <1, or ω^{2} <1/β^{2}.
This a maximum angular velocity condition for stability which makes physical sense to me, as generically the bead should be flung off of the rod.
P.S. How do I fix the LaTex?

 

[haruspex]

How do I fix the LaTex?

Use a pair of hash symbols (#) instead of a single $.
You could also prefix each trig function with a backslash and use \beta etc.
Fixing the latex in post 4:
Working with the displacement from equilibrium, $r= r_{0} + ε(t)$, the equation of motion becomes:
$\ddot{ε(t)} -gcosβ -ε(t)sin^{2}βω^{2} =-gcosβ$,
which simplifies to,
$\ddot{ε(t)} -ε(t)sin^{2}βω^{2} = 0$,
Linearizing this expression gives:
$\ddot{ε(t)} -ε(t)β^{2}ω^{2} =0$.

 

[haruspex]

In order for stability to occur the solution for ε(t) must be a sinusoidal function of time.

Yes.

Due to this β^{2}ω^{2} <1,

That's not what I get, and it's dimensionally inconsistent. Where does that 1 come from?

 

[Ichigo449]

Then it seems impossible for the equilibrium to ever be stable. A sinusoidal displacement from equilibrium would require that $\beta^{2} ω^{2}$ is a negative number, which can't happen since $\beta$ and $\omega$ are presumably positive real numbers.

 

[haruspex]

Then it seems impossible for the equilibrium to ever be stable. A sinusoidal displacement from equilibrium would require that $\beta^{2} ω^{2}$ is a negative number, which can't happen since $\beta$ and $\omega$ are presumably positive real numbers.

Right, and that agrees with my intuition. An outward displacement increase the centrifugal force, accelerating it outwards, and conversely for an inwards displacement. So for part c to make sense, it cannot be in regard to a stable position. We have to accept its being unstable equilibrium.
But then, asking for the minimum rotation rate is strange. There is an exact rotation rate obtained by rewriting your answer to b) into the form ω=... Varying β, seems like any value for ω is possible.

 

[Ichigo449]

Okay, I think I have a pretty good handle on 3/4 of the question then. But I'm still not sure how to even begin part d. Even just a hint for how to get started would be much appreciated.

 

[TSny]

But then, asking for the minimum rotation rate is strange. There is an exact rotation rate obtained by rewriting your answer to b) into the form ω=... Varying β, seems like any value for ω is possible.

I agree. The only thing I can think of is that the tube has a finite length L. So this limits the range for the location of the equilibrium point.

 

[TSny]

Okay, I think I have a pretty good handle on 3/4 of the question then. But I'm still not sure how to even begin part d. Even just a hint for how to get started would be much appreciated.

I ran across this rendition of the problem where it says in part d to assume the mass of the tube is negligible.
https://physics.illinois.edu/academics/graduates/qual/cm/CMSpring95A.pdf

That additional remark helps a lot. But I don't see why they want you to assume that initially there is a very small displacement from equilibrium in part d. Why bother with that? Couldn't the bead be at the equilibrium position when the motor is shut off?

Edit: Maybe I'm not interpreting part d as intended. What does "freely spin" about the end mean? At first I thought that means complete freedom to rotate in any direction. But now I think it means free to rotate about the vertical axis while forced to maintain the angle $\beta$. Then I can see why they want an initial displacement from equilibrium. Interesting result!

 

[Ichigo449]

So how do I solve part d then? I agree it's definitely an interesting situation but can't wrap my head around how to solve it. Do I go back to the Lagrangian, find new equations of motion and then solve them subject to ω being the angular velocity at time 0?

 

[haruspex]

So how do I solve part d then? I agree it's definitely an interesting situation but can't wrap my head around how to solve it. Do I go back to the Lagrangian, find new equations of motion and then solve them subject to ω being the angular velocity at time 0?

Yes, go back to the equation you found at the end of part a), but now ω is a variable. What is another relationship between ω and r now that the motor is off?

 

[Ichigo449]

With the motor turned off and the angle to the vertical fixed at $\beta$ angular momentum should now be conserved (unsure about this).
If it is conserved then the value at time 0 is:
$L = m[r_{0} + \delta]^{2}\omega$, which would be equal to the angular momentum at any subsequent time:
$mr^{2} \dot{\phi}$.

 

[haruspex]

With the motor turned off and the angle to the vertical fixed at $\beta$ angular momentum should now be conserved (unsure about this).
If it is conserved then the value at time 0 is:
$L = m[r_{0} + \delta]^{2}\omega$, which would be equal to the angular momentum at any subsequent time:
$mr^{2} \dot{\phi}$.

Isn't your r the distance up the tube?

 

[Ichigo449]

Yes it is, so the moment arms are $rsin(\beta)$. But the angular momentum argument is correct?

 

[haruspex]

Yes it is, so the moment arms are $rsin(\beta)$. But the angular momentum argument is correct?

Yes.

 

[TSny]

But the angular momentum argument is correct?

You can derive the conservation of angular momentum from the Lagrangian, where $\varphi$ is now a degree of freedom.