1. Jan 21, 2008

### z_offer09

I am trying to solve this problem but am getting nowhere. Perhaps someone could help...
Here goes:

===================
A horizontal disk rotates uniformly with $$\omega$$=const
Polar coordinate system $$r \ , \theta$$ is attached to the disk and rotates with it.
A smooth string is attached to the disk. The string has spiral shape $$r = \theta^2$$.
The string rotates together with the disk.

A point mass (a bead) slides without friction on the string. What is the trajectory of the
bead in the static polar coordinate system F: $$r, \phi$$? There is no gravity.
===================

Relevant equations:
Newton's equations of motion in the rotating frame F':

$$\ddot{r} - r\dot{\theta^2} = 2\omega r \dot{\theta} + \omega^2 r + R_r$$
asimuthal equation:
$$r\ddot{\theta} + 2\dot{r}\dot{\theta} = -2\omega\dot{r} + R_{\theta}$$
where $$R_r, R_{\theta}$$ are the polar components of the "reaction" force
that holds the bead on the string.

ATEMPTS to solve it:

-------------------
ATTEMPT number 1:

Trying it with d'Alembert method (the method of virtual work)
The constraint equation is $$\varphi = r - \theta^2 = 0$$ and then the virtual displacements $$\delta r \ , \ \ \delta\theta$$ must obey

$$\frac{\partial\varphi}{\partial r}\delta r + \frac{\partial\varphi}{\partial\theta}\delta\theta = 0$$

which means

$$\delta r = 2\theta \delta\theta$$

Newton's equations of motion in the rotating frame F' are:

$$\ddot{r} - r\dot{\theta^2} - 2\omega r \dot{\theta} - \omega^2 r = 0$$
asimuthal equation:
$$r\ddot{\theta} + 2\dot{r}\dot{\theta} + 2\omega\dot{r} =0$$

Here I omit the reaction forces of the string, as their virtual work is zero.

Then, applying d'Alembert's principle, I get

$$\{ \ddot{r} - r\dot{\theta^2} - 2\omega r \dot{\theta} - \omega^2 r \}2\theta + \{ r\ddot{\theta} + 2\dot{r}\dot{\theta} + 2\omega\dot{r} \} = 0$$

I also expressed $$\theta$$ through $$r$$ by using the constraint condition and got this equation for $$\theta$$ only.
I am way off here? This differential equation is very hard to solve. I tried it with Maple and got some hack-looking expressions.
So this attempt seems to have failed.

-------------------
ATTEMPT number 2:

Trying it with variational calculus
The time to reach an arbitrary position $$r=R\ ,\ \phi=\phi_1$$ in F is

$$T = \int \frac{ds}{v} = \frac{1}{\omega}\int_0^{\phi_1} \frac{\sqrt{r'^2+r^2}}{r}\ d\phi = \int_0^{\phi_1} J(r,r')\ d\phi$$
I tried to find the trajectory by varying $$J(r,r')$$ with the Euler-Lagrange method and applying the constraint
$$r = \theta^2$$. I applied the constraind and re-wrote $$J(r,r')$$ as $$J(\theta,\dot{\theta})$$ and then did:

$$\frac{\partial J}{\partial \theta} - \frac{d}{dt}\{\frac{\partial J}{\partial \dot{\theta}}\} = 0$$

This gives a differential equation which allows the solution $$\theta = -\omega t$$,
That is, the trajectory seen in F is the straight line $$\phi = 0$$

But this seems to be off, because the integral may be incorrect. It assumes $$v=\omega \ r$$,
which is wrong. Try the same integral on a straight-line constraint. After varying, it should exactly give
the solution $$r = e^{\phi}$$, but it doesn't.

I guess one needs to use properly the energy conservation and get the proper expression for $$v$$,
but I am confused here and couldn't do it.

So this one failed.

-------------------
ATTEMPT number 3:

Trying it with Newton's equations of motion.

I wrote them in the rotating frame F', because the constraint applies in F':

$$\ddot{r} - r\dot{\theta^2} = 2\omega r \dot{\theta} + \omega^2 r + R_r$$
asimuthal equation:
$$r\ddot{\theta} + 2\dot{r}\dot{\theta} = -2\omega\dot{r} + R_{\theta}$$

where $$R_r, R_{\theta}$$ are the polar components of the "reaction" force
that holds the bead on the string.

I tried the solution $$\theta = -\omega t$$,
That is, the trajectory seen in F is the straight line $$\phi = 0$$.

This means

$$R_\theta = 0$$
$$R_r = 2 r_0 \omega^2$$ = const

It's all OK, but it is a "backwards" solution, assuming how the reaction looks like.
So it is a hack and it assumed failed.

Any thoughts, please? I am running out of ideas.

Last edited: Jan 21, 2008