3 attempts to solve rotational problem, all failed. Please help

[z_offer09]

3 attempts to solve rotational problem, all failed. Please help!

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':

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

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

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

 

[Jhero]

Dear scientist,

Thank you for reaching out for help with this rotational problem. It seems like you have put a lot of effort into solving it, but have not yet found a successful solution. Here are a few suggestions that may help:

1. Check your equations: Before attempting any solution method, it is important to make sure that your equations are correct. Double check your equations of motion and constraint equation to ensure that they are accurate.

2. Consider using Lagrangian mechanics: Instead of using Newton's equations of motion, you may want to consider using Lagrangian mechanics. This method can often simplify the equations of motion and make the problem easier to solve.

3. Use energy conservation: As you mentioned in your attempts, energy conservation can be a useful tool in solving this problem. Try using the conservation of energy equation to eliminate one of the unknowns in your equations of motion.

4. Break the problem into smaller parts: Sometimes, it can be helpful to break the problem into smaller parts and solve them separately. In this case, you could first solve for the trajectory of the bead if the disk was not rotating, and then add in the rotation of the disk as a separate step.

5. Seek help from others: Don't be afraid to seek help from other experts or colleagues. They may have a different perspective or approach that could help you solve the problem.

I hope these suggestions are helpful and that you are able to find a successful solution to this rotational problem. Best of luck!