(adsbygoogle = window.adsbygoogle || []).push({}); 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 [tex]\omega[/tex]=const

Polar coordinate system [tex]r \ , \theta[/tex] is attached to the disk and rotates with it.

A smooth string is attached to the disk. The string has spiral shape [tex]r = \theta^2[/tex].

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: [tex] r, \phi [/tex]? There is no gravity.

===================

Relevant equations:

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

radial equation:

[tex] \ddot{r} - r\dot{\theta^2} = 2\omega r \dot{\theta} + \omega^2 r + R_r [/tex]

asimuthal equation:

[tex] r\ddot{\theta} + 2\dot{r}\dot{\theta} = -2\omega\dot{r} + R_{\theta} [/tex]

where [tex] R_r, R_{\theta} [/tex] 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 [tex]\varphi = r - \theta^2 = 0[/tex] and then the virtual displacements [tex]\delta r \ , \ \ \delta\theta [/tex] must obey

[tex]\frac{\partial\varphi}{\partial r}\delta r + \frac{\partial\varphi}{\partial\theta}\delta\theta = 0[/tex]

which means

[tex] \delta r = 2\theta \delta\theta[/tex]

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

radial equation:

[tex] \ddot{r} - r\dot{\theta^2} - 2\omega r \dot{\theta} - \omega^2 r = 0[/tex]

asimuthal equation:

[tex] r\ddot{\theta} + 2\dot{r}\dot{\theta} + 2\omega\dot{r} =0 [/tex]

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

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

[tex] \{ \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[/tex]

I also expressed [tex]\theta[/tex] through [tex]r[/tex] by using the constraint condition and got this equation for [tex]\theta[/tex] 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 [tex]r=R\ ,\ \phi=\phi_1[/tex] in F is

[tex] 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 [/tex]

I tried to find the trajectory by varying [tex]J(r,r')[/tex] with the Euler-Lagrange method and applying the constraint

[tex]r = \theta^2[/tex]. I applied the constraind and re-wrote [tex]J(r,r')[/tex] as [tex]J(\theta,\dot{\theta})[/tex] and then did:

[tex] \frac{\partial J}{\partial \theta} - \frac{d}{dt}\{\frac{\partial J}{\partial \dot{\theta}}\} = 0[/tex]

This gives a differential equation which allows the solution [tex]\theta = -\omega t[/tex],

That is, the trajectory seen in F is the straight line [tex]\phi = 0[/tex]

But this seems to be off, because the integral may be incorrect. It assumes [tex]v=\omega \ r[/tex],

which is wrong. Try the same integral on a straight-line constraint. After varying, it should exactly give

the solution [tex] r = e^{\phi}[/tex], but it doesn't.

I guess one needs to use properly the energy conservation and get the proper expression for [tex]v[/tex],

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:

[tex] \ddot{r} - r\dot{\theta^2} = 2\omega r \dot{\theta} + \omega^2 r + R_r [/tex]

asimuthal equation:

[tex] r\ddot{\theta} + 2\dot{r}\dot{\theta} = -2\omega\dot{r} + R_{\theta} [/tex]

where [tex] R_r, R_{\theta} [/tex] are the polar components of the "reaction" force

that holds the bead on the string.

I tried the solution [tex]\theta = -\omega t[/tex],

That is, the trajectory seen in F is the straight line [tex]\phi = 0[/tex].

This means

[tex] R_\theta = 0 [/tex]

[tex] R_r = 2 r_0 \omega^2 [/tex] = 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.

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# Homework Help: 3 attempts to solve rotational problem, all failed. Please help

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