z_offer09
Jan21-08, 12:05 PM
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.
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.