Is there any chance that $$rtl(\ddot\omega+\ddot\phi)\cdot\sin{(\phi+\omega t)}- gl\dot\phi \cdot \sin{\phi} + ltr\dot\omega(\dot\phi^2-\dot\omega)\cdot \cos{(\phi+\omega t)}-gtr\cos{(\omega t)}+l^2\ddot\phi-t^2r^2\ddot\omega=0$$

is equation of motion for the ball in the picture?

To describe x and y coordinates I wrote $x = l\cdot \sin{\phi} + r\cdot \cos{\omega t} $ and $y = l\cdot \cos{\phi} + r\cdot \sin{\omega t}$

Than express kinetic energy as $E_k = \frac{1}{2} m \cdot (\dot x^2 + \dot y^2)$ and potential energy as $E_p = mgy$.

Next step was Langrangian $\mathcal{L} = E_k - E_p$

I put $\frac{d}{dt}\left( \frac{\partial{\mathcal{L}}}{\partial \dot \phi} \right) - \frac{\partial \mathcal{L}}{\partial \phi } = \frac{d}{dt}\left( \frac{\partial{\mathcal{L}}}{\partial \dot \omega} \right) - \frac{\partial \mathcal{L}}{\partial \omega}$

and that lead to the equation above, but I might do some mistakes. I hope it's ok to put the two equations together like this. Is it?

So if $q$ is only $\phi$ than

$$\dot x = \dot \phi l \cos{(\phi)}-r \omega \sin{(\omega t)}$$ $$\dot y = - \dot \phi l \sin{(\phi)} + r\omega \cos{(\omega t)}$$

$$E_k = \frac{1}{2}m(\dot \phi^2 l^2 \cos^2{(\phi)} - 2r \omega l \dot \phi \cos{(\phi)} \sin{(\omega t)}+r^2 \omega^2 \sin^2{(\omega t)} + r^2 \omega^2 \cos^2{(\omega t)} - 2r \omega l \dot \phi \sin{(\phi)} \cos{(\omega t)}+l^2 \dot \phi^2 \sin^2{(\phi)})$$

$$Ek = \frac12 m(\dot \phi^2 l^2 (\cos^2{(\phi)}+\sin^2{(\phi)}) + r^2 \omega^2 (\sin^2{(\omega t)}+\cos^2{(\omega t)})-2rl \omega \dot \phi (\cos{(\phi)}\sin{(\omega t)}+\sin{(\phi)}\cos{(\omega t)}))$$

$$E_k = \frac12 m \dot\phi^2 l^2 + \frac12 m r^2 \omega^2 - mrl \omega \dot \phi \sin{(\phi + \omega t)}$$

So it would be kinetic energy of the ball. Than I need potential energy.

$$E_p = mgl\cos{(\phi)} + mgr\sin{(\omega t)}$$

Therefor Lagrangian should be:

$$\mathcal{L} = \frac12 m \dot\phi^2 l^2 + \frac12 m r^2 \omega^2 - mrl \omega \dot \phi \sin{(\phi + \omega t)} - mgl\cos{(\phi)} - mgr\sin{(\omega t)}$$

It seems like right time for Lagrange equation. I'll compute every part of equation on the new line.

$$\frac{\partial \mathcal{L}}{\partial \phi} = -mrl\omega \dot \phi^2 \cos{(\phi+ \omega t)} + mgl \dot \phi \sin{(\phi)}$$

$$\frac{\partial \mathcal{L}}{\partial \dot \phi} = ml^2 \dot \phi - mrl \omega \sin{(\phi + \omega t)}$$

$$\frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot \phi} \right) = ml^2\ddot \phi - mrl \omega^2 \dot \phi \cos{(\phi + \omega t)}$$

If this was right, the equation would look like:

$$ml^2\ddot \phi - mrl \omega^2 \dot \phi \cos{(\phi + \omega t)} + mrl\omega \dot \phi^2 \cos{(\phi+ \omega t)} - mgl \dot \phi \sin{(\phi)} = 0$$

If I put cos together and multiply whole equation by $\frac{1}{ml^2}$ I get

$$\ddot \phi - \frac rl \omega \dot \phi \cos{(\phi + \omega t)} (\omega + \dot \phi) - \frac gl \dot \phi \sin{(\phi)}$$

which is not right.


closed as off-topic by Emilio Pisanty, user10851, Waffle's Crazy Peanut, Manishearth Aug 29 '13 at 11:58

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  • 5
    $\begingroup$ Please explain what motion your diagram shows and why you think your equation matches that motion. Nobody is going to do the work needed to decipher both and comment at that level of detail and check-your-work questions aren't accepted here. $\endgroup$ – Brandon Enright Apr 29 '13 at 19:55
  • $\begingroup$ It's a mathematical pendulum on rotating disk. $m$ is its weight. $\omega$ is angular velocity of disk. $\endgroup$ – user50222 Apr 29 '13 at 20:18
  • $\begingroup$ Both terms in $\dot y$ should be positive. If you correct the sign there all $\phi + \omega t$ will become $\phi - \omega t$. Except this your ${\partial L \over \partial \dot \phi}$ is right. The second term in ${d \over dt}{\partial L \over \partial \dot \phi}$ should be $m r l \omega (\dot \phi - \omega) cos (\phi - \omega t)$. First term in ${\partial L \over \partial \phi}$ should be $+m r l \omega \dot \phi cos(\phi - \omega t)$. $\endgroup$ – Azad Apr 30 '13 at 12:27
  • $\begingroup$ Also except the sign case your Lagrangian is correct. $\endgroup$ – Azad Apr 30 '13 at 12:30
  • $\begingroup$ $y$ should be $y = -l\cos{(\phi)} + r\sin{(\omega t)}$ ? Is that right? $\endgroup$ – user50222 Apr 30 '13 at 15:04

Your equation is wrong. Intuitively as either $r$ or $\omega$ vanishes you should recover the equation for a simple pendulum which is not true about your equation.

In the Lagrangian formalism we equate $\frac{d}{dt} \frac{\partial L}{\partial \dot q_i} - \frac{\partial L}{\partial q_i}$ for each generalised coordinates to zero, not to each other.

Usually we take $\omega$ to be constant. If this is not the case you have two generalised coordinates and two independent equations which means a coupled two coordinate oscillator which I think is not the intention of the problem designer here.

If you take $\omega$ to be constant you should arrive at this equation: $\ddot \phi + {g \over l}sin(\phi) + {r \over l} \omega^2 cos(\phi - \omega t) = 0$

You could check that it reduces to simple pendulum formula as $\omega$ and/or $r$ vanishes.

  • $\begingroup$ But you got one equation. So what did you do when you had both $\frac{d}{dt} \frac{\partial L}{\partial \dot q_i} - \frac{\partial L}{\partial q_i}$ ? $\endgroup$ – user50222 Apr 30 '13 at 5:57
  • $\begingroup$ What are $q_i$? I think $\phi$ is right, isn't it? What should I use as the other? $\endgroup$ – user50222 Apr 30 '13 at 6:42
  • $\begingroup$ Or should it be only $\phi$? I'm running out of time, I will try to write a little computation above, to the question. Please, could you tell me if it is anyhow better than before and what are mistakes? $\endgroup$ – user50222 Apr 30 '13 at 7:28
  • $\begingroup$ If you take $\omega$ to be constant you get only one equation involving one variable and its derivatives $\phi$. But if you take $\omega$ as a variable as well (meaning that the disk could oscillate in addition to pendulum) you'll get two equations each of which containing both variable and its derivatives. This is a coupled non-linear ODE system that you can only find the answer with linear approximation or numerical methods. You can't decouple it analytically. $\endgroup$ – Azad Apr 30 '13 at 7:30
  • $\begingroup$ Yes, it should only be $\phi$ and I think your only $q_i$ is $\phi$. $\endgroup$ – Azad Apr 30 '13 at 7:34

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