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 Author Topic: Pendulum in 3D  (Read 28049 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
Fu-Kwun Hwang
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 « Embed this message on: September 05, 2009, 11:25:17 pm » posted from:Taipei,T\'ai-pei,Taiwan

This is a pendulum in 3D.
Let the angle between the pendulum and the vertical line is $\theta$ and the angular velocity $\omega=\frac{d\theta}{dt}$
And the angle of the pendulum (projected to x-y plane) with the x-axis is $\phi$, and it's angular velocity $\dot\phi=\frac{d\phi}{dt}$

The lagrange equation for the system is $L=T-V = \tfrac{1}{2}m (L\dot\theta)^2+\tfrac{1}{2}m (L\sin\theta \dot{\phi})^2- (-mgL\cos\theta)$

The equation of the motion is

$\ddot\theta=\sin\theta\cos\theta\dot{\phi}^2-\frac{g}{L}\sin\theta$ ...... from $\frac{d}{dt}(\frac{\partial L}{\partial \dot{\theta}})-\frac{\partial L}{\partial \theta}=0$
and
$m L^2 \sin\theta^2 \dot{\phi}=const$ Angular momentum is conserved. ...... from $\frac{d}{dt}(\frac{\partial L}{\partial \dot{\phi}})-\frac{\partial L}{\partial \phi}=0$

And the following is the simulation of such a system:
When the checkbox (circular loop) is checked, $\omega=0$. and $\dot{\phi}= \sqrt{\frac{g}{L\cos\theta}}$ It is a circular motion.
The vertical component tangential of the string balanced with the mass m, and the horizontal component tangential provide the centripetal force for circular motion.

You can uncheck it and change the period $T=\frac{2\pi}{\dot{\phi}}$ ,
and you will find out the z-coordinate of the pendulum will change with time when
$\omega\neq 0$ or $\dot{\phi}\neq \sqrt{\frac{g}{L\cos\theta}}$
You can also drag the blue dot to change the length of the pendulum.

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msntito
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 « Embed this message Reply #1 on: September 28, 2009, 05:50:10 pm » posted from:Kanpur,Uttar Pradesh,India

Hi,
I was trying to derive the same equations of motion,
Why have you considered the RHS of second equation = constant ?

I got:-  m(L^2) * [  (sin theta)^2 * phi_dot_dot  +  sin(2*theta) * theta_dot * phi_dot  ]  =  0
as the second eq of Motion.

After this, how to solve?

regards,
tito
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Fu-Kwun Hwang
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 « Embed this message Reply #2 on: September 28, 2009, 09:21:41 pm » posted from:Taipei,T\'ai-pei,Taiwan

The lagrange equation for the system is $L=T-V = \tfrac{1}{2}m (L\dot\theta)^2+\tfrac{1}{2}m (L\sin\theta \dot{\phi})^2- (-mgL\cos\theta)$

from $\frac{d}{dt}(\frac{\partial L}{\partial \dot{\phi}})-\frac{\partial L}{\partial \phi}=0$
you will find $m L^2 \sin\theta^2 \dot{\phi}=const$

The reason is:
since there is no $\phi$ in lagrange equation so $\frac{\partial L}{\partial \phi}=0$
which mean $\frac{d}{dt}(\frac{\partial L}{\partial \dot{\phi}})=0$ ,
so $(\frac{\partial L}{\partial \dot{\phi}})$ must be a constant. i.e. $m L^2 \sin\theta^2 \dot{\phi}=const$
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msntito
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 « Embed this message Reply #3 on: September 29, 2009, 11:16:37 am » posted from:Kanpur,Uttar Pradesh,India

`thanks a lot.
I was wondering if this problem can be solved using Euler angles instead of phi and theta (like in rigid body dynamics).How should I begin,  Do you have any idea?
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Fu-Kwun Hwang
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 « Embed this message Reply #4 on: September 29, 2009, 01:34:11 pm » posted from:Taipei,T'ai-pei,Taiwan

You will need to use Euler's angle if the pendulum is not a sphere,
for example:  when it is a cylinder, then, you can use the following rules to draw it:
1. rotate around z axis by $\phi$
2. rotate around y axis by $-\theta$
3. rotate around z axis by $\phi$

You can check out the following applet.

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• Post questions to be asked to help students to think, to explore.
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diinxcom
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 « Embed this message Reply #5 on: December 14, 2014, 05:54:03 pm » posted from:,,Satellite Provider

-*-
Thanks alot! I enjoy this content
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lookang
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 « Embed this message Reply #6 on: November 19, 2015, 02:19:33 pm » posted from:SINGAPORE,SINGAPORE,SINGAPORE