Pendulum in 3D

Fu-Kwun Hwang

Administrator
Hero Member

Offline

Posts: 3047

1 Pendulum in 3D

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.

Embed a running copy of this simulation

Embed a running copy link(show simulation in a popuped window)

Full screen applet or Problem viewing java?Add http://www.phy.ntnu.edu.tw/ to exception site list
Press the Alt key and the left mouse button to drag the applet off the browser and onto the desktop. This work is licensed under a Creative Commons Attribution 2.5 Taiwan License

Please feel free to post your ideas about how to use the simulation for better teaching and learning.
Post questions to be asked to help students to think, to explore.
Upload worksheets as attached files to share with more users.

Let's work together. We can help more users understand physics conceptually and enjoy the fun of learning physics!

pendulum3d.jpg (21.07 KB, 580x499 - viewed 214 times.)
	Logged

msntito

Newbie

Offline

Posts: 2

2 Re: Pendulum in 3D

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


	Logged

Fu-Kwun Hwang

Administrator
Hero Member

Offline

Posts: 3047

3 Re: Pendulum in 3D

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. $m L^2 \sin\theta^2 \dot{\phi}=const$


	Logged

msntito

Newbie

Offline

Posts: 2

4 Re: Pendulum in 3D

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?


	Logged

Fu-Kwun Hwang

Administrator
Hero Member

Offline

Posts: 3047

5 Re: Pendulum in 3D

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.

Embed a running copy of this simulation

Embed a running copy link(show simulation in a popuped window)

Please feel free to post your ideas about how to use the simulation for better teaching and learning.
Post questions to be asked to help students to think, to explore.
Upload worksheets as attached files to share with more users.

Let's work together. We can help more users understand physics conceptually and enjoy the fun of learning physics!


	Logged

Pages: [1] Go Up

Use resources around us effectively. ...Wisdom

« previous next »

Related Topics
	Subject	Started by	Replies	Views	Last post
	Ballistic pendulum Dynamics	Fu-Kwun Hwang	1	13855	March 14, 2010, 09:27:33 am by eric2010
	Free Oscillations and Rotations of a Rigid Pendulum by Eugene Butikov Ejs simulations from other web sites	Fu-Kwun Hwang	2	20809	January 06, 2009, 08:32:35 pm by Fu-Kwun Hwang
	Two Springs and a Pendulum Model Simulations from other web sites	ahmedelshfie	0	2314	April 14, 2010, 08:16:26 pm by ahmedelshfie
	Driven pendulum (forced oscillation) Dynamics	Fu-Kwun Hwang	0	4798	April 24, 2010, 02:48:14 pm by Fu-Kwun Hwang
	3D rotational pendulum also known as Furuta's pendulum by Francisco Esquembre Dynamics	lookang	2	8444	March 10, 2011, 07:04:23 pm by ahmedelshfie

	Author	Topic: Pendulum in 3D (Read 14595 times)
0 Members and 2 Guests are viewing this topic. Click to toggle author information(expand message area).