NTNUJAVA Virtual Physics Laboratory
Enjoy the fun of physics with simulations!
November 01, 2014, 05:58:13 pm *
Welcome, Guest. Please login or register.
Did you miss your activation email?

Login with username, password and session length
 
   Home   Help Search Login Register  
To be a citizen of the world. ...Wisdom
Google Bookmarks Yahoo My Web MSN Live Netscape Del.icio.us FURL Stumble Upon Delirious Ask FaceBook

Pages: [1]   Go Down
  Print  
Author Topic: Pendulum in 3D  (Read 16076 times)
0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
Fu-Kwun Hwang
Administrator
Hero Member
*****
Offline Offline

Posts: 3057



WWW
«
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 313 times.)
Logged
msntito
Newbie
*
Offline Offline

Posts: 2

«
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 Offline

Posts: 3057



WWW
«
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 Offline

Posts: 2

«
Embed this message
Reply #3 on: September 29, 2009, 11:16:37 am » posted from:Kanpur,Uttar Pradesh,India

 Smiley`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 Offline

Posts: 3057



WWW
«
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)
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!
Logged
Pages: [1]   Go Up
  Print  
To be a citizen of the world. ...Wisdom
 
Jump to:  


Related Topics
Subject Started by Replies Views Last post
Ballistic pendulum
Dynamics
Fu-Kwun Hwang 1 14860 Last post 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 21234 Last post January 06, 2009, 08:32:35 pm
by Fu-Kwun Hwang
Two Springs and a Pendulum Model
Simulations from other web sites
ahmedelshfie 0 2745 Last post April 14, 2010, 08:16:26 pm
by ahmedelshfie
Driven pendulum (forced oscillation)
Dynamics
Fu-Kwun Hwang 0 5185 Last post 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 9638 Last post March 10, 2011, 07:04:23 pm
by ahmedelshfie
Powered by MySQL Powered by PHP Powered by SMF 1.1.13 | SMF © 2006-2011, Simple Machines LLC Valid XHTML 1.0! Valid CSS!
Page created in 1.942 seconds with 22 queries.since 2011/06/15