# NTNUJAVA Virtual Physics LaboratoryEnjoy the fun of physics with simulations! Backup site http://enjoy.phy.ntnu.edu.tw/ntnujava/

## Easy Java Simulations (2001- ) => Dynamics => Topic started by: Fu-Kwun Hwang on September 05, 2009, 11:25:17 pm

 Title: Pendulum in 3D Post by: Fu-Kwun Hwang on September 05, 2009, 11:25:17 pm 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 (http://en.wikipedia.org/wiki/Lagrangian_mechanics) 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. Title: Re: Pendulum in 3D Post by: msntito on September 28, 2009, 05:50:10 pm 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   ]   =    0as the second eq of Motion.After this, how to solve?regards,tito Title: Re: Pendulum in 3D Post by: Fu-Kwun Hwang on September 28, 2009, 09:21:41 pm The lagrange equation (http://en.wikipedia.org/wiki/Lagrangian_mechanics) 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. $m L^2 \sin\theta^2 \dot{\phi}=const$ Title: Re: Pendulum in 3D Post by: msntito on September 29, 2009, 11:16:37 am  :)`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? Title: Re: Pendulum in 3D Post by: Fu-Kwun Hwang on September 29, 2009, 01:34:11 pm 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. Title: Re: Pendulum in 3D Post by: diinxcom on December 14, 2014, 05:54:03 pm -*-Thanks alot! :) I enjoy this content Title: Re: Pendulum in 3D Post by: lookang on November 19, 2015, 02:19:33 pm made this JavaScript version!enjoyhttp://iwant2study.org/ospsg/index.php/interactive-resources/physics/02-newtonian-mechanics/09-oscillations/274-pendulum3d