# 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 March 12, 2010, 11:06:08 pm

 Title: Conservation of Angular momentum and 3D circular motion Post by: Fu-Kwun Hwang on March 12, 2010, 11:06:08 pm A particle with mass $m$ is moving with constant speed $v$ along a circular orbit (radius $r$).The centripetal force $F=m\frac{v^2}{r}$ is provided by gravitation force from another mass $M=F/g$.A string is connected from mass m to the origin then connected to mass $M$.Because the force is always in the $\hat{r}$ direction, so the angular momentum $\vec{L}=m\,\vec{r}\times \vec{v}$ is conserved. i.e. $L=mr^2\omega$ is a constant. For particle with mass m: $m \frac{d^2r}{dt^2}=m\frac{dv}{dt}= m \frac{v^2}{r}-Mg=\frac{L^2}{mr^3}- Mg$ $\omega=\frac{L}{mr^2}$The following is a simulation of the above model.You can change the mass M or the radius r with sliders.The mass M also changed to keep the mass m in circular motion when you change r.However, if you change mass M , the equilibrium condition will be broken. Title: Re: Conservation of Angular momentum and 3D circular motion Post by: macfamous on March 19, 2010, 10:53:15 am there is more simple formula for this simulation ??? Title: Re: Conservation of Angular momentum and 3D circular motion Post by: Fu-Kwun Hwang on March 19, 2010, 01:22:45 pm Quote there is more simple formula for this simulationCould you explain what do you mean in detail? Title: Re: Conservation of Angular momentum and 3D circular motion Post by: lookang on July 07, 2010, 12:32:23 pm question:how is the tension in the string calculated?My understanding:T = m*v*v/r; // added lookang for tension only work for circular motionmy hypothesis:T1 = m*v*v/r + m*dvr/dt  ? // radial acceleration due to tangential change in velocity and radial acceleration due to radial change in velocity ?where dvr/dt = (cst*cst/(m*r*r*r)-M*g)/mMy equation:T1 = m*v*v/r +(cst*cst/(m*r*r*r)-M*g) ;can help take a look if my equation of Tension is correct ?did i get the generalized equation for tension in any elliptical motion?the logic seems correct and i implement the equation looks correct. is it correct conceptually in the absence of frictional forces :) Title: Re: Conservation of Angular momentum and 3D circular motion Post by: Fu-Kwun Hwang on July 07, 2010, 06:05:49 pm If the mass is in circular motion, then the tension $T=m\frac{v^2}{r}$.What if we increase the tension so that $T>m\frac{v^2}{r}$.  Will the mass moving inward($\frac{dv_r}{dt}<0$) or moving outward($\frac{dv_r}{dt}>0$)?Thank about it and then check with your equation! Title: Re: Conservation of Angular momentum and 3D circular motion Post by: lookang on July 07, 2010, 06:16:08 pm Quote from: Fu-Kwun Hwang on July 07, 2010, 06:05:49 pmIf the mass is in circular motion, then the tension $T=m\frac{v^2}{r}$.What if we increase the tension so that $T>m\frac{v^2}{r}$.  Will the mass moving inward($\frac{dv_r}{dt}<0$) or moving outward($\frac{dv_r}{dt}>0$)?Thank about it and then check with your equation!I see I got the sign convention wrong!I forgot the convention for r unit vector is positive outward. Will change it to T1 = m*v*v/r -(cst*cst/(m*r*r*r)-M*g) ;thanks !!! :)your masterful questioning allows to think and figure out the answer myself.Appreciate your superb advise :)my remixed applet is herehttp://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=1883.0enjoy!