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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 simulation Could 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 motion my 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)/m My 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
If the mass is in circular motion, then the tension $T=m\frac{v^2}{r}$. I see I got the sign convention wrong!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 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 here http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=1883.0 enjoy! |