Author Topic: Conservation of Angular momentum and 3D circular motion  (Read 20568 times)

Fu-Kwun Hwang

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Conservation of Angular momentum and 3D circular motion
« on: March 13, 2010, 12:06:08 am »
A particle with mass $m$ is moving with constant speed $v$ along a circular orbit (radius $r$).
The centripetal force $F=mfrac{v^2}{r}$ 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 $vec{L}=m,vec{r}   imes vec{v}$ is conserved. i.e. $L=mr^2omega$ is a constant.
 
For particle with mass m:

 $ m frac{d^2r}{dt^2}=mfrac{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.

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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
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macfamous

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Re: Conservation of Angular momentum and 3D circular motion
« Reply #1 on: March 19, 2010, 11:53:15 am »
there is more simple formula for this simulation ???

Fu-Kwun Hwang

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Re: Conservation of Angular momentum and 3D circular motion
« Reply #2 on: March 19, 2010, 02:22:45 pm »
Quote
there is more simple formula for this simulation

Could you explain what do you mean in detail?

lookang

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Re: Conservation of Angular momentum and 3D circular motion
« Reply #3 on: July 07, 2010, 01:32:23 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 :)
« Last Edit: July 07, 2010, 01:54:54 pm by lookang »

Fu-Kwun Hwang

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Re: Conservation of Angular momentum and 3D circular motion
« Reply #4 on: July 07, 2010, 07:05:49 pm »
If the mass is in circular motion, then the tension $T=mfrac{v^2}{r}$.

What if we increase the tension so that $T>mfrac{v^2}{r}$. 
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!

lookang

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Re: Conservation of Angular momentum and 3D circular motion
« Reply #5 on: July 07, 2010, 07:16:08 pm »
If the mass is in circular motion, then the tension $T=mfrac{v^2}{r}$.

What if we increase the tension so that $T>mfrac{v^2}{r}$.
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 here
http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=1883.0
enjoy!
« Last Edit: July 07, 2010, 07:21:45 pm by lookang »