NTNUJAVA Virtual Physics Laboratory
Enjoy 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 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 $T=m\frac{v^2}{r}$.

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 here
http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=1883.0
enjoy!