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Easy Java Simulations (2001- ) => Dynamics => Topic started by: Fu-Kwun Hwang on May 16, 2005, 07:59:39 am



Title: Condition for stable equilibrium
Post by: Fu-Kwun Hwang on May 16, 2005, 07:59:39 am
Under the influence of gravity: the condition for stable equilibrium is
 the center of gravity [b:22e0df7a63](c.g.)[/b:22e0df7a63]  must be lower than the supporting point.  (potential energy is minimum)


Title: Re: Condition for stable equilibrium
Post by: lookang on February 11, 2010, 05:38:31 pm
great applet! i am trying to re-customize it to allow for 2 mass variables, 2 distances.

i define
d(omega)/dt = -(m1*g*(xl-x) +m2*g*(xr-x))/(inertia1+inertia2)

but it didn't evolution as i thought it would in real life.

i will work on it again.

btw, the time is off by one hour, it is 6.39 pm instead of the time shown 5.39pm. for ur info ;D


Title: Re: Condition for stable equilibrium
Post by: Fu-Kwun Hwang on February 11, 2010, 05:49:54 pm
If the center of gravity for those two objects is (xc,yc) and the supporting point is (x,y)
The conditions for stable equilibrium is (x=xc and y>yc) when supporting bar is horizontal.

The who system can be model like a pendulum, the supporting point is the tip of the pendulum, and assume all the mass is located at the center of the gravity.

For the time offset:
Please click the PROFILE (http://www.phy.ntnu.edu.tw/ntnujava/index.php?action=profile) link,
then click Look and Layout Preferences from the left menu.
You will find "Time Offset:" in the table,  change the value from 0 to 1 (or click auto detect link)
then click Change Profile button , then everything should be fine for your time format.


Title: Re: Condition for stable equilibrium
Post by: lookang on February 12, 2010, 07:53:20 am
If the center of gravity for those two objects is (xc,yc) and the supporting point is (x,y)
The conditions for stable equilibrium is (x=xc and y>yc) when supporting bar is horizontal.

The who system can be model like a pendulum, the supporting point is the tip of the pendulum, and assume all the mass is located at the center of the gravity.

This is amazing insight, thanks for the tips, looks like i need to change quite a bit of evolution equation....LOL
will remember your tips! thx!


Title: Re: Condition for stable equilibrium
Post by: Fu-Kwun Hwang on February 12, 2010, 10:18:43 am
The first applet was model in the same way. Use the center of gravity relative to the supporting point as a pendulum to model the system.  You can check out how it was done.