Velocity of the center of mass for MD simulations

AI Thread Summary
In molecular dynamics (MD) simulations, subtracting the average velocity from each atom's velocity is a method used to ensure the center of mass velocity is zero. This adjustment allows for a more accurate representation of the system's dynamics by removing any translational motion. The mass-weighted average velocity, calculated by dividing the sum of each atom's momentum by the total mass, is equivalent to the center of mass velocity. Understanding this equivalence clarifies why subtracting the average velocity is necessary for achieving a zero center of mass velocity. This approach enhances the accuracy of the simulation's results.
xperrylinn
Messages
3
Reaction score
0
When calculating the center of mass for an MD simulation, I've seen a few codes that subtract the average velocity from each atoms velocity. I think the reasoning behind this is that you want the velocity of center of mass for the simulation to be zero, which I understand, but I don't understand computationally how subtracting the average velocity from each velocity achieves this.
 
Physics news on Phys.org
xperrylinn said:
When calculating the center of mass for an MD simulation, I've seen a few codes that subtract the average velocity from each atoms velocity. I think the reasoning behind this is that you want the velocity of center of mass for the simulation to be zero, which I understand, but I don't understand computationally how subtracting the average velocity from each velocity achieves this.
Can you see that the [mass-weighted] average velocity is equal to the velocity of the center of mass?
 
I don't see that. This is how I'm seeing it:

mass-weighted average velocity = (sum each atom's momentum) / (num atoms)

velocity center of mass = (sum of each atom's momentum) / (total mass)

And if they are equal, how does help me understand my original question?
 
xperrylinn said:
mass-weighted average velocity = (sum each atom's momentum) / (num atoms)
The correct way to compute a weighted average is to divide by the sum of the weights, not the count of items.
 
  • Like
Likes Ibix
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
Back
Top