Question about the velocity of the center of mass reference frame

[gionole]

I'm looking into center of mass and I saw the derivation of:

$V = \frac{\sum\limits_{i = 1}^{n} m_iv_i}{\sum\limits_{i = 1}^{n} m_i}$

I understand how it's derived, so no need to explain this further. It's a velocity of the frame in which total momentum of our objects is zero. Forget about the center of mass as well and let's focus on this $V$ only.

Imagine, we got 2 objects moving both to the right with masses 2 and 3 kg with the speeds: 10 and 25. If there is a train moving to the right by $19m/s$, in this frame, total momentum would be 0, but what's so special about this that is different if train had moved by $20m/s$ ? I know that total momentum wouldn't be 0, but so what ?

Let's analyze this example.

Case 1: train moving by $19$ and see the distances travelled by objects and our tain.

After 1sec, 10 -> 19 -> 25 ( 9 - train - 6)
After 2sec, 20 -> 38 -> 50 ( 18- train - 12)
After 3sec, 30 -> 57 -> 75 (27- train - 18)

Case 2: train moving by $20$ and see the distances travelled by objects and our tain.

After 1sec, 10 -> 20 -> 25 ( 10 - train - 5)
After 2sec, 20 -> 40 -> 50 ( 20- train - 10)
After 3sec, 30 -> 60 -> 75 (30- train - 15)

What's so special in this example $19m/s$ moving, that for $20m/s$ train, it's not special anymore ?

 

[A.T.]

I'm looking into center of mass and I saw the derivation of:

$V = \frac{\sum\limits_{i = 1}^{n} m_iv_i}{\sum\limits_{i = 1}^{n} m_i}$

I understand how it's derived, so no need to explain this further. It's a velocity of the frame in which total momentum of our objects is zero. Forget about the center of mass as well and let's focus on this $V$ only.

Imagine, we got 2 objects moving both to the right with masses 2 and 3 kg with the speeds: 10 and 25. If there is a train moving to the right by $19m/s$, in this frame, total momentum would be 0, but what's so special about this that is different if train had moved by $20m/s$ ? I know that total momentum wouldn't be 0, but so what ?

Let's analyze this example.

Case 1: train moving by $19$ and see the distances travelled by objects and our tain.

After 1sec, 10 -> 19 -> 25 ( 9 - train - 6)
After 2sec, 20 -> 38 -> 50 ( 18- train - 12)
After 3sec, 30 -> 57 -> 75 (27- train - 18)

Case 2: train moving by $20$ and see the distances travelled by objects and our tain.

After 1sec, 10 -> 20 -> 25 ( 10 - train - 5)
After 2sec, 20 -> 40 -> 50 ( 20- train - 10)
After 3sec, 30 -> 60 -> 75 (30- train - 15)

What's so special in this example $19m/s$ moving, that for $20m/s$ train, it's not special anymore ?

1) Do you care about the momentum of the train here, or do you just mean a reference frame used for analysis? If you don't care about the momentum of the train, why introduce a physical train object at all?

2) What do you mean by "special"? All inertial frames are equivalent in terms the laws of motion. But depending on the scenario, the math can be simpler in some frames than in others. For example, when analyzing a two body collision in the CoM frame, the momenta pre and post collision are equal but opposite.

 

[gionole]

1) Do you care about the momentum of the train here, or do you just mean a reference frame used for analysis? If you don't care about the momentum of the train, why introduce a physical train object at all?

2) What do you mean by "special"? All inertial frames are equivalent in terms the laws of motion. But depending on the scenario, the math can be simpler in some frames than in others. For example, when analyzing a two body collision in the CoM frame, in momenta pre and post collision are equal but opposite.

1. No, i don't care about momentum of train, I brought it as an example. The thing is the formula derivation is for the frame in which total momentum is zero. I get that, but why do we care about determining the velocity of such frame ?

2. I showed the example where velocity of the frame is 19 and then 20. We determined 19 is the velocity of such frame where total momentum is 0. Why did we even care about determining velocity of such frame ? Thats the question I guess, what makes 19 differ from 20 in my example(other than total momentum is zero ? )

 

[DrClaude]

@A.T. said it:

All inertial frames are equivalent in terms the laws of motion. But depending on the scenario, the math can be simpler in some frames than in others.

 

[gionole]

@A.T. said it:

Have you read what I said in reply #3 ?

 

[DrClaude]

Have you read what I said in reply #3 ?

Yes, where you basically repeated what you said in #1.

Imagine now that your two masses are connected by a spring (massless, linear). Which frame would you rather use to calculate the motion of the two masses?

 

[gionole]

Yes, where you basically repeated what you said in #1.

Imagine now that your two masses are connected by a spring (massless, linear). Which frame would you rather use to calculate the motion of the two masses?

I don't know. I'm asking why we chose/decided to use the frame specifically in which momentum is zero..

 

[DrClaude]

I don't know. I'm asking why we chose/decided to use the frame specifically in which momentum is zero..

Because this is the reference frame in which you can separate the centre-of-mass motion from the relative motion. In the example I gave above, the relative motion of the two masses reduces to the problem of a single mass connected to a spring. This is easy to solve and one can find the motion of the two masses in any other frame from this solution.

 

[A.T.]

I'm asking why we chose/decided to use the frame specifically in which momentum is zero..

...depending on the scenario, the math can be simpler in some frames than in others.

 

[gionole]

Because this is the reference frame in which you can separate the centre-of-mass motion from the relative motion. In the example I gave above, the relative motion of the two masses reduces to the problem of a single mass connected to a spring. This is easy to solve and one can find the motion of the two masses in any other frame from this solution.

In my example in the #1, $V = 19$ is such frame. Can you explain by example for my example how $19$ makes things easier and $20$ wouldn't do ?

 

[A.T.]

In my example in the #1, $V = 19$ is such frame. Can you explain by example for my example how $19$ makes things easier and $20$ wouldn't do ?

What are "things"? It depends on the scenario and the goal, if the CoM frame makes it easier. I gave you an example in post #2:

For example, when analyzing a two body collision in the CoM frame, in momenta pre and post collision are equal but opposite.

 

[gionole]

For example, when analyzing a two body collision in the CoM frame, in momenta pre and post collision are equal but opposite.

Wouldn't the pre momentum be 0 in the CoM frame ? in which case is this why you mean "simpler" ?

Update: ah, you mean, individual momenta of these 2 bodies.

 

[A.T.]

Update: ah, you mean, individual momenta of these 2 bodies.

Yes.

 

[Dale]

What's so special in this example 19m/s moving, that for 20m/s train, it's not special anymore ?

It isn’t “special”, but it is easier for some calculations. Especially collisions. In perfectly elastic collisions each object keeps the same speed before and after and just changes direction. In perfectly inelastic collisions their final speed is 0. Easy.

 

[gionole]

It isn’t “special”, but it is easier for some calculations. Especially collisions. In perfectly elastic collisions each object keeps the same speed before and after and just changes direction. In perfectly inelastic collisions their final speed is 0. Easy.

Perfectly elastic collisions:

Without zero momentum frame, to find after speeds, one has to do:
1. write conservation of momentum
2. write conservation of kinetic energy
3. solve system of equations

with zero momentum frame, to find after speeds, one has to do:
1. find V of zero momentum frame
2. find initial velocities of each object in this frame
3. reverse the signs from (2) to find after speeds
4. move back to original frame(lab frame) - i.e transform velocities

Question 1: How do you asses that zero momentum frame makes things easy in terms of calculations ?

Question 2: If the collision is not perfect(since not perfect, doesn't matter whether it's elastic or inelastic, there definitely was a loss of kinetic energy and we can't solve/find after velocities even in zero momentum frame. Then what ?

 

[A.T.]

How do you asses that zero momentum frame makes things easy in terms of calculations ?

By doing the calculations. And if you don't find it easier, nobody is forcing you to use a particular frame.

... we can't solve/find after velocities even in zero momentum frame. Then what ?

If you can't solve it then you can't solve it. You need additional information for partially elastic collisions.

 

[gionole]

By doing the calculations. And if you don't find it easier, nobody is forcing you to use a particular frame.If you can't solve it then you can't solve it. You need addition information for partially elastic collisions.

I agree, I'm just asking why you find it easier to do calculations in zero momentum frame ? maybe then I will see why it's easier once you tell me why you find it easier.

 

[Dale]

How do you asses that zero momentum frame makes things easy in terms of calculations ?

I can do a sign reversal in my head. I usually need to open Mathematica to solve a system of equations.

If the collision is not perfect(since not perfect, doesn't matter whether it's elastic or inelastic, there definitely was a loss of kinetic energy and we can't solve/find after velocities even in zero momentum frame. Then what ?

Then the benefit is less.

 

[A.T.]

maybe then I will see why it's easier

Just do the calculations in different frames and decide for yourself.

 

[gionole]

I found one use case. For perfectly elastic collision, If 3 bodies collide, without zero momentum frame, we got 2 equations(kinetic energy conservation + momentum conservation), but for 3 bodies, these 2 equations are not enough and we would need to try to find 3rd(who knows what this 3rd could be). In ZMF though, we don't need this headache. Thoughts ?

 

[Dale]

And if you don't find it easier, nobody is forcing you to use a particular frame.

@gionole this is key. You can use whatever coordinates you like. Sometimes changing coordinates helps a lot. Sometimes it doesn’t. It is your choice what to use for a given problem.

 

[Ibix]

Question 1: How do you asses that zero momentum frame makes things easy in terms of calculations ?

This is a bit like asking why we do long multiplication even though it requires several multiplication steps where directly doing the multiplication only requires one. Sure, but the steps are easy, you don't ever need anything higher than your nine times table, and there's a standard way of handling carried digits.

Similarly, the ZMF approach gives a series of steps that can be carried out relatively easily. Your "solve system of equations" step involves at least one matrix inversion. The ZMF approach requires nothing more complex than division and vector addition.

If you prefer the matrix inversion approach, use that.