How Does the Center of Mass Remain Stationary During Constant Motion?

AI Thread Summary
The center of mass of a system remains stationary when two individuals pull on a rope towards each other on a frictionless surface because the forces they exert are internal to the system. As one person moves closer, they simultaneously pull the other toward them, resulting in no net movement of the center of mass. This principle is supported by Newton's third law and the conservation of momentum, which dictates that the total momentum of the system must remain constant. If both individuals start at rest, their final momenta must also sum to zero, confirming that the center of mass does not change. Understanding this concept can be challenging, but it is fundamental in physics.
Cyrus
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Hi, I have another question on the center of mass this time. I read a problem in my university physics book by sears and zemanski which had a problem in which two guys want a mug of beer on an ice pond. There is a rope between them and each man tugs on the rope pulling themselves towards each other. It says that the center of mass of the system never changes as the men move towards one another. I was wondering if you could show me why the center of mass would remain stationary even though each man is constnantly moving closer and closer to each other, (assuming they can mysteriously pass through each other and not collide when they meet at some point in time.)

Thanks much,

-Cyrus
 
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cyrus said:
Hi, I have another question on the center of mass this time. I read a problem in my university physics book by sears and zemanski which had a problem in which two guys want a mug of beer on an ice pond. There is a rope between them and each man tugs on the rope pulling themselves towards each other. It says that the center of mass of the system never changes as the men move towards one another. I was wondering if you could show me why the center of mass would remain stationary even though each man is constnantly moving closer and closer to each other, (assuming they can mysteriously pass through each other and not collide when they meet at some point in time.)

Thanks much,

-Cyrus
The men are causing only forces which are internal to the system, and therefore cannot move the center of mass of the whole system. When one man tugs on the rope, he pulls himself closer towards the other guy's position (shifting the center of mass towards the other guy), but he cannot do that without also moving the other guy towards him (shifting the center of mass right back).

Note that the example must be contrived to the point that they're both on frictionless ice - if they were to be on a surface with friction, the force between the man's feet and the ground beneath him would be external, rather than internal, to the two-man system, and could indeed change the position of the center of mass.

Try balancing a ruler on two fingers, one from each hand. Gradually move your fingers together - they meet at the center of the ruler (unless it is unevenly weighted) every single time, no matter where they start from or how you move them.
 
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Its a result of Newton's third law. Its sometimes taken as a postulate as Einstein did in 1906 in his famous "photon in a box" experiment. He demanded that the center of mass remain fixed. I think it can also be derived from the principle of the conservation of momentum. Give it a try. Suppose there are two objects of different mass which are given energy. Demand that momentum be conserved. Find the position of each object as a function of time. Then find the center of mass. Does it remain fixed or is it a function of time too? You'll see that its constant in time.

Pete
 
I think (HOPE HOPE HOPE FINGERS CROSSED!) that I got it based on a conservation of momentum application like you said. If momentum is to be conserved, then

m_1i v_1i+m_2i v_2i = m_1f v_2f + m_2f v_2f, but since they are both at rest at the start, their total initial momentum is zero. so the two final momenta must also sum up to zero, i.e.

0=m_1f v_1f + m_2f v_2f

Now, the the first derivative of the center of mass equation gives:

mCM * vCM = m_1 v_1 + m_2 v_2 + ... (CM = center of mass)

Since the right hand side equals zero, and the total mass of the entire system obviously CANNOT equal zero, then the velocity must be the zero vector on the right hand side! Thus the center of mass has no velocity and remains stationary! GOD I HOPE THIS IS CORRECT! You don't know how much of a pain this has been for me to figure out. :-) I talk to my physics teacher on a regular basis but he's a real busy man. I feel bad at times always calling him up, but its the only way ill learn i guess. I tried to work this problem with him over the phone from work, since i don't have time during the day when he's at work, and I am as well. But I always get brushed away because he has to grade papers or help another student. So, although all this is unnecessary, I wanted you to know that I am thankful for your reply Pete, I am hard pressed to find a constant reliable source to explain physics to me. :-)

Cheers,

Cyrus
 
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