Tug Between Two People On Friction-less Ice.

[Bashyboy]

Okay, suppose we have two people, one being 50 kg and the other being 100 kg; they are separated by 600 m, with each person holding on the end of a rope of negligible mass. Why is it, that when one of the two people creates a tension in the string, they meet at their center of mass?

Also, how my friend solved a problem similar to this one, was that he said since the one person is half as heavy, that person will travel twice as far; so, 2x + x = 600 m.

Why do both of these methods work? Could someone fill in the reasoning behind each method, please? I'd like to understand why both work.

 

[Doc Al]

Ask yourself: How can you calculate the location of the center of mass? Can the location of the center of mass change?

 

[Bashyboy]

It would seem like if the two people approached it other, then the center of mass would change.

 

[Doc Al]

It would seem like if the two people approached it other, then the center of mass would change.

Consider it more generally. Imagine a system at rest. For the center of mass to start moving, what is required?

 

[Bashyboy]

Oooh, would it be that one of the people has to start moving, and only one?

 

[Doc Al]

Oooh, would it be that one of the people has to start moving, and only one?

Well, that would do it. But it's not what I was going for. What is required for a system to accelerate?

 

[Bashyboy]

It would be a force.

 

[Doc Al]

It would be a force.

Right. In order for the center of mass of a system to accelerate, there needs to be a net force on the system. An external force. But here there are no such forces available, since the ice is frictionless.

 

[Bashyboy]

Hmm, okay, I see. So, how can I apply that knowledge to the two different methods of solving a problem like this, to understand the reasoning behind them?

 

[Doc Al]

Hmm, okay, I see. So, how can I apply that knowledge to the two different methods of solving a problem like this, to understand the reasoning behind them?

Well, the first method is already done. Since there are no external forces acting, the only place they can meet is at the center of mass.

As far as the second method goes: If the 100 kg person moved 1 m to the left, how far must the 50 kg person move to keep the center of mass unchanged?

 

[Bashyboy]

I would guess two meters. But how can I justify that?

 

[Doc Al]

I would guess two meters. But how can I justify that?

What's the definition of the center of mass? How would you calculate it?

 

[Bashyboy]

For one dimensional motion: $x_{cm}= \Large\frac{m_1x_1+m_2x_2+m_3x_3+ \cdots m_nx_n}{m_1+m_2+m_3+ \cdots m_n}$. And the center of mass would be constant?

 

[Doc Al]

For one dimensional motion: $x_{cm}= \Large\frac{m_1x_1+m_2x_2+m_3x_3+ \cdots m_nx_n}{m_1+m_2+m_3+ \cdots m_n}$. And the center of mass would be constant?

Good. Now write it in terms of Δx for two masses. Then you can relate Δx₁ to Δx₂.

 

[Bashyboy]

Oh, so if I know the center of mass, and I know the change in position of one of the people, then I can find the corresponding change in position of the other person, so that the center of mass remains the same, is that right?

 

[Doc Al]

Oh, so if I know the center of mass, and I know the change in position of one of the people, then I can find the corresponding change in position of the other person, so that the center of mass remains the same, is that right?

Exactly.

 

[TurtleMeister]

You can also think about it in the form of Newton's laws of motion:

$$a_1m_1=F=-a_2m_2$$

If $m_1$ is twice the mass of $m_2$, then $a_2$ must be twice the acceleration of $a_1$ (but in the opposite direction).

 

[Bashyboy]

Okay, I believe I understand. Thank you both for your insight.