How Do You Solve a Center of Mass Problem on Frictionless Ice?

[schern42]

Here is the problem that I'm faced with:
A 52 kg woman and a 80 kg man stand 12.0 m apart on frictionless ice.

(a) How far from the man is their center of mass (CM)?
(b) If they hold onto the two ends of a rope, and the man pulls on the rope so that he moves 2.8 m, then how far from the woman will he be?
(c) How far will the man have moved when he collides with the woman?


I understand how to do (A). Using the equation for location of center of mass, x_{cm} = \frac{1}{M} \sum_{i=1}^n m_i x_i, I found that the man is 4.73 m from the CM.

Now I'm trying to do (B) and realize that I don't even know how to approach it at all. When the man pulls on the rope, does that mean that the woman will also move? I don't know how to take this into account to solve the problem.

I appreciate any help.

 

[Andrew Mason]

Here is the problem that I'm faced with:
A 52 kg woman and a 80 kg man stand 12.0 m apart on frictionless ice.

(a) How far from the man is their center of mass (CM)?
(b) If they hold onto the two ends of a rope, and the man pulls on the rope so that he moves 2.8 m, then how far from the woman will he be?
(c) How far will the man have moved when he collides with the woman?


I understand how to do (A). Using the equation for location of center of mass, x_cm=\frac{1}{M} \sum_{i=1}^n m_i x_i, I found that the man is 4.73 m from the CM.

Now I'm trying to do (B) and realize that I don't even know how to approach it at all. When the man pulls on the rope, does that mean that the woman will also move? I don't know how to take this into account to solve the problem.

I appreciate any help.

Is there any external force on the man, woman and rope? Is it possible for the centre of mass to move if there is no external force?

AM

 

[schern42]

I guess the center of mass remains the same, but how can I use this to calculate the distance between the man and the woman after the man pulls on the rope?

 

[arildno]

Remember:
The position of the C.M remains UNCHANGED at ALL times!

It is conveniently to regard C.M as your origin, and denote the man's and woman's positions respectively with regard to this origin.

Remember that the man&woman may well move relative to the C.M, even if the C.M remains at rest relative to the ground.

 

[schern42]

In the end, a friend gave me the formula to calculate the solution, but did not explain as to how he acheived the equation. Would anyone be kind enough to highlight how he got to this point:
\frac{M_{man} X}{M_{woman}}<br />
And then:
12 - ans - X = d

 

[Andrew Mason]

In the end, a friend gave me the formula to calculate the solution, but did not explain as to how he acheived the equation. Would anyone be kind enough to highlight how he got to this point:
\frac{M_{man} X}{M_{woman}}<br />
And then:
12 - ans - X = d

Placing the centre of mass is at the origin:

m_{m}\vec{x_m} + m_w\vec{x_w} = 0

\vec{x_m} = -\frac{m_w\vec{x_w}}{m_m}

This applies to all changes to positions by pulling on the rope. The centre of mass remains at the origin.

AM