Friction free tug of war - result check

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Two skaters, one weighing 65 kg and the other 42 kg, are on a 9.7 m pole and pull themselves towards each other. The center of mass is calculated to be 3.8 m from the 65 kg skater. Since there are no external forces acting on the system, the center of mass remains stationary. Therefore, the distance the 42 kg skater moves can be determined by subtracting the center of mass distance from the total length of the pole. The conclusion confirms that the skaters will meet at the calculated positions based on their masses.
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Homework Statement


Okay left my book at home - foolishly thinking I had all my equations - oops

Two Skaters, one with mass 65kg and the other with mass 42kg stand on an ice rink holding a pole with a length of 9.7m and a mass that is negligible. Starting from the ends of the pole the scaters pull themselves along the pole until they meet. How far will the 42kg skater move?


Homework Equations



?

& \vec{}Rcm = \Sigma mi\vec{}R/mi


The Attempt at a Solution


Okay I've determined that the center of mass is 3.8m from the 65 kg mass.
Am I correct in thinking that 9.7-3.8 is the answer? I don't have a place to run through a sample and couldn't find one here.
 
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JWDavid said:
Two Skaters, one with mass 65kg and the other with mass 42kg stand on an ice rink holding a pole with a length of 9.7m and a mass that is negligible. Starting from the ends of the pole the scaters pull themselves along the pole until they meet. How far will the 42kg skater move?
…
Okay I've determined that the center of mass is 3.8m from the 65 kg mass.
Am I correct in thinking that 9.7-3.8 is the answer? I don't have a place to run through a sample and couldn't find one here.

Hi JWDavid! :smile:

Yes, there's no external force, so the centre of mass doesn't move!
 
thanks
 

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