Solving for the Centre of Mass in an Astronaut Space Walk

AI Thread Summary
Two astronauts, A and B, are tied together in space, and A pulls the rope, reducing its length. The center of mass of the system does not move because the forces involved are internal; A's pull is countered by the rope's tension. To change the center of mass's position, an external force must act on the system. The discussion emphasizes Newton's third law, where internal forces cancel out in terms of motion. Participants are encouraged to demonstrate their understanding before receiving guidance on solutions.
JenL
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Homework Statement


a) Two astronauts, A of total mass 90kg and B of total mass 110kg are tied together by a light rope 15m long during a space walk. A starts to reel in the rope at her end. After 8.5s she has reduced the length of the rope by 5.5m. What distance does the centre of mass move in this time? Explain your answer.

Homework Equations


x=md/(M+m)

The Attempt at a Solution


a) I thought that because A was pulling the rope the centre of mass would move towards A? As the astronauts are in outer space this is an isolated system but would the pulling of the rope by A be counted as an external force? Or is it an internal force?
 
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JenL said:
a) I thought that because A was pulling the rope the centre of mass would move towards A? As the astronauts are in outer space this is an isolated system but would the pulling of the rope by A be counted as an external force? Or is it an internal force?
It is definitely an internal force, originating inside the system. While A was pulling on the rope, the rope was pulling back :)
 
So if the rope pulls back does that mean it is not moving? Sorry I still do not fully get this
 
JenL said:
So if the rope pulls back does that mean it is not moving?
Indeed. The only way to change the motion of the center of mass of a system is for some external force to act on the system.
 
JenL said:
So if the rope pulls back does that mean it is not moving? Sorry I still do not fully get this
The way to see this is by considering Newton three: action = - reaction.

The center of mass has some coordinate for which (in a simple case)
(m1+ m2) x = m1 x1+ m2 x2
so (m1+ m2) v = m1 v1+ m2 v2
and (m1+ m2) a = m1 a1+ m2 a2
and with Newton two (F = ma), F21 the force by m2 on m1 and F12 the force by m1 on m2:
(m1+ m2) a = F21 + F12 which, according to Newton three is zero !

In short: internal forces cancel for the equations of motion of the center of mass. Cute, isn't it ?
 
So what is the answer ?
 
Nirrjhhar said:
So what is the answer ?
Hi Nirrjhhar, Welcome to PF.

Sorry, but we don't hand out homework results here. You have to show your effort at a solution, then we can point out where its right or wrong, offer advice and other help so that you can solve the problem yourself. See the forum guidelines for more information.
 
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