Center of Mass and Translational Motion

[mmravunac]

Suppose two friends, with masses m2 = 1.3 m1, are on a perfectly smooth, frictionless, frozen lake.
They are both holding the end of a rope of length Lo .
a. Find the position of the center of mass, in terms
of Lo , from the smaller person.
b. If the two pull half of the rope in such that the final length of the rope is L = Lo/2, find the new position of the center of mass from the smaller person.
c. Find the distance each person moves from their original positions.

Homework Equations

3. I've got no clue

 

[haruspex]

What is the definition of centre of mass?

 

[mmravunac]

I am not sure of what exactly you are asking but center of mass is equal to

Xcm= (m1X1 + m2X2) / (m1 + m2)

 

[ehild]

You have to find the distance of the centre of mass from the smaller person. So put the origin at the smaller person. What are X1, X2 then?

ehild

 

[mmravunac]

I believe X1 would be 0 and that would make X2 the whole length making it equal to Lo?

 

[ehild]

I believe X1 would be 0 and that would make X2 the whole length making it equal to Lo?

Yes. So where is the centre of mass?

ehild

 

[mmravunac]

This is that part of the equation that has been tricking me up.

I want to say that
Xcm=(m2X2) / (m1+m2)

So I believe I can take out m2 from each side of the equation giving me
Xcm=(X2)/(m1) (?)

 

[HallsofIvy]

That's just bad arithmetic: 5/(6+ 5) is NOT equal to 1/6.

 

[mmravunac]

So I cannot cancel out the m2.

I am not positive where to go on from Xcm=(m2X2) / (m1+m2)
Would this be my final equation or is there more I can do?

 

[mfb]

That is your final equation for the center of mass, you cannot simplify it any more.

Edit: Oh right. You can use your given ratio for the masses, of course.

 

[ehild]

So I cannot cancel out the m2.

I am not positive where to go on from Xcm=(m2X2) / (m1+m2)
Would this be my final equation or is there more I can do?

X2=Lo and m2=1.3 m1. So what is the position of the CM in terms of Lo?

 

[mmravunac]

I think I have it now.
Xcm= (1.3m1*Lo) / 2.3m1

Xcm=.57Lo (which makes this the distance from the smaller person) ?

And for the second part, would this mean i set it up like

Xcm=(1.3m1*.5L) / 2.3m1

 

[haruspex]

I think I have it now.
Xcm= (1.3m1*Lo) / 2.3m1

Xcm=.57Lo (which makes this the distance from the smaller person) ?

and for the second question of this problem would I just need to divide both sides by 2 since L=Lo/2 (?)

yes.