Determining Position and Center of Mass using Reference Frame

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Homework Statement



mKGZivf.png

KEFvArt.png


Homework Equations



Xcm = (x1m1+x2m2) / (m1+m2)

The Attempt at a Solution



ok so i'm having trouble understanding what's happening in the question... when the penguin moves, is the sled also moving? If it's friction less ice then if the penguins moves L, the sled will move -L. Is this right or am I just not understanding the question?

for the first one, Changed what I think, penguin will move 0 relative to itself, in the sled system it will move L so center of mass at L/2?

second one Changed what I think for this too, since the penguin moves 0 relative to itself and from the sled system it will move L, in the lab frame it moves L/2.

third one seems pretty simple. Since penguin at origin and the center of mass of sled is at L/2, the center of mass would be L/4.

fourth one sled frame moves -L, penguin relative to itself always 0 so the origin is -L/2

fifth one sled moves -L/2 so it's center of mass is at 0, the penguin moves L/2 so center of mass is L/4?

sixth one L/4

Are these right so far?
 
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Answers and Replies

  • #2
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Come to think of it, I say first one would be -3L/4 because the sled moves opposite direction and it's center of mass becomes -L/2.
 
  • #4
Mister T
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Homework Equations



Xcm = (x1m1+x2m2) / (m1+m2)
Can you simplify this equation?

ok so i'm having trouble understanding what's happening in the question... when the penguin moves, is the sled also moving? If it's friction less ice then if the penguins moves L, the sled will move -L. Is this right or am I just not understanding the question?
Are you using your equation to guide your thinking?
 
  • #5
gneill
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Since the COM of the whole system as viewed in the Lab frame doesn't move (right?), it can be convenient to use before and after sketches of each situation, label the resulting COM in the sled frame, then align that sled COM with the Lab frame's original COM. Then make your measurements/calculations from the resulting image.

For example, for part (2) the penguin moves from one end of the sled to the other:
Fig1.png
 
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Can you simplify this equation?



Are you using your equation to guide your thinking?
Yeah for each relevant question about Xcm I would substitute the variables accordingly.

Since the COM of the whole system as viewed in the Lab frame doesn't move (right?), it can be convenient to use before and after sketches of each situation, label the resulting COM in the sled frame, then align that sled COM with the Lab frame's original COM. Then make your measurements/calculations from the resulting image.

For example, for part (2) the penguin moves from one end of the sled to the other:
View attachment 91080
That is incredibly helpful, thanks a lot! I was confused how much the sled actually moves back in the lab frame but that pictures makes it crystal clear.

Using that, I'm pretty confident in these answers except number 6:
1) 3L/4
2) L/2
3) L/4
4) -L/2
5) L/4
6) either L/2 or L/4... It asks for the center of mass of the sled which is always L/2 but it asks for the sled frame. If I include the penguin the combined center of mass is L/4 but it doesn't seem to include the penguin so I'm going with L/2.
 
  • #7
gneill
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Looks good.
 
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  • #9
Mister T
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Yeah for each relevant question about Xcm I would substitute the variables accordingly.
Well, I was thinking of simplifying it before looking at each specific scenario.

Xcm = (x1m1+x2m2) / (m1+m2)
You let ##m=m_1=m_2## and get something that would allow you look at the situation in a different way. Even though you've solved the problem, it might help you understand it better in case you encounter similar situations in the future.
 

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