Displacement of Center of Mass Problem

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Homework Help Overview

The problem involves calculating the displacement of the center of mass of a modified square sandwich after a square section is removed. The original poster presents a scenario where a square of side 2L is removed from a square of side L, leading to questions about the resulting center of mass and its displacement.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the center of mass using mass distribution and coordinates but questions the reasoning behind the distance off from the center being L/4. Other participants raise concerns about the feasibility of removing a square larger than the original and suggest that the distance might actually be L/2. There is also a reference to the importance of visual aids, such as sketches, in understanding the problem.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem setup and questioning the assumptions made in the calculations. Some guidance has been offered regarding the division of the remaining area and the positioning of centers, but no consensus has been reached.

Contextual Notes

Participants are grappling with the implications of the dimensions involved in the problem, particularly regarding the removal of a square and the resulting geometry. There is a reference to external documentation that may provide additional context.

halo168
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1. The problem statement, all variables, and given/known data
A square of side 2 L is removed from one corner of a square sandwich that has sides of length L. The center of mass of the remainder of the sandwich moves from C to C’. The displacement of the y coordinate of the center of mass (from C to C’) is:

Homework Equations


xcm=Σmx/Σm

The Attempt at a Solution


M/4 = mass of the quarter of the main square that is off center
L/4 = distance off from center
3M/4= total mass of the new object

Xcm = (M/4)(L/4)/(3M/4)=(1/12)L
Ycm = (M/4)(L/4)/(3M/4)=(1/12)L

Therefore, the displacement is sqrt(2)/12 by Pythagorean Theorem. I'm not sure why L/4 = distance off from center (what point is L/4 off from the center and how is it found?), but it's the only one that works. Can someone please explain?
 
Last edited:
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How can you remove a square with longer sides than the original, full square? Based on your calculations I guess it should be L/2?
halo168 said:
I'm not sure why L/4 = distance off from center (what point is L/4 off from the center and how is it found?), but it's the only one that works.
Did you draw a sketch and mark the centers?
 
Divide the remaining area into three squares. It should be obvious that their centers are L/4 to the left/right/top/bottom relative to the center.
 

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