The Student Room Group

Mechanics - Centres of mass problem

I have this question:
60DEBCDA-F36E-4B3E-87F3-5A36F10FCD1B.jpg.jpeg

I was able to successfully show the centre of mass of the uniform lamina is as stated.

For part (b) since both squares lie on the same line y=41/13 they must be spaced symmetrically from the centre of mass of the lamina for the centre of mass to remain constant (they have the same ‘mass’ - same area) so by symmetry I did this for part (c):
ABA1207F-548D-4FDF-8C77-5D7A51B2D610.jpg.jpeg

However the textbook answer states otherwise and the solution bank doesn’t provide any solution
I tried to recalculate the centre of mass by considering the uniform lamina as a particle at its original centre of mass and the two cut outs as negative masses:
EC16DECC-F2F8-4DCC-8679-541D66DF226B.jpg.jpeg
This method also yielded the same result for a.
I don’t know where I’m going wrong because intuition tells me the two squares are cut down symmetrically from the original centre of mass so the centre of mass shouldn’t change...

Any help in clearing this misunderstanding would be appreciated, thanks in advance.
Original post by BrandonS15
I have this question:

I was able to successfully show the centre of mass of the uniform lamina is as stated.

For part (b) since both squares lie on the same line y=41/13 they must be spaced symmetrically from the centre of mass of the lamina for the centre of mass to remain constant (they have the same ‘mass’ - same area) so by symmetry I did this for part (c):

However the textbook answer states otherwise and the solution bank doesn’t provide any solution
I tried to recalculate the centre of mass by considering the uniform lamina as a particle at its original centre of mass and the two cut outs as negative masses:
This method also yielded the same result for a.
I don’t know where I’m going wrong because intuition tells me the two squares are cut down symmetrically from the original centre of mass so the centre of mass shouldn’t change...

Any help in clearing this misunderstanding would be appreciated, thanks in advance.


I don't have time to look at this right now, but be careful so that you dont count overlapping area between the two squares twice.

Also, draw a nice sketch because just from looking at the numbers, it seems to me that this is a pretty weird question since you cannot cut out a proper square centred at (5,41/13) with side length 3 without part of this square lying OUTSIDE the lamina in the first place. [i.e.]
(edited 4 years ago)
Reply 2
Original post by BrandonS15
I have this question:
60DEBCDA-F36E-4B3E-87F3-5A36F10FCD1B.jpg.jpeg

I was able to successfully show the centre of mass of the uniform lamina is as stated.

For part (b) since both squares lie on the same line y=41/13 they must be spaced symmetrically from the centre of mass of the lamina for the centre of mass to remain constant (they have the same ‘mass’ - same area) so by symmetry I did this for part (c):
ABA1207F-548D-4FDF-8C77-5D7A51B2D610.jpg.jpeg

However the textbook answer states otherwise and the solution bank doesn’t provide any solution
I tried to recalculate the centre of mass by considering the uniform lamina as a particle at its original centre of mass and the two cut outs as negative masses:
EC16DECC-F2F8-4DCC-8679-541D66DF226B.jpg.jpeg
This method also yielded the same result for a.
I don’t know where I’m going wrong because intuition tells me the two squares are cut down symmetrically from the original centre of mass so the centre of mass shouldn’t change...

Any help in clearing this misunderstanding would be appreciated, thanks in advance.


As RDKGames says, be careful about counting area outside the card and also overlapping areas as the stated center for the 3*3 square is close to the COM and close to the edge of the card.
What did the book give as an answer?
Original post by BrandonS15

Any help in clearing this misunderstanding would be appreciated, thanks in advance.


Based on the wording I would say there is an error in the question.

"Show that the CofM is at ...." and "needs to be changed so that CofM lies at ...same thing" doesn't sound right; I'd expect the new CofM to be different.

Also parts b,c together are awarded as many marks at part a alone, so your simplistic (and correct, IMO - but see next paragraph) working seems too easy.

Further, if you're going to cut out two squares, then they can't overlap or go outside the boundary, as that wouldn't be cutting out two squares.

Question is ****ed!
And for info, here are the alleged answers:

How on earth they can get "9" in the denominator....

And the answer to part b is ****e also.

Untitled.jpg
(edited 4 years ago)
Reply 5
Original post by ghostwalker
And for info, here are the alleged answers:

How on earth they can get "9" in the denominator....

And the answer to part b is ****e also.

Untitled.jpg


Get my argument right ... both of those 3*3 "squares" would be cutting more away above the y-COM than below it so it must change in the y-direction. The x-COM is also suspect ...

Also the answer saying the centers must be placed symmetrically around the y-COM? Must be symmetric about the x-COM (assuming they are really squares)?
(edited 4 years ago)
Original post by mqb2766
Get my argument right ... both of those 3*3 "squares" would be cutting more away above the y-COM than below it so it must change in the y-direction. The x-COM is also suspect ...

Also the answer saying the centers must be placed symmetrically around the y-COM? Must be symmetric about the x-COM (assuming they are really squares)?


Yep, another great <expletive deleted> (in best Watergate style) by Pearson.

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