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    Name:  image.png
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Size:  97.9 KBI'm trying to revise for M2 and I was doing the Jan 15 IAL paper when I saw this question. I have no idea how to do it and the mark scheme has no explanation, could someone help?
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    I also was unable to do this question. Any help would be greatly appreciated
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    Here is my solution

    ask if you don't understand it

    http://imgur.com/wyKE9BB

    (Original post by ctij)
    I also was unable to do this question. Any help would be greatly appreciated
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    (Original post by xyz9856)
    Here is my solution

    ask if you don't understand it

    http://imgur.com/wyKE9BB

    How have you assumed the the COM of the lamina is halfway up?

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    (Original post by ctij)
    How have you assumed the the COM of the lamina is halfway up?

    Thanks for the reply
    ok no problem

    it lies above A since the shape has symmetry

    I thought about this in a weird way, but it makes sense.

    1.) A rhombus is an equilateral parallelogram
    2.) Each rhombus has a line of symmetry, splitting it into two triangles, the COM lies in the centre of the line of symmetry.
    3.) The COM of each rhombus lies at the same point on the (y-axis), therefore the COM of the overall shape is also at this same point on the y axis
    4.) Calculating the whole length and halving it gives you the COM Y co-ord
 
 
 
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