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    I am having real problems. I can't cope with a square that has a line of symmetry diagonally, and a small square near the bottom right.

    This is e and f from M2 (heinemann) - ex 2B - Q2

    Someone please do this for me and explain WHY!!!

    Heinemann has no example I can follow, except some crap about an earring. They need to make mathematics more accessible, and verbose...

    Please help somebody, if not with the question, outline a method for me to follow in this case! For both 1 and 2 holes in the lamina!

    Thanks, Robbie.
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    I don't have the question, so I can only give general advice. But take moments from one edge. You need to consider it in separate shapes: the lamina, the bit cut out, and the large square without any bits cut out. The moment of the lamina and the bit cut out will equal the moment of the large square.
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    That kinda does help, but if anyone has the question, please!
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    Can you post the question on here? I think most people have given their textbooks back for the summer. Or I can go and dig out my answer, if I can find it
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    Can you get stuck on question 8 onwards please? I've found my answers to those. But we did Q2 in rough, so I haven't kept my answers.
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    (Original post by RobbieC)
    I am having real problems. I can't cope with a square that has a line of symmetry diagonally, and a small square near the bottom right.

    This is e and f from M2 (heinemann) - ex 2B - Q2

    Someone please do this for me and explain WHY!!!

    Heinemann has no example I can follow, except some crap about an earring. They need to make mathematics more accessible, and verbose...

    Please help somebody, if not with the question, outline a method for me to follow in this case! For both 1 and 2 holes in the lamina!

    Thanks, Robbie.
    Just break up the object into the respective parts with the centres of mass. I usually draw up a table.

    E)

    Mass 15 1 ] 16
    x x 2.5 ] 2
    y y 1.5 ] 2

    15x + 2.5 = 32
    15x = 29.5
    x = 1.97.

    15y + 1.5 = 32
    15y =30.5
    y=2.03

    It's just an example of working backwards. If you use the mass/centre of mass of a shape without the whole and use the mass/centre of mass of the hole with the mass of the shape without the whole you can calculate the coordinates of the mass without the whole.
    It's an example of working backwards.
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    (Original post by juno_the)
    Can you get stuck on question 8 onwards please? I've found my answers to those. But we did Q2 in rough, so I haven't kept my answers.
    Stuck on 15 now!!! Lol..

    Need to do it for absolute completion of the exercise! Post the solution if you can.

    Thanks for help everyone, but I had a breakthrough today and did alright with it all now!

    Except damn 15!
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    (Original post by RobbieC)
    Stuck on 15 now!!! Lol..

    Need to do it for absolute completion of the exercise! Post the solution if you can.

    Thanks for help everyone, but I had a breakthrough today and did alright with it all now!

    Except damn 15!
    It's been a while since my M2 days, but I'll give it a shot...

    The centre of mass (CoM) of the semicircle lies on the axis of symmetry, at a distance of (2r.sin(alpha))/(3.alpha) from A, where r is the radius of the semicircle and 2.alpha is the angle at the centre. This formula is given to you on page 57.

    2.alpha is pi radians, so alpha is pi/2; sin(alpha) is therefore 1, so the CoM is 4r/(3.pi) from A, along the axis of symmetry. If you define the bottom-left corner of the circle to be at (0, 0), then the CoM has coordinates (r, 4r/(3.pi))

    The CoM of the particle attached to the lamina is obviously at B, which has coordinates (r, r).

    The CoM of the whole system is halfway between A and B, so it is at (r, ½r).

    So (M x 4r/(3.pi)) + mr = ½(M + m)r

    Do a bit of rearranging, and you get:

    M(1 - 8/(3.pi)) = m

    which will give you the ratio of M:m as

    1: 1 - 8/(3.pi)
    or 3pi : 3pi - 8
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    OH! I thought I had to add the load to the arched side of the semicricle and then treat it as 6.2 something times the mass of the flat side. I didn't know it had given the COM for the loaded semicircle!

    *Is ashamed*

    Once again, thanks for the save Squish!
 
 
 
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