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    OK, can someone help me with this, because sometimes it looks right, and sometimes it doesn't.

    Consider a Hemisphere radius r and a Cylinder radius r and height h. Find the centre of mass of the hemisphere and cylinder (glued on top of each other). If this object was on a slope of thirty degrees, how big could h be before the object toppled (considering no tendency to slip)

    So for the first section, we know:

    Hemi: 3/8 r at 2/3 (pi)r^3m
    Cylin: 1/2 h at (pi)r^2hm

    Total mass: (pi)r^2m(2/3 r + h)

    So:

    (pi)r^2m(2/3 r + h) X = 2/3(pi)r^3m * 3r/8 + (pi)r^2h * 1/2 h

    which cancels to:

    X = 3/4 (r + h)

    Presumably the object topples when the 'vertical' line through the centre of mass lies outside the shape? But I could only find this in arbitary terms of f, the lengh of the perpendicular from a corner to the centre of mass at 3/4 (r + h) Which seems fairly pointless...
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    I'm guessing the point of tipping happens when the centre of mass is directly above the bottom edge (lowest point of the shape). Any further and the object tips. So at which value of h does this happen. I don't know how to do centre of mass questions (yet).
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    (Original post by mik1a)
    I'm guessing the point of tipping happens when the centre of mass is directly above the bottom edge (lowest point of the shape). Any further and the object tips. So at which value of h does this happen. I don't know how to do centre of mass questions (yet).
    Yeah thats what I was trying to say in my mushed up english. It will tip when the vertical line through the lowest point of the shape passes through the centre of mass.

    I figure, through simple trig, taking the length of the perp line to be an arbitary number, f you can get (f is hypotoneuse)

    fcos30 = 3/4 (r+h)
    fsin30 = r

    hence,

    fcos30 = 3/4 (fsin30 + h)
    4fcos30 = 3fsin30 + 3h
    h = 4/3 fcos30 - fsin30
    h = 4root3/6 - 1/2 f

    but this doesn't actually sell me anything, only that if I know the distance from lowest corner to centre of mass it will be so high. Plus this should work for those lines who do not pass through the CofG as well...

    Cheers anyway Mik1a
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    Can you not just model it in 2d, because it is symmetrical?
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    Yeah, thats kinda what I've done. On your diagram the hypotenuse of the triange is my arbitary length f. You cna find h in terms of f because you can replace r (the base) with fsin30.
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    I can do this, but I can't explain it without attaching a diagram. But I'm new and don't know how.
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    Follow link for diagram (I'm not great at drawing):

    http://img78.photobucket.com/albums/...he/moments.bmp

    In the diagram, the angle shaded blue is 30 degrees. The tan of this angle is opposite (green line) over adjacent (brown line).

    Now, you know that the green line is r, and the brown line is 3/4 (r+h)

    So tan 30 = r/ (3/4 (r+h))

    Which you can rearrange to get h
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    Hmm, I'd not thought about it that way, here I was trying to resolve f in the most complicated fashion, when I could use tan and get it in terms of r. Thanks, preumably both these methods work -- as, after all, if we knew length f we could work out height h, not that it necessarily makes any sense
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    Presumably most methods will work, but mine is definitely easier.
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    I'm afraid you took the moments wrong!
    When you take moments of the masses about the base, the cog of the hemisphere is 3r/8 from the flat bit plus h from the base of the cylinder.
    If you think about it, imagine the hemisphere atop the cylinder is instead another cylinder, of base radius = r and height = r. Then you will have a composite cylinder, of base radius = r and height = (h+r). The cog of this cylinder is ½(h+r), which is less than your value of X=(3/4)(r+h) when in fact it should be greater since there is now more mass at the top end.

    So, taking moments about the base,

    pi.r²(2r/3 + h).X = (2/3)pi.r³(h + 3r/8) + pi.r²h(h/2)
    (2r/3 + h).X = (2r/3)(h + 3r/8) + h²/2
    (2r/3 + h).X = 2rh/3 + r²/4 + h²/2
    (1/3)(2r + 3h).X = (1/12)(3r² + 8rh + 6h²)
    X = (3r² + 8rh + 6h²)/[4(2r + 3h)]
    =======================

    You can use tan30 = r/X

    1/√3 = 4r(2r + 3h)/(3r² + 8rh + 6h²)
    3r² + 8rh + 6h² = 4√(3).r(2r+3h)
    3r² + 8rh + 6h² = 8√(3).r² + 12√(3).rh
    6h² + (8r - 12√(3).r)h + (3r² - 8√(3).r²) = 0

    h = (12√3 - 8)r ± √{(12√3 - 8)² - 4.6.(3 - 8√3)} / 12
    h = (12√3 - 8)r + r.√(424) /12
    h = (12.785r + 20.591r)/12
    h = 33.756r/12
    h = 2.78r
    ======

    and since X = √3.r

    X = √3*h/2.78
    X = 0.623h
    =======
 
 
 
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