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Where the centre of gravity of solid hemisphere of radius 'a'?
Wiki User
∙ 14y ago
This is incorrect. The problem with the answer starts here:
"Also, the point will lie on the plane which divides the top and bottom such that the volumes are equal."
This would only be true if the cg's for both halves were the same distance from that plane. But they aren't.
The solution is pi*int( t*(a2-t2)dt,0 to a)/(volume of hemisphere=2/3*pi*a3)
pi's cancel.
cg at ([a2t2/2-t4/4],0,a)/(2/3*a3)
= (a4/4)/(2/3*a3)
= 3/8 a from the base.
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Whoops! I did not mean to wipe out the original answer, and am a bit surprised that I apparently have the power to do that. Sorry!
The original answer found the distance from the base to a plane that divided the hemisphere into two equal halves, at t= ~0.347a. That might be useful information but is not the cg.
Wiki User
∙ 14y ago
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