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Mat666
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#1
Report Thread starter 14 years ago
#1
Could someone please work out the following question with method so I can check I've got it right? Cheers, Mat.

A solid cone has a Solid hemisphere glued to its base. The cone is of vertical height h, radius r, and the hemisphere is also radius r. The axis AB runs through the centre of the object.

a) Find the position of the centre of gravity of the whole object.
b) Find h (in terms of r) if the object will right itself from the axis AB being horizontal.
c) Find h if the object will right itself from wherever it is placed.
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shift3
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#2
Report 14 years ago
#2
I'm assuming that the hemisphere is glued to the bottom of the cylinder.

a)
Taking A, the vertix of the cone, to be (0,0):
m1 = pi*r²*h
cm1 = (0,-0.5h)
m2 = (2/3)*pi*r³
cm2 = (0,-h - (3/8)r)

CM = (m1*cm1 + m2*cm2) / (m1 + m2)
CM = [pi*r²*h*-0.5h + (2/3)*pi*r³*(-h - (3/8)r)] / [pi*r²*h + (2/3)*pi*r³]
CM = -[(h²/2) - (2rh/3) - (r²/4)] / (3h + 2r)
CM = -(6h²+8rh+3r²)/4(3h+2r)
That's the y-coordinate of the center of mass from A. (The x-coord is 0.)

b) & c) What exactly do you mean by "righting"?
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Mat666
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#3
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#3
As in, if its laid on its side, then it will pop up to stand vertical. My answer for A) is the same I think but mine's positive. That's just an axis thing.
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Fermat
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#4
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#4
(Original post by shift3)
I'm assuming that the hemisphere is glued to the bottom of the cylinder.

a)
Taking A, the vertix of the cone, to be (0,0):
m1 = pi*r²*h
...
The volume of a cone is (1/3)*pi*r²*h

Anyway, I got cm = (8r² - pi.h²)/(6pi.r + 3pi.h) from the base of the cone - inwards to the hemisphere.

and for part b, the condition for righting was,
h² < 8r²/pi
========

For part c) is that question right - The way I tried it, it was impossible. If the object lay on it's cone slant side, then it would never right itself !!!
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BCHL85
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#5
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#5
Having the table:
Cone Hemisphere Object
Distance from A 3h/4 3r/8 + h x
Mass ~ volume pi*r²h/3 (2/3)*pi*r³ (pi*r²/3)(h +2r)Taking moment about A
(pi*r²/3)(h+ 2r)*x = (pi*r²/3)(3h²/4 + 3r²/4 + 2hr)
So x = (3h²+ 3r²+ 8hr)/4(h +2r)

b, The object is righting (it will pop up to stand vertical)
x = h ( the centre of mass lines in the base)
So 4h² + 8hr = 3h² + 3r² +8hr
h² = 3r²
h = r√3

c, No value for h.
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BCHL85
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#6
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#6
Ah, no, part c) has the same answer with part b)
h = r√3. (if righting means keeping its position)
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Mat666
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#7
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#7
Yes Yes YES! BCHL85 that is what I got. Surely in the Centre of Mass bit all the pi's should cancel Fermat?
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Fermat
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#8
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#8
(Original post by Mat666)
Yes Yes YES! BCHL85 that is what I got. Surely in the Centre of Mass bit all the pi's should cancel Fermat?
My apologies. I used cog for laminae, not solids
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shift3
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#9
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#9
Oh, it was a cone... For some reason I thought it was a cylinder.
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