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.
I'm assuming that the hemisphere is glued to the bottom of the cylinder.
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"?
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.
Ah, no, part c) has the same answer with part b)
h = r√3. (if righting means keeping its position)
Yes Yes YES! BCHL85 that is what I got. Surely in the Centre of Mass bit all the pi's should cancel Fermat?