Question

How to proof that the centre of gravity of a cone is a quarter of the height

Answer 1

Suppose the cylindrical symmetry of the problem to note that the center of mass must lie along the z axis (x = y = 0). The only issue is how high does it lie.

If the uniform density of the cone is
ρ
, then first compute the mass of the cone. If we slice the cone into circular disks of area
pi
r^2 and height dz, the mass is given by the integral:

M=∫ρdV=ρ∫0hpir2dz


However, we know that the radius r starts at a for z = 0, and goes linearly to zero when z = h. This means that r = a(1 -z/h), so that:

M=ρ∫0hpia2(1−z/h)2dz=pia2ρ∫0h(1−2z/h+z2/h2)dz=1/3pia2hρ


Noe that this is simply indicates that the volume of the cone is given by
V=1/3pia2h
.
To find the height of the center of mass, we then compute:
zcm=1/M∫ρzdV=ρ/M∫0hpir2zdz=pia2ρ/M∫0h(1−z/h)2zdz


=pia2ρ/M∫0h(z−2z2/h+z3/h)dz=1/12pia2h2rh÷M=1/4h


As a result, the center of mass of the cone is along the symmetry axis, one quarter of the way up from the base to the tip