Center of Mass?

Question

Here is the question:

Find the ' y ' component of the center of mass for a uniform solid cone of radius ' R ' and height ' h '. The answer should be expressed as a (numeric) fraction of ' h ', measured from the base of the cone.

If someone could show me a step by step process on how to conquer this problem, it would really help.

Answer 1

If the cone is uniform (and 'right' -- meaning it is a cone where the vertical axis is perpendicular to the circular base), then the centre of mass is somewhere on the vertical axis.

Assuming that the density is uniform, then the height of the centre of mass will be where there is an equal quantity of mass below the centre as there is above. Because density is uniform, this will be where the volume above the centre of gravity is exactly half the volume below the centre.

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Total volume of the cone =
(1/3)*pi*R^2*h

R varies as h.

For example, if you are to slice the cone halfway up (at h/2), then the radius at that height would be half of the Radius at the bottom (R/2)

Volume of 'top half' = (1/3)*pi*(R/2)^2*(h/2)
Volume of top half = (1/3)*pi*R^2*h*(1/8) which is one eight of the total volume.

We want to have the 'top portion' to be half in volume. Therefore the height from the top has to be h divided by the cube root of 2 (almost 1.26). Then the new radius will be R divided by the cube root of 2.

Volume of the top portion = (1/3)*pi*[(R/CubeRoot(2))^2]*[h/CubeRoot(2)] = half of the total volume.

So the centre of mass is on the vertical axis, and located at h/CubeRoot(2) from the top.

Beware, this is the new h measured from the top; the question asks for the new h measured from the bottom
new height = h - h/CubeRoot(2) = h(1-1/CubeRoot(2))
= [(CubeRoot(2) -1) / CubeRoot(2)]*h