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Question: Calculate the centre of mass of a uniform, squarebased pyramid of height H and base length B (taking the centre of the base as the origin). Hence, or otherwise, derive an expression for the centre of mass of such a pyramid with a pyramidshaped hollow cavity (of height h and base length b) inside, as shown:
Key equation: $$CoM = \frac {\int z \, dm} {\int\, dm} = \frac {\int zρ \, dV} {\int\, dV}$$
Attempt: By symmetry, the CoM must lie on the line connecting the apex and the centre of the base, let's call this the z axis. By considering horizontal slices parallel to the base, of thickness dz and area xy, ##dV = xydz = \frac {B^2} {H^2} (H  z)^2 dz,## therefore
$$CoM = \frac {\int_0^H zρ \frac{B^2}{H^2} (H  z)^2 \, dz} {\int_0^H ρ \frac{B^2}{H^2} (H  z)^2 \, dz} = \frac {H} {4}$$
However I don't know how to get an expression for the hollow/cavity pyramid. I was told to treat the hollow as being a pyramid of negative mass but I'm unsure as to how to go from there.
Key equation: $$CoM = \frac {\int z \, dm} {\int\, dm} = \frac {\int zρ \, dV} {\int\, dV}$$
Attempt: By symmetry, the CoM must lie on the line connecting the apex and the centre of the base, let's call this the z axis. By considering horizontal slices parallel to the base, of thickness dz and area xy, ##dV = xydz = \frac {B^2} {H^2} (H  z)^2 dz,## therefore
$$CoM = \frac {\int_0^H zρ \frac{B^2}{H^2} (H  z)^2 \, dz} {\int_0^H ρ \frac{B^2}{H^2} (H  z)^2 \, dz} = \frac {H} {4}$$
However I don't know how to get an expression for the hollow/cavity pyramid. I was told to treat the hollow as being a pyramid of negative mass but I'm unsure as to how to go from there.
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