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Center of mass of cone, cylindrical coordinate system

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We are given a cone of height H and angle alpha with constant density. We want to calculate the center of mass using triple integrals in cylindrical coordinates. This requires a description of the solid in such coordinates and the use of the element of volume in the same system of coordinates:

a. Write the element of volume dV in a cylindrical coordinate system.

b. Calculate the Center of Mass of a cone with mass M and height H, and the angle at its head is alpha (or 2 alphas to cover the entire angle of the top). Assume constant density.

c. Now we remove the top from the cone, and its height is now h. What is the center of the mass now?

Please see the attached figures.

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Solution Summary

The center of mass of a cone and a truncated cone are calculated, assuming that they have constant density. The triple integrals are set up in cylindrical coordinates.

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Please see attachment for properly formatted copy. There is a bit more in c below compared to the pdf, so check it below.

a. The element of volume in cylindrical coordinates is dV=r dr d theta dz, where theta is the polar angle, r is the polar
radius (r=(x^2+y^2)^(1/2) for a point (x,y,z) in space) and z is the hight from the xy-plane.

b. The coordinates of the center of mass x', y', z' are given by the expressions

x'=(1/M)*(Triple integral over E of) x*rho dV,

y'=(1/M)*(Triple integral over E of) y*rho dV,

z'=(1/M)(Triple integral over E of) z*rho dV,

where rho is the (posible non-constant) density of the solid E, in this case a cone. We ...

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