Moment of Inertia of a Solid Cylinder With a Wedge Removed

AI Thread Summary
The discussion focuses on calculating the moment of inertia of a solid cylinder with a wedge removed, specifically questioning the correct axis for this calculation. Participants clarify whether the axis should be the z-axis through the origin or through the mass center. There is confusion regarding the formula used, particularly the expression r2=ρ2+z2, which is deemed incorrect for both axes. The conversation emphasizes the need for proper definitions and formulas to accurately determine the moment of inertia. Accurate calculations are essential for understanding the physical properties of the modified cylinder.
cedoty1989
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Homework Statement
Imagine a solid cylinder height h able to rotate vertically around z-axis (centered at x=0 y=0). There is a wedge cut out so that when looking down with z hat pointing out of the page there is an angle 2a formed such that their is symmetry with respect to reflection over the x axis (See picture). The question is to calculate the moment of inertia.
Relevant Equations
I=∫∫∫ dm r^2 -> Cylindrical Coords: (r from 0 to R) (z from zero to h) (theta from -Pi + a to Pi-a)

Uniform density -> dm=(M/V)dV

dV = (dr)r(dtheta)(dz)

The rest of the equations are in the picture...
CamScanner 11-23-2020 21.50-1.jpg
 

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I know there should be an alpha in there somewhere but I cannot see where I'm going wrong. Thank you in advance.
 
cedoty1989 said:
The question is to calculate the moment of inertia.
About what axis? z axis through origin or through mass centre?
Your r2=ρ2+z2 doesn't make sense for either.
 

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