Mass Moment of inertia of a cylinder with 4 holes

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The discussion focuses on calculating the mass moment of inertia for a cylinder with four symmetrically drilled holes. The user outlines their approach, which includes converting weight density to mass density, calculating the mass of the solid cylinder and the holes, and applying the parallel axis theorem for the holes. A suggestion is made to retain the specific weight until the final calculation to enhance accuracy. Overall, the proposed method is validated as correct, encouraging the user to proceed with their calculations.
qpham26
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Hi I was reviewing for my final and I came across this problem:

Problem:
Basically there is a 6in long cylinder with dia = 24in
Given weight density: 490 lb/ft^3

Each Hole is drilled symmetrically, each has 6in dia and equally space around a 10in dia circle concentric with the cylinder.

This is the picture of the frontview of the cylinder: https://lh4.googleusercontent.com/-8m1r4npC7cg/T11OqGF6qhI/AAAAAAAAABY/4v2HCrEqhd4/s333/cyinder.png
Approach
So what I would do is:
-First get that weight density into mass density by dividing it by 32.2
-Then find the mass of the cylinder without holes.
-Find mass of each holes.
-calculate the M of I of each hole about the center axis (parallel axis theorem)
-calculate the M of I of the whole solid cylinder (no holes)
-subtract the M of I of the holes from the whole cylinder

Will I get the correct answer based on the technique above?
Is there any mistake?

Thanks for your time and consideration.
Appreciate it!
 
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What you are given is a specific weight, 490 lb/ft^3, and I would suggest that you work with that value until the last step, converting to mass only in the last step. This will preserve a little bit of accuracy (less rounding).

Otherwise, your procedure is fine; go to it!
 
yes,,,, go ahead
 
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