Mass Moment of inertia of a cylinder with 4 holes

  • Thread starter qpham26
  • Start date
  • #1
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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!!!
 

Answers and Replies

  • #2
OldEngr63
Gold Member
732
51
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!
 
  • #3
4
0
yes,,,, go ahead
 

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