Moment of inertia of a leftover square

  • #1
Krushnaraj Pandya
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Homework Statement


A square plate is of mass M and length of edge 2a. Its M.I about its centre of mass axis, perpendicular to its plane is equal to I(1). Four identical disks of diameter a are cut from the plane. The MI of leftover square about the same axis?

Homework Equations


1) MI of square plate of edge length a- Ma^2/6
2) MI of disk of radius R- MR^2/2
3) I about a point P, x distance from COM is I about COM + Mx^2

The Attempt at a Solution


I first calculated MI of original square plate using formula 1 as (2/3)Ma^2=I(1). Mass of one disc as (pi/16)M=M(d). MI of one disc about the COM of disk=M(d)a^2/8=I(disc). MI of one disk about CM of square=I(disc)+(M(d)*a^2)/2=I(2). The MI of leftover square should be I(1)-4I(2) but I'm getting the wrong value every time, can you point out my mistake and write out the calculations in case I'm making a mistake somewhere?
 

Answers and Replies

  • #2
Nathanael
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Assuming the plate is uniform and we take the only non-overlapping configuration of circles (in each corner) then I see no mistakes in your post.
 
  • #3
Krushnaraj Pandya
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Assuming the plate is uniform and we take the only non-overlapping configuration of circles (in each corner) then I see no mistakes in your post.
I must be making a calculation mistake somewhere then...I'll give it a go again. Thanks a lot for the help, I'm very grateful
 
  • #5
Nathanael
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The sides are length 2a, not a.
He prefaced that with “for edge length a.” (A bit confusing, I agree. He should’ve used an L or something.)

If you look into his calculation though, he does it right:
... using formula 1 as (2/3)Ma^2=I(1).

Every other step also seemed correct, but maybe I missed something.
 
  • #7
Krushnaraj Pandya
Gold Member
697
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Please post the remainder of your calculations.

Ah yes. Thanks.
He prefaced that with “for edge length a.” (A bit confusing, I agree. He should’ve used an L or something.)

If you look into his calculation though, he does it right:


Every other step also seemed correct, but maybe I missed something.
Sorry, I forgot to post yesterday- since I solved it late at night. I was indeed making a silly calculation mistake over and over again but the entire method and the calculations I posted here are correct, I verified this by cross-checking my answer with the one given in my textbook. I'm sorry for wasting your time and am really grateful for your help, Thank you.
 

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