Moment of inertia of a leftover square

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Homework Help Overview

The problem involves calculating the moment of inertia (M.I) of a leftover square plate after cutting out four identical disks from it. The original square plate has a mass M and an edge length of 2a, and the discussion centers around the application of formulas for M.I of both the square and the disks.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of the M.I for the original square plate and the disks, questioning the correctness of the original poster's calculations and assumptions about the configuration of the disks.

Discussion Status

Some participants have offered guidance by confirming the original poster's method appears correct, while others have pointed out potential confusion regarding the dimensions used in the calculations. There is an ongoing exploration of the calculations and assumptions involved.

Contextual Notes

Participants note that the original poster may have made a calculation mistake, and there is a discussion about the uniformity of the plate and the configuration of the disks being cut out. The dimensions of the square plate are clarified as being 2a rather than a.

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?
 
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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.
 
Nathanael said:
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
 
Krushnaraj Pandya said:
Ma^2/6
The sides are length 2a, not a.
 
haruspex said:
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:
Krushnaraj Pandya said:
... using formula 1 as (2/3)Ma^2=I(1).

Every other step also seemed correct, but maybe I missed something.
 
Krushnaraj Pandya said:
but I'm getting the wrong value every time,
Please post the remainder of your calculations.
Nathanael said:
He prefaced that with “for edge length a.”
Ah yes. Thanks.
 
haruspex said:
Please post the remainder of your calculations.

Ah yes. Thanks.
Nathanael said:
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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