Moment of inertia of a leftover square

In summary, the homework statement is that if you take the only non-overlapping configuration of circles (in each corner) then the MI of the leftover square should be I(1)-4I(2). However, the student is getting the wrong value every time.
  • #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?
 
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  • #2
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
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
 
  • #4
Krushnaraj Pandya said:
Ma^2/6
The sides are length 2a, not a.
 
  • #5
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.
 
  • #6
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.
 
  • #7
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.
 

FAQ: Moment of inertia of a leftover square

What is the moment of inertia of a leftover square?

The moment of inertia of a leftover square is a measure of its resistance to rotational motion about an axis passing through its center of mass. It is affected by the mass distribution and shape of the square.

How is the moment of inertia of a leftover square calculated?

The moment of inertia of a leftover square can be calculated using the formula I = (1/12) * m * s^2, where I is the moment of inertia, m is the mass of the square, and s is the length of one side of the square.

What is the significance of the moment of inertia of a leftover square?

The moment of inertia of a leftover square is important in understanding the rotational behavior of the square. It can help determine how easily the square will rotate and how much torque is required to change its rotational motion.

How does the moment of inertia of a leftover square differ from other shapes?

The moment of inertia of a leftover square is unique to its shape and mass distribution. Other shapes, such as circles or triangles, have different moments of inertia due to their varying mass distributions and distances from the axis of rotation.

Can the moment of inertia of a leftover square be changed?

Yes, the moment of inertia of a leftover square can be changed by altering its mass distribution or its shape. For example, adding weight to one side or changing the square into a rectangle will affect its moment of inertia.

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