Calculating Moment of Inertia for Rectangular Plate | Center & Corner Axis Proof

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Homework Help Overview

The discussion revolves around calculating the moment of inertia for a rectangular plate with mass M and sides A and B, specifically focusing on two axes: one through the center of mass and another through a corner of the plate.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to derive the moment of inertia for both axes, referencing known formulas and seeking clarification on the application of the parallel axis theorem.

Discussion Status

Participants are exploring the relationship between the moment of inertia about the center of mass and that about the corner axis, with some providing guidance on using the parallel axis theorem. There is an ongoing examination of the displacement involved in the calculations.

Contextual Notes

Participants mention the need to prove the moment of inertia about the corner axis and discuss the implications of the parallel axis theorem in this context.

johnnyb
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I have to show what the moment of inertia of a rectangular plate with mass M and sides A and B is about its centre of mass. I have come up with
[tex]\frac{1}{12}M(a^2 + b^2)[/tex]

Now I have to show what the moment of inertia of the same plate is except this time about an axis perpedicular to the plate and passes through one corner. I know it is:
[tex]\frac{1}{3}M(a^2 + b^2)[/tex] But having some problems proving it
 
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Yes it is.

I = I (at centre of mass) + [tex]Md^2[/tex]

But I can't see how that can get me from the first to second equation
 
Easy,

[tex]I_{cm} = \frac{1}{12}M(a^2 + b^2)[/tex]

The displacement is half the diagonal, that is,

[tex]\frac{a^2 + b^2}{4}[/tex]

So add them up and you get:

[tex]I = \frac{1}{3}M(a^2 + b^2)[/tex]

:smile:
 
Steiner's theorem.That's the name i learned once with the theorem itself...

Daniel.
 

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