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

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SUMMARY

The moment of inertia for a rectangular plate with mass M and sides A and B about its center of mass is calculated as I = (1/12)M(a^2 + b^2). To find the moment of inertia about an axis perpendicular to the plate and passing through one corner, the parallel axis theorem is applied, resulting in I = (1/3)M(a^2 + b^2). The displacement used in the theorem is half the diagonal of the plate, specifically (a^2 + b^2)/4, which confirms the transition from the center of mass to the corner axis moment of inertia.

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  • Understanding of moment of inertia concepts
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  • Basic knowledge of geometry related to rectangles
  • Ability to manipulate algebraic expressions
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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
\frac{1}{12}M(a^2 + b^2)

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:
\frac{1}{3}M(a^2 + b^2) But having some problems proving it
 
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Yes it is.

I = I (at centre of mass) + Md^2

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

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

The displacement is half the diagonal, that is,

\frac{a^2 + b^2}{4}

So add them up and you get:

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

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

Daniel.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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