Parallel Axis Theorem and sheet of metal

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The discussion focuses on calculating the moment of inertia of a thin rectangular sheet of metal using the parallel-axis theorem. The moment of inertia for the sheet about an axis through one corner is derived as I = 1/3M(a^2 + b^2). Participants clarify the need to identify the center of mass (Icm) and the distance (d) to apply the theorem correctly. The correct formula for Icm is noted as (1/12)M(a^2 + b^2). The conversation emphasizes understanding the rotational inertia of the rectangular sheet and the significance of the chosen axis for accurate calculations.
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Homework Statement


A thin, rectangular sheet of metal has a mass M and sides of length a and b. Use the parallel-axis theorem to calculate the moment of inertia of the sheet for an axis that is perpendicular to the plane of the sheet that passes through one corner of the sheet

Homework Equations





The Attempt at a Solution



I'm not really sure what axis the problem is saying that the sheet will be rotating around I know the answer is 1/3M(a^2 + b^2) because this is an odd problem in my book, but I'm not sure how to go about it. I know the parallel axis theorem is I = Icm +Md^2
 
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Just try step by step to find all the ingredients to apply the parallel axis theorem.
Like: what is Icm? What is it in this case? Can you look it up, or calculate it?
Then, what is d (draw a picture, and then try to get a formula from that)?
 
i was thinking that the Icm was 1/3Ma^2 the distance I'm not sure about because I am not sure hwat axis it is rotating about
 
anyone?
 
physstudent1 said:
i was thinking that the Icm was 1/3Ma^2
No. Look up the rotational inertia of a rectangular sheet.
the distance I'm not sure about because I am not sure hwat axis it is rotating about
Pick any corner. The distance from the cm will be the same.
 
For rectangular sheet
I[CM]=(1/12)M(a^2+b^2)

I(corner) means parallel to x-axis or y axis.

I=I[x]+Md^2
I=(1/12)Mb^2+M(b/2)^2=(1/3)Mb^2
 
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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