Calculating Rotational Inertia of Flat Plate

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Discussion Overview

The discussion focuses on deriving the rotational inertia for a flat plate, exploring the differences in approach compared to one-dimensional bodies like thin rods. The scope includes theoretical derivation and mathematical reasoning related to rotational dynamics.

Discussion Character

  • Technical explanation, Mathematical reasoning

Main Points Raised

  • One participant questions the symmetry of the flat plate regarding the axis of rotation, contrasting it with a thin rod.
  • Another participant provides the definition of rotational inertia as an integral, noting the need to evaluate it in two dimensions for a flat plate.
  • A request for clarification on the variable r in the integral is made, highlighting confusion over the use of antiderivatives versus definite integrals.
  • A clarification is offered that r represents the distance from the point being evaluated to the axis of revolution, and that the integral should be evaluated over the volume of the body.
  • Another participant suggests using a rectangular element of the plate and applying the parallel axis theorem to derive the moment of inertia, proposing an integration approach to arrive at the answer.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and approach to the derivation, with no consensus reached on a single method or interpretation of the integral involved.

Contextual Notes

Some participants note the complexity of integrating over two dimensions and the implications of the geometry of the flat plate, which may affect the assumptions made in the derivation.

Who May Find This Useful

This discussion may be useful for students and professionals interested in rotational dynamics, particularly those looking to understand the derivation of rotational inertia for two-dimensional bodies.

azizlwl
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How to derive the Rotational Inertia for a flat plate.
I do not see any symmetry about the axis of rotation as in a thin rod.http://imageshack.us/a/img845/1134/53433231.jpg
 
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Well, you just calculate it from the definition:

I = \int r^{2} \rho \mathrm{d}V

The difference here is that the integral has to be evaluated in two dimensions. For a thin rod you can eliminate two coordinates (it's a one-dimensional body); this is a two-dimensional body.
 
Thank you.
Can you elaborate more.
Can you tell me what r represents in your equation?
Your equation is antiderivatives not definite integral.
 
Last edited:
Okay, r is the distance of the point you're evaluating to the axis of revolution. So \rho is generally a function of r too.

It is a definite integral (I should have written \int_V instead of \int. You integrate over the volume of the body. That's why you only evaluate in one dimension for a thin rod, but two dimensions for a thin plane.
 
Hey Azizlwl,

Take a rectangular element of length a and height as given at a distance r from the axis. Make the width very small i.e. dr. What is the moment of inertia of this element about the axis? (Use parallel axis theorem). Now integrate. you will get your answer.
 

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