Roational Inertia of a triangle

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Homework Help Overview

The discussion revolves around calculating the moment of inertia (I) of an equilateral triangle about the midpoint of its base. Participants are exploring different methods and considerations related to this topic in the context of rotational inertia.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to split the triangle into two right triangles to find the rotational inertia about a specific point. Some participants suggest calculating the moment of inertia about the center of mass first and then applying the Principal Axis Theorem. Others express uncertainty about the correctness of their calculations and question the assumptions made regarding the values obtained.

Discussion Status

The discussion is ongoing, with participants offering different approaches and questioning the validity of their results. There is a mix of exploration of methods and clarification of concepts, but no consensus has been reached regarding the correct approach or final answer.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the information they can share. There is also a discussion about the known values of moment of inertia for different axes, which may influence their calculations.

joserse46
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I want to find the I of an equialteral triangle about the middle of it's base.
 

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Welcome to PF!

Hi joserse46! Welcome to PF! :wink:

Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 
Well what I am doing now is splitting the triangle up into two right triangles and finding the RI of on about it's tip (the tip that makes the right anlge), from there I could use the rod about the tip the intergrate from 0 to root 3 over 2A (the length og the triangle) But I sure there is an easier way to do this>
 
First, calculate the moment of inertia about its center of mass, which is at the intersection of the three medians (drawn from each vertex to the center of the opposite side). Now add to this moment of inertia the product of the triangle's mass times 2/3 the length of the median from the vertex to the base. This latter term is called the Principal Axis Theorem.
 
well i came to 7/12 MA^2 but that doesn't seem right since I know for a fact that it's 5/12MA^2 about it's tip and 1/12MA^2 about it's center. so shouldn't it be between that or I'm assuming too much
Hope someone can confirm this
 

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