Calculate Moment of Inertia of Rods | Solution Included

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Homework Help Overview

The discussion revolves around calculating the moment of inertia of two rods about their center of mass, with each rod having mass m and length L. The original poster provides an equation for the moment of inertia of a single rod and seeks to determine the combined moment of inertia for both rods.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to find the center of mass before calculating the moment of inertia. There are attempts to apply the parallel axis theorem and questions about how to adjust calculations if the configuration changes, such as if the rods were arranged in a square.

Discussion Status

Participants are actively engaging with the problem, exploring different approaches to find the center of mass and applying the moment of inertia formulas. Some guidance has been provided regarding the use of the parallel axis theorem, and there is an ongoing exploration of how different configurations might affect the calculations.

Contextual Notes

There is a mention of a picture attached to the original post, which may contain relevant information about the arrangement of the rods. The discussion also reflects uncertainty about the calculations and the implications of changing the configuration of the rods.

supercali
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[SOLVED] moment of inertia

Homework Statement



what is the moment of interia of these 2 rods (picture attached) about their center of mass
each rod mass is m with length L

Homework Equations


one rod i know is [tex]\frac{L^2m}{12}[/tex]


The Attempt at a Solution

 

Attachments

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First things first: Find the center of mass.
 
good question

i thought the answer is [tex]\frac{L^2m}{12}*2+\frac{L^2m}{4}*2[/tex]
obviously it isnt

i think center of mass is at [tex]\frac{L}{4}\hat{x}-\frac{L}{4}\hat{y}[/tex]
 
Last edited:
supercali said:
i think center of mass is at [tex]\frac{L}{4}\hat{x}-\frac{L}{4}\hat{y}[/tex]
Good--it's right in the middle between the two rod centers. So what's the distance from the center of each rod to that center of mass? To find the rotational inertia of the rods about that center of mass, use the parallel axis theorem.
 
yeah i got it it is [tex]\frac{L^2m}{12}*2+\frac{L^2m}{8}*2[/tex] and one more thing if it was a square or rods i would have to multiply by 4?
 
supercali said:
yeah i got it it is [tex]\frac{L^2m}{12}*2+\frac{L^2m}{8}*2[/tex]
Good.
and one more thing if it was a square or rods i would have to multiply by 4?
If it were a square of rods, the distance from each rod to the center of mass would be different. (But there would be a factor of 4 involved.)
 

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