- #1
rakshit gupta
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I am unable to find it.
What is [s]motion[/s] moment of inertia of an equilateral triangle?
- Context: Undergrad
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SUMMARY
The discussion focuses on calculating the moment of inertia for an equilateral triangle, specifically for a thin lamina. The integral definition of moment of inertia, represented as $$I \equiv \int r^2 \, dm$$, is emphasized as a crucial method for deriving the value. Participants suggest using the parallel axis theorem to simplify calculations when positioning the axes appropriately. The conversation clarifies the distinction between the second moment of area and mass moment of inertia, guiding users to specify their area of interest.
PREREQUISITES- Understanding of moment of inertia and its mathematical definition
- Familiarity with integrals and calculus
- Knowledge of the parallel axis theorem
- Basic geometry of equilateral triangles
- Study the derivation of moment of inertia for various shapes, focusing on equilateral triangles
- Learn about the application of the parallel axis theorem in different contexts
- Explore the differences between the second moment of area and mass moment of inertia
- Practice solving integrals involving mass distribution for thin laminae
Students and professionals in physics and engineering, particularly those studying mechanics, structural analysis, or materials science, will benefit from this discussion.
- #2
BvU
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Hello Rakshit, welcome to PF 
!
Here is an example. But you are well advised to make your own drawing and work out the integral from the definition of moment of inertia $$I \equiv \int r^2 \, dm$$
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!Here is an example. But you are well advised to make your own drawing and work out the integral from the definition of moment of inertia $$I \equiv \int r^2 \, dm$$
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- #3
Meir Achuz
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It is easier to do the integrals if you place the x-axis along the base of the pyramid, and the y-axis going vertically through the top apex.
Then you can use the parallel axis theorem to get the moment of inertia about the center of the pyramid.
Then you can use the parallel axis theorem to get the moment of inertia about the center of the pyramid.
- #4
SteamKing
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I'm not surprised. What does "motion of inertia" even mean?rakshit gupta said:
You can determine the second moment of area for a region which is an equilateral triangle, or the mass moment of inertia for a plate or thin lamina which has the shape of an equilateral triangle.
So which are you interested in finding?
- #5
rakshit gupta
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BvU said:Hello Rakshit, welcome to PF
Here is an example. But you are well advised to make your own drawing and work out the integral from the definition of moment of inertia $$I \equiv \int r^2 \, dm$$
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For a thin laminaSteamKing said:
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