Trying to understand moment of inertia; example of a rod

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SUMMARY

The moment of inertia for a uniform rod is defined by two key formulas: for a thin rod of length L about one end, it is given by Iend = 1/3 M L2, and about the center of mass, it is Icm = 1/12 M L2. The discussion emphasizes the importance of understanding the integration process to derive these results, specifically using the integral ∫ r2 dm. The focus is on a uniform rod, denoted as AB, with respect to its ends A or B.

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teng125
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i found smtg that i don't understand.moment of inertia = 1/3 m(r^2) stands for which type pf rod??or reference point??
 
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The moment of inertia of a thin rod of length L about one end is:
[tex]I_{end} = \frac{1}{3} M L^2[/tex]

About the middle, it would be:
[tex]I_{cm} = \frac{1}{12} M L^2[/tex]

(Set up the integral [itex]\int r^2 dm[/itex] and see for yourself!)
 
Don't understand something like that. Do the integration by yourself and derive the result. Anyway it is for the moment of inertia of a uniform rod, suppose AB, with respect to its ends A or B.
There is no complication in the integral. Just think about relating dm with dr.
 

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