Trying to understand moment of inertia; example of a rod

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:
I_{end} = \frac{1}{3} M L^2

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

(Set up the integral \int r^2 dm 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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