Finding Moment of Inertia of Rod

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SUMMARY

The discussion focuses on calculating the moment of inertia for a rod with a linear density of 4.0 kg/m and a total length of 4 m. The correct formula for the moment of inertia about the center of mass is I = (1/12) ML². The total mass M is determined by the product of linear density and total length, yielding M = (4.0 kg/m) * (4.0 m) = 16.0 kg. The confusion arises from the misunderstanding of the center of mass and the application of linear density in the calculation.

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Alem2000
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I had a problem with finding the moment of inertia. I have a rod and it has linear density[tex]\lambda=4.0kg[/tex] and the equation for its moment of inertia is [tex]I=\frac{1}{12} ML^2[/tex] now I have an axis at the center of mass of a rod the rod has a total length [tex]L=4m[/tex] so in my equation I would have to substute [tex]\lambda L=M[/tex] now the thing that I don't understand is if my center of mass horizontal cordinate is [tex]2m[/tex] the length I multipy by lambda should be 2m correct?
 
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Kg is not the unit for linear density. I'm assuming you meant [itex]4.0\frac{kg}{m}[/itex]. If each meter of the rod has a mass of 4.0kg, why would you only multiply by half the length of the rod to get its mass? I think [itex]M=(4.0\frac{kg}{m})(4.0m)[/itex]. Does that give you the right answer?
 

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