Finding Moment of Inertia of Rod

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To find the moment of inertia of a rod with linear density λ=4.0 kg/m and length L=4 m, the total mass M is calculated as M=λL, resulting in M=16 kg. The moment of inertia about the center of mass is given by the formula I=1/12 ML². Substituting the values, I becomes I=1/12 (16 kg)(4 m)². The confusion arises from the interpretation of the center of mass; the entire length of the rod should be used for mass calculation, not half. Therefore, the correct approach confirms that M should be 16 kg for accurate moment of inertia calculation.
Alem2000
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I had a problem with finding the moment of inertia. I have a rod and it has linear density\lambda=4.0kg and the equation for its moment of inertia is I=\frac{1}{12} ML^2 now I have an axis at the center of mass of a rod the rod has a total length L=4m so in my equation I would have to substute \lambda L=M now the thing that I don't understand is if my center of mass horizontal cordinate is 2m 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 4.0\frac{kg}{m}. 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 M=(4.0\frac{kg}{m})(4.0m). Does that give you the right answer?
 
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