What is the moment of inertia of a rod with a mass of $3 k g$ $3 kg$ and length of $7 m$ $7 m$ that is spinning around its center?

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Answer 1 · 2016-02-05T21:25:21

$12.25 k g . m^{2}$ $12.25kg.m^2$

The formula for moment of inertia of a thin rod of mass M and length L about an axis through its centre is given by

$I = \frac{1}{12} M L^{2}$ $I=1/12ML^2$

$= \frac{1}{12} \times 3 \times 7^{2}$ $=1/12xx3xx7^2$

$= 12.25 k g . m^{2}$ $=12.25kg.m^2$

What is the moment of inertia of a rod with a mass of 3kg3 kg and length of 7m7 m that is spinning around its center?