What is the moment of inertia of a rod with a mass of 6 kg and length of 1 m that is spinning around its center?

Mar 26, 2016

Answer:

$0.5 k g . {m}^{2}$

Explanation:

In this case, $I = \frac{1}{12} M {L}^{2}$

$= \frac{1}{12} \times 6 \times {1}^{2}$

$= 0.5 k g . {m}^{2}$.

(Let me know if you want me to show you the derivation of the equation for $I = \frac{1}{12} M {L}^{2}$ from definition/first principles then I will)