moment of inertia - 5

[logistic_guy]

Find the moment of inertia of a thin uniform hoop of radius $\displaystyle R$ and mass $\displaystyle M$ when the rotation is through its center.

 

[logistic_guy]

$\displaystyle I = \int R^2 \ dm = \int R^2 \lambda \ dL = \int_{0}^{2\pi} R^2 \lambda R \ d\theta$

Solving this integral gives:

$\displaystyle I = 2\pi R^3 \lambda = 2\pi R^3 \frac{M}{2\pi R} = \textcolor{blue}{MR^2}$