How to proof mass moment of inertia formula for a hoop with axis across the diameter?

Moment of inertia of the hoop is given by: $\frac{1}{2} m {r}^{2}$ How to proof that? You may attach a hyperlink or write down the derivation from $I = m {r}^{2}$. Thanks

Jul 20, 2017

See the proof below

Explanation:

The volume is $= 2 \pi R t$

The thickness is $= t$

The radius of the hoop is $= R$

The density is $\rho = \frac{M}{2 \pi R t}$

The moment of inertia is

$I = \int {r}^{2} \mathrm{dm}$

As the axis is across the diameter

The distance from the differential mass $\mathrm{dm}$ is $= R \sin \theta$

$\mathrm{dm} = \rho R t d \theta$

$\cos 2 \theta = 1 - 2 {\sin}^{2} \theta$

${\sin}^{2} \theta = \frac{1}{2} - \frac{1}{2} \cos 2 \theta$

Therefore, substituting in the integral, we integrate from $0$ to $\pi$ and multiply by $2$

$I = 2 {\int}_{0}^{\pi} {R}^{2} {\sin}^{2} \theta \rho R t d \theta$

$= 2 {R}^{3} \rho t {\int}_{0}^{\pi} {\sin}^{2} \theta d \theta$

$= 2 {R}^{3} \rho t \cdot {\left[\frac{\theta}{2} - \frac{1}{4} \sin \left(2 \theta\right)\right]}_{0}^{\pi}$

$= 2 {R}^{3} \rho t \left(\frac{\pi}{2}\right)$

$= \frac{M}{2 \pi R t} \cdot 2 {R}^{3} \cdot t \cdot \frac{\pi}{2}$

$= \frac{1}{2} M {R}^{2}$