Mass Moment of Inertia

The mass moment of inertia $I$ measures a rigid body’s resistance to angular acceleration about a given axis.
$I={\int }_m{r}^{2}dm$
Units: $\text{kg}\cdot {\text{m}}^2$

Alternatively, for a point mass, mass moment of inertia is equal to:
$I=m{r}^2$

Mass moment of inertia $I$ is related to angular acceleration by:
$T=I\alpha$

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Radius of Gyration

Instead of quoting $I$ directly, many problems (including textbook wheel/drum problems) give a radius of gyration $k$:

$I=m{k}^2$

$k$ is the distance from the axis at which the entire mass could be concentrated to give the same $I$. It is not a physical dimension of the body.

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Parallel-Axis Theorem

$I=\bar{I}+m{d}^2$

• $\bar{I}$ $-$ moment of inertia about an axis through the mass center G, parallel to the axis of interest (also notated as $I_G$)
• $d$ $-$ perpendicular distance between the two parallel axes
• $m$ $-$ total mass of the body

Non-Centroidal Axis

This theorem only ever moves you to or from the centroidal axis. To go from one non-centroidal axis to another, you must go through $G$: subtract $m{d}_1^2$ to get $\bar{I}$, then add $m{d}_2^2$.

Geometric vs Mass Centre

$I_O$ is the geometric centre.
$I_G$ is the mass centre.

$I_O\ne I_G$ unless $O$ and $G$ coincide.

Moment of Inertia for Common Shapes

Shape	Axis	$\bar{I}$
Slender rod, length $l$	$\perp$, through centre	$\frac{1}{12}m{l}^2$
Slender rod, length $l$	$\perp$, through end	$\frac{1}{3}m{l}^2$
Thin ring/hoop, radius $r$	Through centre, $\perp$ to plane	$m{r}^2$
Circular disk, radius $r$	Through centre, $\perp$ to plane	$\frac{1}{2}m{r}^2$
Rectangular plate, sides $a,b$	Through centre, $\perp$ to plane	$\frac{1}{12}m(a^2+b^2)$
Solid sphere, radius $r$	Through centre	$\frac{2}{5}m{r}^2$

Composite bodies

For a body built from several simple shapes, find $I$ (about the same axis) for each part separately. Apply the parallel-axis theorem to each part individually as needed and then sum them. Subtract for holes/cut-outs.

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