My name is Alexander FufaeV and here I write about:

Moment of Inertia of a Rigid Body

Important Formula

Formula: Moment of inertia (continuous mass distribution)

$$\class{brown}{I} ~=~ \int_{V} r_{\perp}^{~~2} \, \rho(\boldsymbol{r}) \, \text{d}v$$ $$\class{brown}{I} ~=~ \int_{V} r_{\perp}^{~~2} \, \rho(\boldsymbol{r}) \, \text{d}v$$

What do the formula symbols mean?

Moment of inertia

$$ \class{brown}{I} $$ Unit $$ \mathrm{kg} \, \mathrm{m}^2 $$

Moment of inertia indicates how inert a body is when the direction of its axis of rotation or angular velocity is changed. The further away a mass element of the body is from the axis of rotation, the more it contributes to the moment of inertia.

Distance

$$ r_{\perp} $$ Unit $$ \mathrm{m} $$

Magnitude of the vector $ \boldsymbol{r}_{\perp} $. This vector is the part of the position vector $ \boldsymbol{r} $ perpendicular to the axis of rotation. Therefore, $r_{\perp}$ is the distance from the axis of rotation to an infinitesimal mass element.

Volume

$$ V $$ Unit $$ \mathrm{m}^3 $$

Volume of the body under consideration.

Mass Density

$$ \rho(\boldsymbol{r}) $$ Unit $$ \frac{ \mathrm{kg} }{ \mathrm{m}^3 } $$

Mass per volume at the location $ \boldsymbol{r} $.

We want to calculate the moment of inertia $ I $ of a rigid system that rotates around a fixed axis. You must, of course, define the axis of rotation. The moment of inertia counteracts the rotary motion and depends on the mass distribution of the system, which is distributed around the axis of rotation. Here we have to distinguish whether the system has a discrete mass distribution (e.g. planets orbiting the sun) or whether it has a continuous mass distribution (e.g. a rotating cylinder).

In the case of a discrete mass distribution, you have $N$ masses that are distributed in space and have different distances $r$ from the axis of rotation. To determine the moment of inertia $ I $ for this discrete rotating system, you need to know the individual masses $ \class{brown}{m_i } $ and their distances $ r_{\perp i} $ perpendicular to the axis of rotation:

$$ \begin{align} I ~&=~ \class{brown}{m_1} \, {r_{\perp 1}}^2 ~+~ \class{brown}{m_2} \, {r_{\perp 2}}^2 ~+~ ... ~+~ \class{brown}{m_N} \, {r_{\perp N}}^2 \\\\
~&=~ \underset{i~=~1}{\overset{N}{\boxed{+}}} \, \class{brown}{m_i} \, {r_{\perp i}}^2 \end{align} $$

In the case of a continuous mass distribution, you do not have individual masses, but the total mass is "smeared" over a certain area of space. A hollow cylinder, for example, has a continuous mass distribution in which its mass is smeared on its surface. To calculate the moment of inertia $ I $ of such a system, you have to replace discrete summation in 1 by an integral and the masses $\class{brown}{m_i}$ by the mass density $\rho(\boldsymbol{r})$. Then you integrate over the volume $V$ of the rotating system:

$$ \begin{align} I ~=~ \int_V \, r_{\perp}^2 \, \rho(\boldsymbol{r}) \, \text{d}v \end{align} $$

The given mass density $\rho(\boldsymbol{r})$ depends on the position vector $\boldsymbol{r} $ in a three-dimensional system. The volume element $\text{d}v$, in which the mass $\rho(\boldsymbol{r}) \text{d}v$ is located, has the perpendicular distance $ r_{\perp} $ to the axis of rotation.

If the theoretical calculation using the integral is difficult (because the system has no symmetries), then the moment of inertia can be determined experimentally. For this purpose, the total torque $M$ and the angular acceleration $\alpha$ are measured and the moment of inertia $I$ is calculated from their ratio:

$$ \begin{align} I ~=~ \frac{M}{\alpha} \end{align} $$

Torque and angular acceleration of a disk (continuous mass distribution).