Calculus &
Linear Algebra II
Chapter 33
33 Moments of inertia (second moments)
By the end of this section, you should be able to answer the following questions:
- How do you locate the centre of mass of a solid object using a triple integral?
- How do you calculate the moments of inertia about the three coordinate axes?
33 Moments of inertia (second moments)
The moment of inertia of a particle of mass $m$ about an axis $(x, y, \text{ or } z)$ is defined to be $mr^2$ where $r$ is the distance from the particle to the axis.
It is sometimes referred to as rotational inertia and can be thought of as the rotational analogue of mass for linear motion. For example, linear kinetic energy can be expressed as $\frac{1}{2}mv^2,$ and the rotational kinetic energy as $\frac{1}{2}I\omega^2.$ Linear momentum is determined by the formula $p= mv,$ while angular momentum is given by $L=I\omega.$ In these examples, $I$ is the moment of inertia and $\omega$ the angular velocity.
33 Moments of inertia (second moments)
As we have seen from previous examples, the mass of a solid with density $\rho(x, y, z)$ occupying a region $R$ in $\R^3$ is given by \[ m = \iiint_R \rho(x,y,z)~dV. \]
The moments about each of the three coordinate planes are
$\displaystyle M_{yz} = \iiint_R x~\rho(x,y,z)~dV,\;\; $ $ \displaystyle M_{xz} = \iiint_R y~\rho(x,y,z)~dV,$
$\displaystyle M_{xy} = \iiint_R z~\rho(x,y,z)~dV$
33 Moments of inertia (second moments)
The centre of mass is then located at the point $\left(\overline{x}, \overline{y}, \overline{z}\right)$ where \[ \overline{x} = \frac{M_{yz}}{m}, \quad \overline{y} = \frac{M_{xz}}{m}, \quad\overline{z} = \frac{M_{xy}}{m} \]
The moments of inertia about each of the three coordinate axes work out to be
$\displaystyle I_{x} = \iiint_R \left(y^2+z^2\right) \rho(x,y,z)~dV,$
$\displaystyle I_{y} = \iiint_R \left(x^2+z^2\right) \rho(x,y,z)~dV,$
$\displaystyle I_{z} = \iiint_R \left(x^2+y^2\right) \rho(x,y,z)~dV.$
33.1 Derive the integral formulas for $M_{xy}$ and $I_x$
For $M_{xy}$, consider the following diagram:
33.1 Derive the integral formulas for $M_{xy}$ and $I_x$
We break down the solid (mass) into tiny boxes and compute the moment of one box about the $xy$-plane. First, we know that \[ \Delta m = \rho(x^*,y^*,z^*) \times \Delta V. \]
Then the (first) moment of the box about the $xy$-plane is
$\displaystyle \Delta m \times \text{distance (to the $xy$-plane)}$ $\approx \rho(x^*,y^*,z^*) \times \Delta V\times z^* $
Now, the (first) moment of the solid about the $xy$-plane is approximately
$ \displaystyle\sum z^*~\rho(x^*,y^*,z^*) ~\Delta V.\;\; $ If $\displaystyle \Delta V \to 0,\,$ then $\displaystyle \iiint_R z~ \rho(x,y,z)~dV$.
33.1 Derive the integral formulas for $M_{xy}$ and $I_x$
For $I_{x},$ consider the following diagram:
33.1 Derive the integral formulas for $M_{xy}$ and $I_x$
Again we break down the solid (mass) into tiny boxes. But now we consider the distance of the box to the $x$-axis. That is
$\displaystyle r = \sqrt{\left(x^*-x^*\right)^2+\left(y^*-0\right)^2+\left(z^*-0\right)^2}$ $\displaystyle = \sqrt{\left(y^*\right)^2+\left(z^*\right)^2}$
Then the moment of inertia of the box about the $x$-axis is approximately
$\displaystyle \Delta m \times r^2$ $\displaystyle = \rho(x^*,y^*,z^*) \times \Delta V\times \left(\left(y^*\right)^2+\left(z^*\right)^2\right) $
Now the moment of inertia of the solid about the $x$-axis is approximately
$ \displaystyle\sum \left(\left(y^*\right)^2+\left(z^*\right)^2\right) \rho(x^*,y^*,z^*) ~\Delta V.\;\; $
If $\displaystyle \Delta V \to 0,\,$ then $\displaystyle \iiint_R \left(y^2+z^2\right) \rho(x,y,z)~dV$.
33.2 Example: locate the centre of mass
of a solid hemisphere of radius $a$ with density proportional to the distance from the centre of the base. Find its moment of inertia about the $z$-axis.
33.2 Example: locate the centre of mass
Consider the upper hemisphere centred at the origin. We will use the spherical coordinates. We know that $\rho \propto r$, with $r$ the distance to the origin.
Take $\rho = kr,$ with $k$ a constant. Let's begin computing:
\[ \overline{x} = \frac{M_{yz}}{m}\;\; \text{ and } \;\;\overline{y} = \frac{M_{xz}}{m} \]
33.2 Example: locate the centre of mass
Consider the upper hemisphere centred at the origin. We will use the spherical coordinates. We know that $\rho \propto r$, with $r$ the distance to the origin.
Take $\rho = kr$, with $k$ a constant. Let's begin computing:
$ \displaystyle \overline{x} = \frac{\iiint_V x ~\rho ~dV}{\iiint_V \rho ~dV}\;\; \text{ and } \;\;\overline{y} = \frac{\iiint_V y~ \rho ~dV}{\iiint_V \rho ~dV} $
33.2 Example: locate the centre of mass
By simmetry of the solid about $z$-axis, we have that
$\displaystyle \iiint_V x~ \rho ~dV = 0 = \iiint_V y ~\rho ~dV.\;\;$
Then $\;\;\overline{x} = \overline{y} = 0$.
So we just need to compute $\displaystyle \;\overline{z} = \frac{\iiint_V z ~\rho ~dV}{\iiint_V \rho ~dV}$.
$\displaystyle V = \left\{ \left(r, \theta, \phi\right)~|~ 0 \leq r \leq a, 0\leq \theta \leq 2 \pi, 0 \leq \phi \leq \pi/2 \right\}$.
33.2 Example: locate the centre of mass
| $\displaystyle m$ | $\displaystyle = \iiint_V \rho ~dV $ |
| $\displaystyle = \int_0^a \int_{0}^{2\pi} \int_0^{\pi/2} (k~r) ~ r^2\sin \phi ~d\phi ~d\theta ~dr$ | |
| $\displaystyle = k \left( \int_0^a r^3~dr \right) \left( \int_0^{2\pi} ~d\theta \right) \left( \int_0^{\pi/2} \sin \phi~d\phi \right)$ | |
| $\displaystyle = k \left( \frac{a^4}{4} \right) \left(2\pi \right) \left( 1 \right)$ $\displaystyle = \large \frac{\pi k a^4}{2}$. |
33.2 Example: locate the centre of mass
| $\displaystyle \iiint_V z ~\rho ~dV$ | $\displaystyle = \iiint_V (r \cos \phi) ~ (k ~r) ~dV $ $\displaystyle = \iiint_V k ~r^2 \cos \phi ~dV $ |
| $\displaystyle = \int_0^a \int_{0}^{2\pi} \int_0^{\pi/2}k~r^2\cos \phi ~ r^2\sin \phi ~d\phi ~d\theta ~dr$ | |
| $\displaystyle = k \left( \int_0^a r^4~dr \right) \left( \int_0^{2\pi} ~d\theta \right) \left( \int_0^{\pi/2} \cos \phi \sin \phi~d\phi \right)$ | |
| $\displaystyle = k \left( \frac{a^5}{5} \right) \left(2\pi \right) \left( \frac{1}{2} \right)$ $\displaystyle = \large \frac{\pi ~k ~a^5}{5}$. |
33.2 Example: locate the centre of mass
Then we have that
$\displaystyle \overline{z}= \dfrac{\dfrac{\pi k a^5}{5}}{\dfrac{\pi k a^4}{2}}$ $\displaystyle = \frac{2}{5}a$.
Therefore \[ \left(\overline{x}, \overline{y}, \overline{z}\right) = \left(0, 0, \frac{2}{5}a\right). \]
33.2 Example: Find the moment of inertia about the $z$-axis
We need to compute \[ I_z = \iiint_V\left(x^2+y^2\right)\rho~ dV. \]
In spherical coordinates we have
$x^2+y^2$ $=r^2 \cos^2 \theta \sin^2 \phi + r^2 \sin^2 \theta \sin^2 \phi$ $=r^2 \sin^2 \phi$.
33.2 Example: Find the moment of inertia about the $z$-axis
| $\displaystyle I_z$ | $\displaystyle =\iiint_V\left(x^2+y^2\right)\rho~ dV$ |
| $\displaystyle = \int_0^a \int_{0}^{2\pi} \int_0^{\pi/2}\left( r^2\sin^2 \phi \right) \left(k ~r\right) ~r^2\sin \phi ~d\phi ~d\theta ~dr$ | |
| $\displaystyle = k \left( \int_0^a r^5~dr \right) \left( \int_0^{2\pi} ~d\theta \right) \left( \int_0^{\pi/2} \sin^3 \phi~d\phi \right)$ | |
| $\displaystyle = k \left( \frac{a^6}{6} \right) \left(2\pi \right) \left( \frac{2}{3} \right)$ $\displaystyle = \frac{2~\pi ~k ~a^6}{9}$. |