# Moments of inertia

The moment of inertia of a body, written $I_{P,\hat{a}}$, is measured about a rotation axis through point $P$ in direction $\hat{a}$. The moment of inertia expresses how hard it is to produce an angular acceleration of the body about this axis. That is, a body with high moment of inertia resists angular acceleration, so if it is not rotating then it is hard to start a rotation, while if it is already rotating then it is hard to stop.

The moment of inertia plays the same role for rotational motion as the mass does for translational motion (a high-mass body resists is hard to start moving and hard to stop again).

Moment of inertia about axis $\hat{a}$ through point $P$.

\[\begin{gathered} I_{P,\hat{a}} = \iiint_\mathcal{B} \rho r^2 \,dV \\ \text{(units: $\rm kg\ m^2$)} \end{gathered}\] |

The distance $r$ is the perpendicular distance to $dV$ from the axis through $P$ in direction $\hat{a}$.

Warning: Mass moments of inertia are different to area moments of inertia.

Be advised that the "moment of inertia" encountered in Statics is not the same as the moment of inertia used in Dynamics. Strictly speaking, the "moment of inertia" from Statics shouldn't even be called "moment of inertia." What it really is is the "second moment of area." Below are the definitions of two such second moments of area:

\[J_{xx}=\iint_{A}{y^{2}dA}\]

\[J_{yy}=\iint_{A}{x^{2}dA}\]

In contrast, the moment of inertia (about the $z$-axis) is defined as stated above.

For example, a rectangle of base $b$ and height $h$ has the following moments about its centroid $C$:

\[J^{C}_{xx}=\frac{1}{12}bh^{3}\]

\[J^{C}_{yy}=\frac{1}{12}b^{3}h\]

\[I^{C}_{zz}=\frac{1}{12}m(b^{2}+h^{2})\]

Notice that the dimensions of the two quantities are different. While the dimension of second moment of area is $(\text{length})^{4}$, the dimension of moment of inertia is $(\text{mass})(\text{length})^{2}$. When doing dynamics problems with moments of inertia, you should not use the formulas you remember for second moment of area instead. You will get the wrong answer!

Observe that the moment of inertia is proportional to the mass, so that doubling the mass of an object will also double its moment of inertia. In addition, the moment of inertia is proportional to the square of the size of the object, so that doubling every dimension of an object (height, width, etc) will cause it to have four times the moment of inertia.

Moments of inertia about coordinate axes through point $P$.

\[\begin{aligned} I_{P,x} &= I_{P,xx} = I_{P,\hat\imath} = \iiint_\mathcal{B} \rho (y^2 + z^2) \,dx\,dy\,dz \\ I_{P,y} &= I_{P,yy} = I_{P,\hat\jmath} = \iiint_\mathcal{B} \rho (z^2 + x^2) \,dx\,dy\,dz \\ I_{P,z} &= I_{P,zz} = I_{P,\hat{k}} = \iiint_\mathcal{B} \rho (x^2 + y^2) \,dx\,dy\,dz \end{aligned}\]

The coordinates $(x,y,z)$ in the body are measured from point $P$.

The distance $r$ in the moment of inertia integral #rem-ei is the *perpendicular*
distance from the axis of rotation to the infinitesimal
volume $dV$. If we are using rectangular coordinates
$(x,y,z)$ measured from point $P$, and the axis of
rotation is one of the coordinate axes, then the
perpendicular distance is very simple.

As an example, consider the case when the axis of rotation $\hat{a}$ is the $\hat{k}$ axis, as shown in the figure. Then the perpendicular distance $r$ satisfies $r^2 = x^2 + y^2$, giving the coordinate expression above. The other two coordinate axes are similar.

Example Problem: Moment of inertia of a square plate.

A solid square uniform plate has mass $m$, side-length $\ell$, and thickness $h$. What is the moment of inertia about the $z$-axis through the center of mass $C$?

We will use the coordinate formula #rem-ec. To do this, we measure the position from the point $C$ about which we are computing the moment of inertia, so the coordinate origin is at $C$. The figure to the right shows the 3D configuration of the body, where we see that the $x$-coordinate range of the body is $-\ell/2$ to $\ell/2$, and the same for the $y$-coordinate, while the $z$-coordinate ranges from $-h/2$ to $h/2$.

Taking the density of the plate to be $\rho$, we thus have: \[\begin{aligned} I_{C,z} &= \iiint_\mathcal{B} \rho (x^2 + y^2) \,dx\,dy\,dz \\ &= \int_{-h/2}^{h/2} \int_{-\ell/2}^{\ell/2} \int_{-\ell/2}^{\ell/2} \rho (x^2 + y^2) \,dx\,dy\,dz \\ &= \int_{-h/2}^{h/2} \int_{-\ell/2}^{\ell/2} \left[ \rho \left(\frac{1}{3}x^3 + y^2 x\right)\right]_{x = -\ell/2}^{x = \ell/2} \,dy\,dz \\ &= \int_{-h/2}^{h/2} \int_{-\ell/2}^{\ell/2} \rho \left(\frac{1}{12}\ell^3 + y^2 \ell \right) \,dy\,dz \\ &= \int_{-h/2}^{h/2} \left[ \rho \left(\frac{1}{12}\ell^3 y + \frac{1}{3} y^3 \ell \right) \right]_{y = -\ell/2}^{y = \ell/2} \,dz \\ &= \int_{-h/2}^{h/2} \rho \left(\frac{1}{12}\ell^4 + \frac{1}{12} \ell^4 \right) \,dz \\ &= \int_{-h/2}^{h/2} \rho \frac{1}{6}\ell^4 \,dz \\ &= \left[ \rho \frac{1}{6}\ell^4 z \right]_{z = -h/2}^{z = h/2} \\ &= \rho \frac{1}{6}\ell^4 h. \end{aligned}\] The total mass of the plate is $m = \rho \ell^2 h$, so we can write the moment of inertia as \[I_{C,z} = \frac{1}{6} m \ell^2.\] The square plate moment of inertia is actually a special case of the rectangular prism formula #rem-er with $\ell_y = \ell_z = \ell$.

We are always considering the moment of inertia to be a
scalar value $I$, which is valid for rotation about a
fixed axis. For more complicated dynamics with tumbling
motion about multiple axes simultaneously, it is necessary
to consider the full 3 × 3 *moment of inertia
matrix*:
\[
I = \begin{bmatrix}
I_{xx} & I_{xy} & I_{xz} \\
I_{yx} & I_{yy} & I_{yz} \\
I_{zx} & I_{zy} & I_{zz}
\end{bmatrix}
\]
The three scalar moments of inertia
from #rem-ec appear on the
diagonal. The off-diagonal terms are called
the *products of inertia* and are given by
\[
I_{xy} = -\iiint_{\mathcal{B}} \rho xy \,dx\,dy\,dz,
\]
and similarly for the other terms. The angular momentum of
a rigid body is given by $\vec{H} = I \vec\omega$, which
is the matrix product of the moment of inertia matrix with
the angular velocity vector. This is important in advanced
dynamics applications such as unbalanced rotating shafts
and the precession of gyroscopes.

## Transforming and combining moments of inertia

Parallel axis theorem.

\[I_{P,\hat{a}} = I_{C,\hat{a}} + m \, d^2\]

Here $d$ is the perpendicular distance between the axes through $P$ and $C$ in direction $\hat{a}$, so $d = \| \operatorname*{Comp}(\vec{r}_{CP}, \hat{a})\|$.

To easily compute the moments of inertia relative to axes through $P$ and $C$ it is easiest to choose a coordinate system aligned with the rotation axis direction $\hat{a}$. We will choose coordinates $(x,y,z)$ measured from the center of mass $C$ and with the $z$-axis $\hat{k}$ in the direction of $\hat{a}$, as shown in the figure.

As we integrate over the body with infinitesimal volume $dV$ at position $(x,y,z)$ from $C$, we will write the distance from the axis through $C$ as $r_c$ and the distance from the axis through $P$ as $r_P$, as illustrated. The distance between the two rotation axes is $d$. In the chosen coordinate system this means that: \[\begin{aligned} {r_C}^2 &= x^2 + y^2 \\ {r_P}^2 &= (x - x_P)^2 + (y - y_P)^2 \\ d^2 &= {x_P}^2 + {y_P}^2, \end{aligned}\] where $P$ has position $(x_P, y_P, z_P)$ in these coordinates.

Computing the integral #rem-ec to find the moment of inertia about the axis through $P$ in direction $\hat{a}$ now gives: \[\begin{aligned} I_{P,\hat{a}} &= \int_{\mathcal{B}} \rho {r_P}^2 \,dV \\ &= \iiint_{\mathcal{B}} \rho \left((x - x_P)^2 + (y - y_P)^2\right) \,dx\,dy\,dz \\ &= \iiint_{\mathcal{B}} \rho \left(x^2 - 2 x x_P + {x_P}^2 + y^2 - 2 y y_P + {y_P}^2\right) \,dx\,dy\,dz \\ &= \iiint_{\mathcal{B}} \rho (x^2 + y^2) \,dx\,dy\,dz - 2 x_P \iiint_{\mathcal{B}} \rho x \,dx\,dy\,dz \\ &\quad - 2 y_P \iiint_{\mathcal{B}} \rho y \,dx\,dy\,dz + ({x_P}^2 + {y_P}^2) \iiint_{\mathcal{B}} \rho \,dx\,dy\,dz \\ &= \int_{\mathcal{B}} \rho {r_C}^2 \,dV + d^2 \int_{\mathcal{B}} \rho \,dV \\ &= I_{C,\hat{a}} + m d^2. \end{aligned}\] Here we used the coordinate representation of the center of mass to realize that the $x$ coordinate of $C$ is $x_C = \frac{1}{m} \int_{\mathcal{B}} \rho x \,dV$, but because our coordinates are measured from $C$ we must have $x_C = 0$ and so $\int_{\mathcal{B}} \rho x \,dV = 0$. The integral of $\rho y$ is similarly zero.

Warning: Parallel axis theorem must start from center of mass $C$.

The parallel axis theorem does not apply to any two parallel rotation axes. The rotation axis on the right-hand-side of the equation must be through the center of mass $C$, although the other axis can be through any point $P$.

Warning: Parallel axis theorem $d$ is perpendicular distance, not $r_{CP}$.

The distance $d$ in the parallel axis theorem is the
*perpendicular distance* between the axes through
points $P$ and $C$. This will normally not be the same
as the distance $r_{CP}$ between $P$ and $C$, except in
2D problems or if it happens that $\vec{r}_{CP}$ is
already perpendicular to the rotation axis
$\hat{a}$. The distance $d$ can be computed by taking
the orthogonal complement of $\vec{r}_{CP}$ with respect
to $\hat{a}$, so $d =
\|\operatorname{Comp}(\vec{r}_{CP}, \hat{a})\|$.

Example Problem: Moment of inertia of a square plate about a corner.

A solid uniform square plate has mass $m$, side-length $\ell$, and thickness $h$. What is the moment of inertia about the $z$-axis through the corner $P$?

Use the answer to Example Problem #rem-xs and the parallel axis theorem #rem-el.

In Example Problem #rem-xs we computed the moment of inertia of a square place about the center to be $I_{C,z} = \frac{1}{6} m \ell^2$. The parallel axis theorem #rem-el now gives: \[\begin{aligned} I_{P,z} &= I_{C,z} + m \, {r_{CP}}^2 \\ &= \frac{1}{6} m \ell^2 + m \left( \left(\frac{\ell}{2}\right)^2 + \left(\frac{\ell}{2}\right)^2 \right) \\ &= \frac{2}{3} m \ell^2. \end{aligned}\]

To check our answer, we can compute the corner moment of inertia directly by integrating with formula #rem-ec. To do this, we put the coordinate origin at the point $P$ about which we wish to find the moment of inertia, as shown in 3D to the right. From this we see that the $x$ range is $0$ to $\ell$, the same for the $y$ range, and the $z$ range is $-h/2$ to $h/2$.

The integral formula now gives: \[\begin{aligned} I_{P,z} &= \iiint_\mathcal{B} \rho (x^2 + y^2) \,dx\,dy\,dz \\ &= \int_{-h/2}^{h/2} \int_0^{\ell} \int_{0}^{\ell} \rho (x^2 + y^2) \,dx\,dy\,dz \\ &= \int_{-h/2}^{h/2} \int_{0}^{\ell} \left[ \rho \left(\frac{1}{3}x^3 + y^2 x\right)\right]_{x = 0}^{x = \ell} \,dy\,dz \\ &= \int_{-h/2}^{h/2} \int_{0}^{\ell} \rho \left(\frac{1}{3}\ell^3 + y^2 \ell \right) \,dy\,dz \\ &= \int_{-h/2}^{h/2} \left[ \rho \left(\frac{1}{3}\ell^3 y + \frac{1}{3} y^3 \ell \right) \right]_{y = 0}^{y = \ell} \,dz \\ &= \int_{-h/2}^{h/2} \rho \left(\frac{1}{3}\ell^4 + \frac{1}{3} \ell^4 \right) \,dz \\ &= \int_{-h/2}^{h/2} \rho \frac{2}{3}\ell^4 \,dz \\ &= \left[ \rho \frac{2}{3}\ell^4 z \right]_{z = -h/2}^{z = h/2} \\ &= \rho \frac{2}{3}\ell^4 h. \end{aligned}\] The total mass of the plate is $m = \rho \ell^2 h$, so we can write the final expression for the moment of inertia is \[I_{P,z} = \frac{2}{3} m \ell^2.\] This is the same as the expression we obtained above using the parallel axis theorem, but clearly the parallel axis theorem version is much easier.

One consequence of the parallel axis theorem is that the moment of inertia can only increase as we move the rotation point $P$ away from the center of mass $C$. This means that the point with the lowest moment of inertia is always the center of mass itself.

A second consequence of the parallel axis theorem is that moving the point $P$ along the direction $\hat{a}$ doesn't change the moment of inertia, because the axis of rotation is not changing as the point moves along the axis itself.

Adding moments of inertia.

\[I^{\mathcal{B}}_{P,\hat{a}} = I^{\mathcal{B}_1}_{P,\hat{a}} + I^{\mathcal{B}_2}_{P,\hat{a}}\] |

The whole body $\mathcal{B}$ is composed of sub-bodies $\mathcal{B}_1$ and $\mathcal{B}_2$.

The addition of moments of inertia for sub-bodies to give the full moment of inertia follows directly from the fact that the integral over the whole body is the sum of the integrals over the sub-bodes. That is, for $\mathcal{B} = \mathcal{B}_1 \cup \mathcal{B}_2$ (and $\mathcal{B}_1$ not overlapping with $\mathcal{B}_2$), we have: \[\begin{aligned} I^{\mathcal{B}}_{P,\hat{a}} &= \int_{\mathcal{B}} \rho r^2 \,dV \\ &= \int_{\mathcal{B_1} \cup \mathcal{B}_2} \rho r^2 \,dV \\ &= \int_{\mathcal{B_1}} \rho r^2 \,dV + \int_{\mathcal{B}_2} \rho r^2 \,dV \\ &= I^{\mathcal{B}_1}_{P,\hat{a}} + I^{\mathcal{B}_2}_{P,\hat{a}}. \end{aligned}\]

Warning: To add moments of inertia they must be about the same axis.

It is only valid to add together moments of inertia for different sub-bodies if each moment of inertia is computed about the same axis of rotation. For example, consider $I^{\mathcal{B}_1}_{C_1,z}$ and $I^{\mathcal{B}_2}_{C_2,z}$, which are the moments of inertia of bodies $\mathcal{B}_1$ and $\mathcal{B}_2$ about each of their individual centers of mass $C_1$ and $C_2$ (with axis direction $z$). Now it is physically meaningless to form the sum $I^{\mathcal{B}_1}_{C_1,z} + I^{\mathcal{B}_2}_{C_2,z}$, because each moment of inertia is about a different axis. We must first shift both moments of inertia to a common axis point $P$ (perhaps by using the parallel axis theorem), and then we can meaningfully add them together.

Example Problem: Moment of inertia of an L-shaped plate.

A solid uniform L-shaped plate has mass $m$ and dimensions as shown. What is the moment of inertia about the $z$-axis through the corner $P$?

Use the answer to Example Problem #rem-es together with the addition theorem #rem-ea and the parallel axis theorem #rem-el.

Recall that in Example Problem #rem-xs we computed the moment of inertia of a square place about the center to be $I_{C,z} = \frac{1}{6} m \ell^2$.

To use this together with the parallel axis theorem #rem-el and the additive theorem #rem-ea, we must first realize that the L-shape can be decomposed into four square plates, as shown. Each small square body has the same mass and the same moment of inertia about its center, so: \[\begin{aligned} m_1 &= \frac{1}{4} m \\ I^{\mathcal{B}_1}_{C_1,z} &= \frac{1}{6} m_1 (2d)^2 = \frac{1}{6} \left(\frac{1}{4} m\right) 4d^2 = \frac{1}{6} m d^2, \end{aligned}\] and similarly for each other small square body.

Now individual moments of inertia about the corner $P$ can be found using the parallel axis theorem: \[\begin{aligned} I^{\mathcal{B}_1}_{P,z} &= I^{\mathcal{B}_1}_{C_1,z} + m_1 \, {r_{C_1P}}^2 = \frac{1}{6} m d^2 + \left(\frac{1}{4} m\right) \left(d^2 + (3d)^2\right) = \frac{8}{3} m d^2 \\ I^{\mathcal{B}_2}_{P,z} &= I^{\mathcal{B}_2}_{C_2,z} + m_2 \, {r_{C_2P}}^2 = \frac{1}{6} m d^2 + \left(\frac{1}{4} m\right) \left(d^2 + d^2\right) = \frac{2}{3} m d^2 \\ I^{\mathcal{B}_3}_{P,z} &= I^{\mathcal{B}_1}_{P,z} = \frac{8}{3} m d^2 \\ I^{\mathcal{B}_4}_{P,z} &= I^{\mathcal{B}_4}_{C_4,z} + m_4 \, {r_{C_4P}}^2 = \frac{1}{6} m d^2 + \left(\frac{1}{4} m\right) \left((5d)^2 + d^2\right) = \frac{20}{3} m d^2. \end{aligned}\] Here we observed that body $\mathcal{B}_3$ will have the same moment of inertia as $\mathcal{B}_1$, saving some computation.

Combining the individual moments of inertia now gives us the total: \[\begin{aligned} I_{P,z} &= I^{\mathcal{B}_1}_{P,z} + I^{\mathcal{B}_2}_{P,z} + I^{\mathcal{B}_3}_{P,z} + I^{\mathcal{B}_4}_{P,z} \\ &= \frac{8}{3} m d^2 + \frac{2}{3} m d^2 + \frac{8}{3} m d^2 + \frac{20}{3} m d^2\\ &= \frac{38}{3} m d^2. \end{aligned}\] We could also directly compute the total moment of inertia using the integral formula #rem-ec, which would be quite complex.

## Basic shapes

The moments of inertia listed below are all computed directly from the integrals #rem-ec.

Point mass: moments of inertia

\[\begin{aligned} I_{P,z} &= m r^2 \end{aligned}\] |

Rectangular prism: moments of inertia

\[\begin{aligned} I_{C,x} &= \frac{1}{12} m ({\ell_y}^2 + {\ell_z}^2) \\ I_{C,y} &= \frac{1}{12} m ({\ell_z}^2 + {\ell_x}^2) \\ I_{C,z} &= \frac{1}{12} m ({\ell_x}^2 + {\ell_y}^2) \end{aligned}\] |

The calculation for any of the axes is the same, so we will only write out the derivation of $I_{C,z}$ here. We use #rem-ec, which gives: \[\begin{aligned} I_{C,z} &= \iiint_\mathcal{B} \rho (x^2 + y^2) \,dx\,dy\,dz \\ &= \int_{-\ell_z/2}^{\ell_z/2} \int_{-\ell_y/2}^{\ell_y/2} \int_{-\ell_x/2}^{\ell_x/2} \rho (x^2 + y^2) \,dx\,dy\,dz \\ &= \int_{-\ell_z/2}^{\ell_z/2} \int_{-\ell_y/2}^{\ell_y/2} \left[ \rho \left(\frac{1}{3}x^3 + y^2 x\right)\right]_{x = -\ell_x/2}^{x = \ell_x/2} \,dy\,dz \\ &= \int_{-\ell_z/2}^{\ell_z/2} \int_{-\ell_y/2}^{\ell_y/2} \rho \left(\frac{1}{12}{\ell_x}^3 + y^2 \ell_x \right) \,dy\,dz \\ &= \int_{-\ell_z/2}^{\ell_z/2} \left[ \rho \left(\frac{1}{12}{\ell_x}^3 y + \frac{1}{3} y^3 \ell_x \right) \right]_{y = -\ell_y/2}^{y = \ell_y/2} \,dz \\ &= \int_{-\ell_z/2}^{\ell_z/2} \rho \left(\frac{1}{12}{\ell_x}^3 \ell_y + \frac{1}{12} \ell_x {\ell_y}^3 \right) \,dz \\ &= \int_{-\ell_z/2}^{\ell_z/2} \rho \ell_x \ell_y \frac{1}{12}\left( {\ell_x}^2 + {\ell_y}^2\right) \,dz \\ &= \left[ \rho \ell_x \ell_y \frac{1}{12}\left( {\ell_x}^2 + {\ell_y}^2\right) z \right]_{z = -\ell_z/2}^{z = \ell_z/2} \\ &= \rho \ell_x \ell_y \ell_z \frac{1}{12}\left( {\ell_x}^2 + {\ell_y}^2\right). \end{aligned}\] The total mass of the plate is $m = \rho \ell_x \ell_y \ell_z$, so we can write the moment of inertia as \[I_{C,z} = \frac{1}{12} m \left( {\ell_x}^2 + {\ell_y}^2\right).\]

Cylindrical thick shell: moments of inertia

\[\begin{aligned} I_{C,x} &= I_{C,y} = \frac{1}{12} m (3({r_1}^2 + {r_2}^2) + \ell^2) \\ I_{C,z} &= \frac{1}{2} m ({r_1}^2 + {r_2}^2) \end{aligned}\] |

To compute the integrals in #rem-ec it is convenient to switch to cylindrical coordinates: \[\begin{aligned} x &= r \cos\theta \\ y &= r \sin\theta \\ z &= z. \end{aligned}\] The Jacobian of this coordinate transformation is: \[\begin{aligned} J &= \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial z} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} & \frac{\partial y}{\partial z} \\ \frac{\partial z}{\partial r} & \frac{\partial z}{\partial \theta} & \frac{\partial z}{\partial z} \end{vmatrix} = \begin{vmatrix} \cos\theta & -r \sin\theta & 0 \\ \sin\theta & r \cos\theta & 0 \\ 0 & 0 & 1 \end{vmatrix} = r \cos^2\theta + r \sin^2\theta = r. \end{aligned}\] Starting with the $z$ axis, the moment of inertia is thus: \[\begin{aligned} I_{C,z} &= \iiint_\mathcal{B} \rho (x^2 + y^2) \,dx\,dy\,dz \\ &= \int_{-\ell/2}^{\ell/2} \int_0^{2\pi} \int_{r_1}^{r_2} \rho \left( (r \cos\theta)^2 + (r \sin\theta)^2 \right) \,J \,dr\,d\theta\,dz \\ &= \int_{-\ell/2}^{\ell/2} \int_0^{2\pi} \int_{r_1}^{r_2} \rho r^3 \,dr\,d\theta\,dz \\ &= \int_{-\ell/2}^{\ell/2} \int_0^{2\pi} \left[ \rho \frac{1}{4} r^4 \right]_{r = r_1}^{r = r_2} \,d\theta\,dz \\ &= \int_{-\ell/2}^{\ell/2} \int_0^{2\pi} \frac{1}{4} \rho \left( {r_2}^4 - {r_1}^4 \right) \,d\theta\,dz \\ &= \int_{-\ell/2}^{\ell/2} \left[ \frac{1}{4} \rho \left( {r_2}^4 - {r_1}^4 \right) \theta \right]_{\theta = 0}^{\theta = 2\pi} \,dz \\ &= \int_{-\ell/2}^{\ell/2} \frac{\pi}{2} \rho \left( {r_2}^4 - {r_1}^4 \right) \,dz \\ &= \left[ \frac{\pi}{2} \rho \left( {r_2}^4 - {r_1}^4 \right) z \right]_{z = -\ell/2}^{z = \ell/2} \\ &= \frac{\pi}{2} \rho \ell \left( {r_2}^4 - {r_1}^4 \right). \end{aligned}\] The total mass of the cylindrical shell is $m = \rho (\pi {r_2}^2 - \pi {r_1}^2) \ell$, so we can write the moment of inertia as \[\begin{aligned} I_{C,z} &= \frac{1}{2} \frac{m}{{r_2}^2 - {r_1}^2} \left( {r_2}^4 - {r_1}^4 \right) \\ &= \frac{1}{2} \frac{m}{{r_2}^2 - {r_1}^2} \left( {r_2}^2 + {r_1}^2 \right) \left( {r_2}^2 - {r_1}^2 \right) \\ &= \frac{1}{2} m \left( {r_2}^2 + {r_1}^2 \right). \end{aligned}\] Due to rotational symmetry of the cylinder, the moment of inertia will be the same about any axis orthogonal to $z$, so we will just write out the derivation for $I_{C,x}$ here. We again use #rem-ec in cylindrical coordinates, giving: \[\begin{aligned} I_{C,x} &= \iiint_\mathcal{B} \rho (y^2 + z^2) \,dx\,dy\,dz \\ &= \int_{-\ell/2}^{\ell/2} \int_0^{2\pi} \int_{r_1}^{r_2} \rho \left( (r \sin\theta)^2 + z^2 \right) \,J \,dr\,d\theta\,dz \\ &= \int_{-\ell/2}^{\ell/2} \int_0^{2\pi} \int_{r_1}^{r_2} \rho \left( r^3 \sin^2\theta + r z^2 \right) \,dr\,d\theta\,dz \\ &= \int_{-\ell/2}^{\ell/2} \int_0^{2\pi} \left[ \rho \left( \frac{1}{4} r^4 \sin^2\theta + \frac{1}{2} r^2 z^2 \right) \right]_{r = r_1}^{r = r_2} \,d\theta\,dz \\ &= \int_{-\ell/2}^{\ell/2} \int_0^{2\pi} \rho \left( \frac{1}{4} ({r_2}^4 - {r_1}^4) \sin^2\theta + \frac{1}{2} ({r_2}^2 - {r_1}^2) z^2 \right) \,d\theta\,dz \\ &= \int_{-\ell/2}^{\ell/2} \int_0^{2\pi} \rho \left( \frac{1}{8} ({r_2}^4 - {r_1}^4) (1 - \cos 2\theta) + \frac{1}{2} ({r_2}^2 - {r_1}^2) z^2 \right) \,d\theta\,dz \\ &= \int_{-\ell/2}^{\ell/2} \left[ \rho \left( \frac{1}{8} ({r_2}^4 - {r_1}^4) \left(\theta - \frac{1}{2} \sin 2\theta\right) + \frac{1}{2} ({r_2}^2 - {r_1}^2) z^2 \theta \right) \right]_{\theta = 0}^{\theta = 2\pi} \,d\theta\,dz \\ &= \int_{-\ell/2}^{\ell/2} \rho \left( \frac{\pi}{4} ({r_2}^4 - {r_1}^4) + \pi ({r_2}^2 - {r_1}^2) z^2 \right) \,d\theta\,dz \\ &= \left[ \rho \left( \frac{\pi}{4} ({r_2}^4 - {r_1}^4) z + \pi ({r_2}^2 - {r_1}^2) \frac{1}{3} z^3 \right) \right]_{z = -\ell/2}^{z = \ell/2} \\ &= \rho \left( \frac{\pi}{4} ({r_2}^3 - {r_1}^3) \ell + \pi ({r_2}^2 - {r_1}^2) \frac{1}{12} \ell^3 \right) \\ &= \frac{1}{12} \rho \pi ({r_2}^2 - {r_1}^2) \ell \left( 3 ({r_2}^2 + {r_1}^2) + \ell^2 \right). \end{aligned}\] In the last line we again used the factorization $({r_2}^4 - {r_1}^4) = ({r_2}^2 + {r_1}^2) ({r_2}^2 - {r_1}^2)$. Now we can substitute in the total mass $m = \rho \pi ({r_2}^2 - {r_1}^2) \ell$ to obtain: \[\begin{aligned} I_{C,x} &= \frac{1}{12} m \left( 3 ({r_2}^2 + {r_1}^2) + \ell^2 \right). \end{aligned}\]

Spherical thick shell: moments of inertia

\[\begin{aligned} I_C &= \frac{2}{5} m \left(\frac{{r_2}^5 - {r_1}^5}{{r_2}^3 - {r_1}^3}\right) \end{aligned}\] |

All axes through $C$ have the same moment of inertia.

To compute the integrals in #rem-ec it is convenient to switch to spherical coordinates: \[\begin{aligned} x &= r \cos\theta \sin\phi \\ y &= r \sin\theta \sin\phi \\ z &= r \cos\phi. \end{aligned}\] To find the Jacobian of this coordinate transformation we use the coordinate order $(r,\phi,\theta)$ to give a right-handed spherical system. Then: \[\begin{aligned} J &= \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \phi} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \phi} & \frac{\partial y}{\partial \theta} \\ \frac{\partial z}{\partial r} & \frac{\partial z}{\partial \phi} & \frac{\partial z}{\partial \theta} \end{vmatrix} \\ &= \begin{vmatrix} \cos\theta \sin\phi & r \cos\theta \cos\phi & -r \sin\theta \sin\phi \\ \sin\theta \sin\phi & r \sin\theta \cos\phi & r \cos\theta \sin\phi \\ \cos\phi & -r \sin\phi & 0 \end{vmatrix} \\ &= \cos\phi (r^2 \cos^2\theta \cos\phi \sin\phi + r^2 \sin^2\theta \cos\phi \sin\phi) \\ &\quad + r \sin\phi (r \cos^2\theta \sin^2\phi + r \sin^2\theta \sin^2\phi) \\ &= r^2 \cos^2\phi \sin\phi + r^2 \sin^3\phi \\ &= r^2 \sin\phi. \end{aligned}\] Due to spherical symmetry, all axes through $C$ will have the same moment of inertia. We will compute $I_{C,z}$, which is: \[\begin{aligned} I_{C,z} &= \iiint_\mathcal{B} \rho (x^2 + y^2) \,dx\,dy\,dz \\ &= \int_0^{\pi} \int_{-\pi}^{\pi} \int_{r_1}^{r_2} \rho \left( (r \cos\theta \sin\phi)^2 + (r \sin\theta \sin\phi)^2 \right) \,J \,dr\,d\theta\,d\phi \\ &= \int_0^{\pi} \int_{-\pi}^{\pi} \int_{r_1}^{r_2} \rho r^4 \sin^3\phi \,dr\,d\theta\,d\phi \\ &= \int_0^{\pi} \int_{-\pi}^{\pi} \left[ \rho \frac{1}{5} r^5 \sin^3\phi \right]_{r = r_1}^{r = r_2} \,d\theta\,d\phi \\ &= \int_0^{\pi} \int_{-\pi}^{\pi} \rho \frac{1}{5} ({r_2}^5 - {r_1}^5) \sin^3\phi \,d\theta\,d\phi \\ &= \int_0^{\pi} \left[ \rho \frac{1}{5} ({r_2}^5 - {r_1}^5) \sin^3\phi \, \theta \right]_{\theta = -\pi}^{\theta = \pi} \,d\phi \\ &= \int_0^{\pi} \rho \frac{2\pi}{5} ({r_2}^5 - {r_1}^5) \sin^3\phi \,d\phi \\ &= \int_0^{\pi} \rho \frac{2\pi}{5} ({r_2}^5 - {r_1}^5) \frac{1}{4} (3\sin\phi - \sin 3\phi) \,d\phi \\ &= \left[ \rho \frac{2\pi}{5} ({r_2}^5 - {r_1}^5) \frac{1}{4} \left(\frac{1}{3} \cos 3\phi - 3\cos\phi\right) \right]_{\phi = 0}^{\phi = \pi} \\ &= \rho \frac{2\pi}{5} ({r_2}^5 - {r_1}^5) \frac{1}{4} \left(-\frac{2}{3} + 6\right) \\ &= \frac{8}{15} \rho \pi ({r_2}^5 - {r_1}^5). \end{aligned}\] The total mass of the spherical shell is $m = \rho \left(\frac{4}{3} \pi {r_2}^3 - \frac{4}{3} \pi {r_1}^3\right)$, so we can write the moment of inertia as \[\begin{aligned} I_{C,z} &= \frac{8}{15} m \frac{3}{4} \frac{1}{{r_2}^3 - {r_1}^3} ({r_2}^5 - {r_1}^5) \\ &= \frac{2}{5} m \left(\frac{{r_2}^5 - {r_1}^5}{{r_2}^3 - {r_1}^3}\right). \end{aligned}\]

## Simplified shapes

The moments of inertia listed below are all special cases of the basic shapes given in Section #rem-sb. Other special cases can be easily obtained by similar methods.

Rod: moments of inertia

\[\begin{aligned} I_{C,z} &= \frac{1}{12} m \ell^2 \\ I_{P,z} &= \frac{1}{3} m \ell^2 \\ I_{C,x} &= I_{P,x} = 0 \end{aligned}\] |

A rod is assumed to be a shape with infinitesimally small cross-section. We can use the formula #rem-ey for a cylinder with zero radii $r_1 = r_2 = 0$. The coordinates here are set up so that $z$ is orthogonal to the rod axis and $x$ is along the axis, so #rem-ey gives: \[\begin{aligned} I_{C,z} &= \frac{1}{12} m (3({r_1}^2 + {r_2}^2) + \ell^2) = \frac{1}{12} m \ell^2 \\ I_{C,x} &= \frac{1}{2} m ({r_1}^2 + {r_2}^2) = 0. \end{aligned}\] To find the moment of inertia about the end point $P$ we can use the parallel axis theorem #rem-el. This results is: \[\begin{aligned} I_{P,z} &= I_{C,z} + m {r_{CP}}^2 \\ &= \frac{1}{12} m \ell^2 + m \left(\frac{\ell}{2}\right)^2 \\ &= \frac{1}{3} m \ell^2. \end{aligned}\] The moment of inertia $I_{P,x}$ is still zero, because $\vec{r}_{CP}$ is parallel to $x$.

Solid cylinder or disk: moments of inertia

\[\begin{aligned} I_{C,x} &= I_{C,y} = \frac{1}{12} m (3r^2 + \ell^2) \\ I_{C,z} &= \frac{1}{2} m r^2 \end{aligned}\] |

The solid cylinder expressions are simply the cylindrical thick shell formulas #rem-ey with the inner radius set to $r_1 = 0$ and the outer radius $r_2 = r$.

Hollow cylinder or hoop: moments of inertia

\[\begin{aligned} I_{C,x} &= I_{C,y} = \frac{1}{12} m (6 r^2 + \ell^2) \\ I_{C,z} &= m r^2 \end{aligned}\] |

The hollow cylinder expressions can be found from cylindrical thick shell formulas #rem-ey by taking the same value for both inner and outer radii, so that $r_1 = r_2 = r$.

Solid ball: moments of inertia

\[\begin{aligned} I_C &= \frac{2}{5} m r^2 \end{aligned}\] |

All axes through $C$ have the same moment of inertia.

The solid ball moment of inertia can be directly obtained from the spherical thick shell expression #rem-es with inner radius $r_1 = 0$ and outer radius $r_2 = r$.

Hollow sphere: moments of inertia

\[\begin{aligned} I_C &= \frac{2}{3} m r^2 \end{aligned}\] |

All axes through $C$ have the same moment of inertia.

A hollow sphere moment of inertia is the same as that for the spherical thick shell #rem-es with inner radius and outer radius both set to $r_1 = r_2 = r$. Some care is needed here, however, as a simple substitution into the spherical thick shell expression would give the undefined value $0 / 0$.

Instead we need to set $r_2 = r$ and then to take the limit as $r_1 \to r$ using l'Hôpital's rule: \[\begin{aligned} I_C &= \lim_{r_1 \to r} \frac{2}{5} m \left(\frac{r^5 - {r_1}^5}{r^3 - {r_1}^3}\right) \\ &= \lim_{r_1 \to r} \frac{2}{5} m \left(\frac{\frac{d}{d r_1}\left(r^5 - {r_1}^5\right)}{ \frac{d}{d r_1}\left(r^3 - {r_1}^3\right)}\right) \\ &= \lim_{r_1 \to r} \frac{2}{5} m \left(\frac{5 {r_1}^4}{3 {r_1}^2}\right) \\ &= \lim_{r_1 \to r} \frac{2}{3} m {r_1}^2 \\ &= \frac{2}{3} m r^2. \end{aligned}\]