10.5: Moment of inertia

10.5.1 Inertia tensor

In Sec. 10.3.2 we have introduced the moment of inertia to simplify the description of the angular momentum of rigid bodies. Let’s discuss it in more detail now.

Definition: Inertia tensor

The inertia tensor \(\mathbf{I}_{G}\) of a rigid body with respect to its CoM is defined by the equation:

\[\overrightarrow{\boldsymbol{L}}_{G}=\mathbf{I}_{G} \overrightarrow{\boldsymbol{\omega}}=\left[\begin{array}{ccc} I_{x x} & I_{x y} & I_{x z} \tag{10.30} \label{10.30}\\[4pt] I_{y x} & I_{y y} & I_{y z} \\[4pt] I_{z x} & I_{z y} & I_{z z} \end{array}\right] \overrightarrow{\boldsymbol{\omega}}\]

The diagonal components of the inertia tensor are called the moments of inertia, and as will be derived later in Equation 10.79 they can be calculated as follows:

\[I_{G, z z}=\sum_{i} m_{i} \rho_{i / G}^{2} \tag{10.31} \label{10.31}\]

The off-diagonal terms of the inertia tensor are called products of inertia. As an example \(I_{G, z z}\) can be written as:

\[I_{G, x z}=-\sum_{i} m_{i} x_{i / G} z_{i / G} \tag{10.32} \label{10.32}\]

It can be seen that if the rigid body has mirror symmetry in the \(z_{i / G}=0\) plane, or if it is axisymmetric around the \(z\)-axis, the products of inertia \(I_{G, z x}\) and \(I_{G, z y}\) are zero. E.g. because the contributions for \(z_{i / G}\) and \(-z_{i / G}\) cancel each other. In this textbook we focus only on rigid bodies with either of those properties, such that we only need to calculate the moment of inertia \(I_{z z}\).

Figure 10.6: FBD of a rigid body \(C\) on which a force \(\overrightarrow{\boldsymbol{F}}_{B}\) acts.

10.5.2 Calculation of the moment of inertia

The moment inertia of a rigid body can be calculated by summing the contribution \(m_{i} \rho_{i / G}^{2}\) of every point mass \(i\), however often we are dealing with continuous materials with a constant mass density \(\rho_{m}\) in \(\mathrm{kg} / \mathrm{m}^{3}\). In those cases the body can be subdivided into infinitesimal volumes \(\mathrm{d} V\) with mass \(\mathrm{d} m=\rho_{m} \mathrm{~d} V\) and contribution to the moment of inertia \(\rho_{i / G}^{2} \mathrm{~d} m\). The moment of inertia can then be calculated by integration over the total volume \(V\) of the rigid body.

\[I_{G, z z}=\sum_{i} m_{i} \rho_{i / G}^{2}=\int_{V} \rho_{m} \rho_{i / G}^{2} \mathrm{dV} \tag{10.33} \label{10.33}\]

Note that \(\rho_{i / G}\) is the distance of a point \(i\) from the axis parallel to the \(z\)-axis that goes through \(G\). Let us give a few examples of how this integral can be performed, where the CoM is positioned in the origin of the coordinate system.

Example 10.3: Moment of inertia of a block

For a block with dimensions \(L_{x}, L_{y}, L_{z}\) along the three axes, we use \(\rho^{2}=x^{2}+y^{2}\) to determine its moment of inertia by integrating over its volume along each of the three axes.

Solution

\[\begin{align} I_{G, z z} & =\rho_{m} \int_{-L_{y} / 2}^{L_{y} / 2} \mathrm{~d} y \int_{-L_{x} / 2}^{L_{x} / 2}\left(x^{2}+y^{2}\right) \mathrm{d} x \int_{-L_{z} / 2}^{L_{z} / 2} \mathrm{~d} z \tag{10.34} \label{10.34}\\[4pt] & =\rho_{m} L_{z} \int_{-L_{y} / 2}^{L_{y} / 2}\left[\frac{1}{3} x^{3}+y^{2} x\right]_{-L_{x} / 2}^{L_{x} / 2} \mathrm{~d} y \tag{10.35} \label{10.35}\\[4pt] & =\rho_{m} L_{z}\left[\frac{1}{12} L_{x}^{3} y+\frac{1}{3} y^{3} L_{x}\right]_{-L_{y} / 2}^{L_{y} / 2} \tag{10.36} \label{10.36}\\[4pt] & =\frac{1}{12} \rho_{m}\left(L_{z} L_{x} L_{y}\right)\left[L_{x}^{2}+L_{y}^{2}\right]=\frac{1}{12} m\left[L_{x}^{2}+L_{y}^{2}\right] \tag{10.37} \label{10.37}\end{align}\]

Note that the triple integral is solved in three steps, starting with the innermost integral going from right to left, first integrate over \(z\), then over \(x\) and finally over y. Multiplicative terms over which we don’t integrate can be moved to the left, out of the integral. In the last step we used that mass is density times volume \(m=\rho_{m} L_{z} L_{x} L_{y}\).

Moment of inertia of axisymmetric bodies

When an object is axisymmetric, it is easiest to determine its moment of inertia by using cylindrical coordinates. An infinitesimal volume element in these coordinates has a volume \(\mathrm{d} V=\rho \mathrm{d} \phi \mathrm{d} \rho \mathrm{d} z\), where the size in the \(\hat{\phi}\) direction is given by the infinitesimal circle segment \(\rho \mathrm{d} \phi\). The outer surface of an
axisymmetric object can be described by a function \(\rho_{d}(z)\) or alternatively by \(z_{h}(\rho)\). We will show two examples to illustrate integration of \(\rho_{m} \rho_{i / G}\) over either volume to obtain the moment of inertia of a cone with radius \(R\) and total height \(h\).

Example 10.4: Moment of inertia of a cone method 1

The height of the cone at each \(\rho\) can be described by \(z_{h}(\rho)=h\left(1-\frac{\rho}{R}\right)\), such that \(I_{z z}\) becomes:

\[\begin{align} I_{G, z z} & =\rho_{m} \int_{0}^{R} \int_{0}^{z_{h}(\rho)} \rho^{2} \int_{0}^{2 \pi} \rho \mathrm{d} \phi \mathrm{d} \rho \mathrm{d} z=2 \pi \rho_{m} \int_{0}^{R} \rho^{3}[z]_{0}^{h\left(1-\frac{\rho}{R}\right)} \mathrm{d} \rho \\[4pt] & =2 \pi \rho_{m} \int_{0}^{R} \rho^{3} h\left(1-\frac{\rho}{R}\right) \mathrm{d} \rho=2 \pi \rho_{m} h\left[\frac{1}{4} \rho^{4}-\frac{1}{5 R} \rho^{5}\right]_{0}^{R} \\[4pt] & =\frac{2}{20} \pi \rho_{m} h R^{4}=\frac{3}{10} m R^{2} \tag{10.38} \label{10.38}\end{align}\]

where we used in the last step that the volume of a cone is \(V=\frac{h}{3} \pi R^{2}\).

Example 10.5: Moment of inertia of a cone method 2

Alternatively, the radius of the disks of the cone depends on height as \(\rho_{d}(z)=R\left(1-\frac{z}{h}\right)\). We now again perform the integral:

\[\begin{align} I_{G, z z} & =\rho_{m} \int_{0}^{h} \int_{0}^{\rho_{d}(z)} \rho^{2} \int_{0}^{2 \pi} \rho \mathrm{d} \phi \mathrm{d} \rho \mathrm{d} z=2 \pi \rho_{m} \int_{0}^{h} \int_{0}^{R\left(1-\frac{z}{h}\right)} \rho^{3} \mathrm{~d} \rho \mathrm{d} z \\[4pt] & =2 \pi \rho_{m} \int_{0}^{h}\left[\frac{1}{4} \rho^{4}\right]_{0}^{R\left(1-\frac{z}{h}\right)} \mathrm{d} z=\frac{1}{2} \pi \rho_{m} \frac{R^{4}}{h^{4}} \int_{0}^{h}(h-z)^{4} \mathrm{~d} z \\[4pt] & =\frac{1}{2} \pi \rho_{m} \frac{R^{4}}{h^{4}}\left[-\frac{1}{5}(h-z)^{5}\right]_{0}^{h} \mathrm{~d} z \tag{10.39} \label{10.39}\\[4pt] & =\frac{1}{10} \pi \rho_{m} h R^{4}=\frac{3}{10} m R^{2} \tag{10.40} \label{10.40}\end{align}\]

We obtain the same result, as expected. Determining the moment of inertia of rigid bodies is a good practice in volume integration skills.

10.5.3 Parallel axis theorem

We have now a method to calculate the moment of inertia of a rigid body with respect to its CoM. But often a reference point \(P\) is chosen that is different from \(G\). In that case one can redo the integrals again and obtain \(I_{P, z z}\), but there is an easier way to determine \(I_{P, z z}\) from \(I_{G, z z}\), because as will be shown in Equation 10.72, the angular momentum with respect to a point \(P\) can be written as
the sum of two parts, namely a component representing the angular momentum a point mass \(m_{\text {tot }}\) at position \(G\) and a component representing the angular momentum with respect to \(G\) :

\[\begin{align} \overrightarrow{\boldsymbol{L}}_{C / P} & =\overrightarrow{\boldsymbol{L}}_{G / P}+\sum_{i} \overrightarrow{\boldsymbol{L}}_{i / G} \tag{10.41} \label{10.41}\\[4pt] \overrightarrow{\boldsymbol{L}}_{C / P, 2 D} & =\left(m_{\mathrm{tot}} \rho_{G / P}^{2}+I_{G, z z}\right) \overrightarrow{\boldsymbol{\omega}}_{C} \tag{10.42} \label{10.42}\\[4pt] \overrightarrow{\boldsymbol{L}}_{C / P, 2 D} & =I_{P, z z} \overrightarrow{\boldsymbol{\omega}}_{C} \tag{10.43} \label{10.43}\end{align}\]

Now from the last two equations we can determine \(I_{P, z z}\), this method is called the parallel axis theorem or Steiner’s theorem.

Concept: Parallel axis theorem

The moment of inertia of a rigid body around an axis through a point \(P\), at a distance \(\rho_{G / P}\) from its CoM, equals the moment of inertia \(m_{\mathrm{tot}} \rho_{G / P}^{2}\) of its CoM, plus the moment of inertia of the rigid body around a parallel axis through its CoM.

\[I_{z z, P}=m_{\mathrm{tot}} \rho_{G / P}^{2}+I_{G, z z} \tag{10.44} \label{10.44}\]

Example 10.6

Determine the moment inertia \(I_{z z, P}\) of the rectangular rigid body with mass \(m\) with respect to reference point \(P\) at the origin.

Since the rigid body lies in the \(x y\)-plane we have \(\rho_{G / P}=\left|\overrightarrow{\boldsymbol{r}}_{G / P}\right|\). The moment of inertia \(I_{G, z z}\) of a rectangle can be determined using Equation 10.37. Then with the parallel axis theorem we find:

\[I_{P, z z}=m \rho_{G / P}^{2}+\frac{1}{12} m\left(L_{x}^{2}+L_{y}^{2}\right) \tag{10.45} \label{10.45}\]

10.5.4 Adding and subtracting moments of inertia

Since the moment of inertia is a sum or integral over individual contributions for every point mass, the moment of inertia of a rigid body \(C\) that is made up out of two (or more) other rigid bodies \(A\) and \(B\) can be calculated by summing the moments of inertia around the same axis \(P\):

\[I_{C, P, z z}=I_{A, P, z z}+I_{B, P, z z} \tag{10.46} \label{10.46}\]

Make sure you use the same reference point \(P\) when making such additions. Similarly, it is also possible to subtract moments of inertia from one another, for instance to determine the moment of inertia of objects that have a hole.

Figure 10.7: With the parallel axis theorem the moment of inertia with respect to an arbitrary axis \(P\) that is not the CoM of the rigid body can be determined.