10.10: Derivation of Euler's second law
After having introduced the equations for angular momentum and moment of inertia, we are now ready to analyse the effect of moments on the rotation of rigid bodies, and will show how Euler’s second law that was introduced in Equation 10.21, can be derived.
10.10.1 Euler’s second law
We first extend Equation 10.15 to a system of point masses \(C\), with respect to the same reference point \(P\). The resultant moment \(\overrightarrow{\boldsymbol{M}}_{C / P}\) on \(C\) is then given by:
\[\overrightarrow{\boldsymbol{M}}_{C / P}=\sum_{i} \overrightarrow{\boldsymbol{M}}_{i / P}=\sum_{i} \overrightarrow{\boldsymbol{r}}_{i / P} \times\left(m_{i} \overrightarrow{\boldsymbol{a}}_{i}\right) \tag{10.84} \label{10.84}\]
To see how the right term of this equation is related to the properties of the total angular momentum \(\overrightarrow{\boldsymbol{L}}_{C / P}\) of \(C\), we compare it to the time derivative of Equation 10.66:
\[\begin{align} \frac{\mathrm{d}}{\mathrm{d} t} \overrightarrow{\boldsymbol{L}}_{C / P} & =\frac{\mathrm{d}}{\mathrm{d} t} \sum_{i} \overrightarrow{\boldsymbol{L}}_{i / P}=\frac{\mathrm{d}}{\mathrm{d} t} \sum_{i} \overrightarrow{\boldsymbol{r}}_{i / P} \times\left(m_{i} \overrightarrow{\boldsymbol{v}}_{i}\right) \tag{10.85} \label{10.85}\\[4pt] & =\sum_{i}\left[m_{i} \overrightarrow{\boldsymbol{v}}_{i / P} \times \overrightarrow{\boldsymbol{v}}_{i}+\overrightarrow{\boldsymbol{r}}_{i / P} \times\left(m_{i} \overrightarrow{\boldsymbol{a}}_{i}\right)\right] \tag{10.86} \label{10.86}\end{align}\]
To simplify this equation further, we would like to eliminate the terms \(\overrightarrow{\boldsymbol{v}}_{i / P} \times \overrightarrow{\boldsymbol{v}}_{i}\). There are two ways to ensure this:
- Use a fixed reference point \(P\) that has zero velocity and acceleration \(\overrightarrow{\boldsymbol{v}}_{P}=\) \(\overrightarrow{\boldsymbol{a}}_{P}=\overrightarrow{\mathbf{0}}\) with respect to the origin of an IRF. Then \(\overrightarrow{\boldsymbol{v}}_{i}=\overrightarrow{\boldsymbol{v}}_{i / P}+\overrightarrow{\boldsymbol{v}}_{P}=\overrightarrow{\boldsymbol{v}}_{i / P}\).
Substitution into Equation 10.85 results in a term \(\overrightarrow{\boldsymbol{v}}_{i / P} \times \overrightarrow{\boldsymbol{v}}_{i / P}=\overrightarrow{\mathbf{0}}\). Thus for a fixed point \(P\) we obtain from Equation 10.86:
\[\frac{\mathrm{d}}{\mathrm{d} t} \sum_{i} \overrightarrow{\boldsymbol{L}}_{i / P}=\sum_{i} \overrightarrow{\boldsymbol{r}}_{i / P} \times\left(m_{i} \overrightarrow{\boldsymbol{a}}_{i}\right) \tag{10.87} \label{10.87}\]
- Choose the CoM as the reference point. Then we substitute \(\overrightarrow{\boldsymbol{v}}_{i}=\overrightarrow{\boldsymbol{v}}_{i / G}+\overrightarrow{\boldsymbol{v}}_{G}\) and get \(\sum_{i} m_{i} \overrightarrow{\boldsymbol{v}}_{i / G} \times\left(\overrightarrow{\boldsymbol{v}}_{i / G}+\overrightarrow{\boldsymbol{v}}_{G}\right)\), where the terms \(\overrightarrow{\boldsymbol{v}}_{i / G} \times \overrightarrow{\boldsymbol{v}}_{i / G}=\overrightarrow{\mathbf{0}}\) are zero and where the term \(\left[\sum_{i} m_{i}\left(\overrightarrow{\boldsymbol{v}}_{i}-\overrightarrow{\boldsymbol{v}}_{G}\right)\right] \times \overrightarrow{\boldsymbol{v}}_{G}\) is also zero because \(\sum_{i} m_{i} \overrightarrow{\boldsymbol{v}}_{i}=m_{\text {tot }} \overrightarrow{\boldsymbol{v}}_{G}\). Then we obtain from Equation 10.86 and that equation is even valid if \(G\) has a nonzero velocity or acceleration:
\[\frac{\mathrm{d}}{\mathrm{d} t} \sum_{i} \overrightarrow{\boldsymbol{L}}_{i / G}=\sum_{i} \overrightarrow{\boldsymbol{r}}_{i / G} \times\left(m_{i} \overrightarrow{\boldsymbol{a}}_{i}\right) \tag{10.88} \label{10.88}\]
Since the right side of Equation 10.84 is identical to that of Equation 10.87 and Equation 10.88, we obtain Euler’s second law.
Euler’s second law states that the total external moment acting on \(C\) equals the time derivative of the total angular momentum of \(C\), as long as a reference point \(P\) is chosen that is fixed in an IRF or the CoM \(G\) is used as reference point.
\[\begin{align} \overrightarrow{\boldsymbol{M}}_{C / P, \text { ext }} & =\frac{\mathrm{d}}{\mathrm{d} t} \overrightarrow{\boldsymbol{L}}_{C / P} \tag{10.89} \label{10.89}\\[4pt] \overrightarrow{\boldsymbol{M}}_{C / G, \mathrm{ext}} & =\frac{\mathrm{d}}{\mathrm{d} t} \overrightarrow{\boldsymbol{L}}_{C / G} \tag{10.90} \label{10.90}\end{align}\]
This equation relates moment and angular momentum in a similar way as Newton’s second law \(\overrightarrow{\boldsymbol{F}}=\frac{\mathrm{d} \overrightarrow{\boldsymbol{p}}}{\mathrm{d} t}=m \overrightarrow{\boldsymbol{a}}\) relates force and momentum. The subscript ext indicates that only external moments contribute to the resultant moment on the rigid body.
10.10.2 Euler’s second law for a rigid body
Euler’s second law can be applied to rigid bodies by using the properties of the angular momentum of rigid bodies that we have derived earlier in this chapter in Equation 10.71 and the inertia tensor Equation 10.30, which can be combined to obtain the total angular momentum of a rigid body \(C\) :
\[\overrightarrow{\boldsymbol{L}}_{C, P}=\overrightarrow{\boldsymbol{r}}_{G / P} \times\left(m_{\mathrm{tot}} \overrightarrow{\boldsymbol{v}}_{G / P}\right)+\mathbf{I}_{C, G} \overrightarrow{\boldsymbol{\omega}}_{C} \tag{10.91} \label{10.91}\]
This equation is substituted in Equation 10.89. By taking the time derivative of Equation 10.91, using that \(\overrightarrow{\boldsymbol{a}}_{P}=\overrightarrow{\mathbf{0}}\) in an IRF and using that \(\overrightarrow{\boldsymbol{v}}_{G / P} \times \overrightarrow{\boldsymbol{v}}_{G / P}=\overrightarrow{\mathbf{0}}\) we obtain:
\[\sum_{i} \overrightarrow{\boldsymbol{M}}_{C, i / P, \mathrm{ext}}=\overrightarrow{\boldsymbol{r}}_{G / P} \times\left(m_{\mathrm{tot}} \overrightarrow{\boldsymbol{a}}_{G}\right)+\frac{\mathrm{d}}{\mathrm{d} t}\left(\mathbf{I}_{C, G}\right) \overrightarrow{\boldsymbol{\omega}}_{C}+\mathbf{I}_{C, G} \overrightarrow{\boldsymbol{\alpha}}_{C} \tag{10.92} \label{10.92}\]
Equation 10.92 shows that the resultant moment acting on a rigid body equals the sum of the of three components. We will discuss all three of them, in particular their relevance for the analysis of the planar kinetics of rigid bodies.
First there is the term \(\overrightarrow{\boldsymbol{r}}_{G / P} \times\left(m_{\mathrm{tot}} \overrightarrow{\boldsymbol{a}}_{G}\right)\), which is due to the angular momentum of the centre of mass \(\overrightarrow{\boldsymbol{L}}_{G / P}\) and becomes zero if the CoM is chosen as a reference point because \(\overrightarrow{\boldsymbol{r}}_{G / G}=\overrightarrow{\mathbf{0}}\). Note that from Equation 8.11 we have \(\sum_{i} \overrightarrow{\boldsymbol{F}}_{i, \text { ext }}=m_{\mathrm{tot}} \overrightarrow{\boldsymbol{a}}_{G}\), which can help to evaluate \(\overrightarrow{\boldsymbol{a}}_{G}\) and the contribution of this term. Note that if we are dealing with point masses, we only have this first term.
The second term is \(\frac{\mathrm{d}}{\mathrm{d} t} \mathbf{I}_{C, G} \overrightarrow{\boldsymbol{\omega}}_{C}\) is relevant when the inertia tensor changes in time. In the previous section we calculated \(\mathbf{I}_{C, G}\) with respect to a Cartesian coordinate system. When an object changes orientation with respect to that system, its inertia tensor can change because the moment and product of inertia integrals change. However, in 2D planar kinetics all objects move in the \(x y\)-plane and all forces are tangential to the \(x y\)-plane, such that all angular velocity and angular acceleration acceleration vectors point in the \(z\)-direction, and the same holds for the moment vectors. In planar kinetics, which is the focus of this textbook, the orientation of the rigid body therefore does not change such that we always have that the time derivative of the inertia tensor \(\frac{\mathrm{d}}{\mathrm{d} t} \mathbf{I}_{C, G}\) is zero, which substantially simplifies the dynamics, but eliminates the occurrence of special dynamic effects like precession that will be dealt with when treating 3D dynamics. Note that another way to ensure that the orientation of the rigid body with respect to the coordinate system does not change is to fix the coordinate system to the rigid body, and have it rotate along with it. This is often the method of choice for analysing more advanced rigid body dynamics problems but the drawback of that is that you are not in an IRF anymore.
Finally we have the term \(\mathbf{I}_{C, G} \overrightarrow{\boldsymbol{\alpha}}_{C}\). As discussed in the previous section, in planar kinetics we only deal with rigid bodies with diagonal inertia tensors, with both \(\overrightarrow{\boldsymbol{M}}_{C, P}\) and \(\overrightarrow{\boldsymbol{\alpha}}_{C}\) pointing along the \(z\) axis, such that \(I_{z z}\) is the only relevant term in the inertia tensor, such that this term becomes \(I_{z z} \alpha_{C} \hat{\boldsymbol{k}}\). From
these considerations, we can significantly simplify Euler’s second law for the case of \(2 \mathrm{D}\) motion in the \(x y\)-plane:
\[\sum_{i} M_{C, i / P, 2 D} \hat{\boldsymbol{k}}=\overrightarrow{\boldsymbol{r}}_{G / P} \times\left(m_{\mathrm{tot}} \overrightarrow{\boldsymbol{a}}_{G}\right)+I_{G, z z} \alpha_{C} \hat{\boldsymbol{k}} \tag{10.93} \label{10.93}\]
This important equation is much easier to deal with than Equation 10.92 and can be used to analyse many dynamic systems, as long as we keep in mind that it is only valid for \(2 \mathrm{D}\) planar systems.