10.6: Euler's second law for special cases

We now discuss a few special types of motion that can be analysed with Euler’s second law.

10.6.1 Pure translation

In the case of pure translation, we have no angular acceleration \(\overrightarrow{\boldsymbol{\alpha}}=\overrightarrow{\mathbf{0}}\), which is the case if the moment of all external forces with respect to the CoM is zero \(\left(\overrightarrow{\boldsymbol{M}}_{C, G}=\overrightarrow{\mathbf{0}}\right)\). This situation is for instance valid for the moments generated in a constant gravitational field, or when constraint equations prevent rotations. Then Equation 10.22 becomes:

\[\begin{align} & \text { If : } \sum_{i} \overrightarrow{\boldsymbol{M}}_{C, i / G}=I_{G, z z} \alpha_{C} \hat{\boldsymbol{k}}=\overrightarrow{\mathbf{0}} \tag{10.47} \label{10.47}\\[4pt] & \sum_{i} M_{C, i / P, 2 D} \hat{\boldsymbol{k}}=\overrightarrow{\boldsymbol{r}}_{G / P} \times\left(m_{\mathrm{tot}} \overrightarrow{\boldsymbol{a}}_{G}\right)=\overrightarrow{\boldsymbol{r}}_{G / P} \times\left(\sum_{i} \overrightarrow{\boldsymbol{F}}_{i, \mathrm{ext}}\right)\end{align}\]

It can be seen that once it is known that \(\overrightarrow{\boldsymbol{\alpha}}_{C}=\overrightarrow{\boldsymbol{0}}\), Euler’s second law on the moments does not provide additional useful information beyond Euler’s first law that determines the acceleration of the CoM of the rigid body via \(\sum_{i} \overrightarrow{\boldsymbol{F}}_{i, \mathrm{ext}}=m_{\mathrm{tot}} \overrightarrow{\boldsymbol{a}}_{G}\).

10.6.2 Pure rotation around the centre of mass

It is interesting to consider the situation where the term \(m_{\text {tot }} \overrightarrow{\boldsymbol{a}}_{G}=\overrightarrow{\boldsymbol{0}}\) is zero because the sum of forces on the rigid body is zero, since it significantly simplifies Equation 10.22, with the disappearance of the term \(\overrightarrow{\boldsymbol{r}}_{G / P} \times\left(m_{\mathrm{tot}} \overrightarrow{\boldsymbol{a}}_{G}\right)\).

\[\begin{align} \text { If : } \sum_{i} \overrightarrow{\boldsymbol{F}}_{\mathrm{i}, \mathrm{ext}} & =m_{\mathrm{tot}} \overrightarrow{\boldsymbol{a}}_{G}=\overrightarrow{\mathbf{0}} \tag{10.48} \label{10.48}\\[4pt] \text { Then }: \sum_{i} M_{C, i / P, 2 D} \hat{\boldsymbol{k}} & =I_{G, z z} \alpha_{C} \hat{\boldsymbol{k}} \tag{10.49} \label{10.49}\end{align}\]

We thus find in Equation 10.49 that the CoM of the rigid body does not accelerate and the acting moments purely cause an angular acceleration \(\alpha_{C}\) of the rigid body.

10.6.3 Couple

We saw in the previous subsection that a special situation arises if the sum of the forces acting on a system is zero, since then \(\overrightarrow{\boldsymbol{a}}_{G}=\overrightarrow{\mathbf{0}}\). Such a combination of forces is called a couple, and if there are only two forces it is called a simple couple.

Concept: Simple couple

\(A\) simple couple consists of two forces \(\overrightarrow{\boldsymbol{F}}_{A}=-\overrightarrow{\boldsymbol{F}}_{B}\) that are equal in magnitude, and opposite in direction such that \(\overrightarrow{\boldsymbol{F}}_{A}+\overrightarrow{\boldsymbol{F}}_{B}=\overrightarrow{\mathbf{0}}\), while the forces have different points of action, such that their resultant moment is not zero.

Besides simple couples, that consist of two forces, there is also a more general definition of a couple, that can consist of more than two forces.

Concept: Couple

A couple consists of a number of \(N\) forces for which the sum of forces is zero, that generate a resultant moment on a system. So \(\sum_{i=1}^{N} \overrightarrow{\boldsymbol{F}}_{i, \mathrm{ext}}=\overrightarrow{\mathbf{0}}\).

According to Euler’s second law, a couple only influences the angular acceleration \(\alpha\) of a rigid body, it does not contribute to the acceleration \(\overrightarrow{\boldsymbol{a}}_{G}\) of the CoM because the sum of the forces is zero. The effect of a couple is therefore also called a pure moment or couple moment.

Another useful property of a couple is that its resultant pure moment is independent of the choice of the location of the reference point \(P\), as we will derive now.

Derivation: Moment of a couple

Consider a simple couple generated by two force vectors with \(\overrightarrow{\boldsymbol{F}}_{A}=-\overrightarrow{\boldsymbol{F}}_{B}\) acting on points \(A\) and \(B\). The moment of this couple with respect to a reference point \(P\) is given by:

\[\begin{align} \overrightarrow{\boldsymbol{M}}_{\text {couple }} & =\overrightarrow{\boldsymbol{r}}_{A / P} \times \overrightarrow{\boldsymbol{F}}_{A}+\overrightarrow{\boldsymbol{r}}_{B / P} \times \overrightarrow{\boldsymbol{F}}_{B} \tag{10.50} \label{10.50}\\[4pt] & =\overrightarrow{\boldsymbol{r}}_{A / P} \times \overrightarrow{\boldsymbol{F}}_{A}-\overrightarrow{\boldsymbol{r}}_{B / P} \times \overrightarrow{\boldsymbol{F}}_{A} \tag{10.51} \label{10.51}\\[4pt] & =\left(\left(\overrightarrow{\boldsymbol{r}}_{A}-\overrightarrow{\boldsymbol{r}}_{P}\right)-\left(\overrightarrow{\boldsymbol{r}}_{B}-\overrightarrow{\boldsymbol{r}}_{P}\right)\right) \times \overrightarrow{\boldsymbol{F}}_{A} \tag{10.52} \label{10.52}\\[4pt] & =\overrightarrow{\boldsymbol{r}}_{A / B} \times \overrightarrow{\boldsymbol{F}}_{A} \tag{10.53} \label{10.53}\end{align}\]

From this equation we see that the couple vector \(\overrightarrow{\boldsymbol{M}}_{\text {couple }}\) is independent of the choice of the reference point \(P\) and only depends on the force vectors of the couple and their points of action \(\overrightarrow{\boldsymbol{r}}_{A}\) and \(\overrightarrow{\boldsymbol{r}}_{B}\).

10.6.4 Pure rotation around a fixed axis

Let us now consider pure rotation of a rigid body \(C\) about a fixed axis \(P\) parallel to the \(z\)-axis, which does not translate \(\left(\overrightarrow{\boldsymbol{a}}_{P}=\overrightarrow{\boldsymbol{v}}_{P}=\overrightarrow{\boldsymbol{0}}\right)\). We choose point \(P\) as reference point. There is no translation of this point, only the angular acceleration of the rigid body \(\alpha\). As a consequence we have from Equation 9.29 that \(\overrightarrow{\boldsymbol{a}}_{G}=\overrightarrow{\boldsymbol{\alpha}} \times \overrightarrow{\boldsymbol{r}}_{G / P}-\omega^{2} \overrightarrow{\boldsymbol{r}}_{G / P}\). Substituting this into Equation 10.22 we obtain:

\[\begin{align} & \text { If : } \overrightarrow{\boldsymbol{a}}_{P}=\overrightarrow{\boldsymbol{v}}_{P}=\overrightarrow{\mathbf{0}} \tag{10.54} \label{10.54}\\[4pt] \sum_{i} M_{C, i / P, 2 D} \hat{\boldsymbol{k}}= & m_{\mathrm{tot}} \overrightarrow{\boldsymbol{r}}_{G / P} \times\left(\overrightarrow{\boldsymbol{\alpha}} \times \overrightarrow{\boldsymbol{r}}_{G / P}-\omega^{2} \overrightarrow{\boldsymbol{r}}_{G / P}\right)+I_{G, z z} \alpha_{C} \hat{\boldsymbol{k}} \\[4pt] = & m_{\mathrm{tot}} \rho_{G / P} \hat{\boldsymbol{\rho}} \times\left(\alpha \hat{\boldsymbol{k}} \times \rho_{G / P} \hat{\boldsymbol{\rho}}\right)+I_{G, z z} \alpha_{C} \hat{\boldsymbol{k}} \\[4pt] = & \left(m_{\mathrm{tot}} \rho_{G / P}^{2}+I_{G, z z}\right) \alpha \hat{\boldsymbol{k}} \tag{10.55} \label{10.55}\\[4pt] = & I_{P, z z} \alpha \hat{\boldsymbol{k}} \tag{10.56} \label{10.56}\end{align}\]

In the last step we used the parallel axes theorem \(I_{z z, P}=m_{\mathrm{tot}} \rho_{G / P}^{2}+I_{G, z z}\), where \(\rho_{G / P}\) is the smallest distance between the parallel axes through \(G\) and \(P\). The conclusion from Equation 10.56 is that for a rigid body \(C\) that rotates in-plane around a fixed axis \(P\) : the sum of the moments with respect to the fixed axis \(P\) equals the product of the moment of inertia \(I_{z z, P}\) and the angular acceleration \(\alpha_{C}\) of the rigid body \(C\).