10.4: Solving the planar EoM for rigid bodies
After having discussed Euler’s second law, we will now show how it is used to predict rotational motion of rigid bodies. Together with Euler’s first law \(\left(\sum_{i} \overrightarrow{\boldsymbol{F}}_{i, \mathrm{ext}}=m_{\mathrm{tot}} \overrightarrow{\boldsymbol{a}}_{G}\right.\), Equation 8.11), it provides three scalar equations of motion (EoM) to analyse the planar dynamics of rigid bodies.
With Euler’s laws, the translation and rotation of rigid bodies can be solved for, using very similar methods as were outlined in Sec. 6.4 for the kinetic analysis of point masses. The main novel aspect is the determination and solution of the rotational EoM that follows from Euler’s second law, including the moments and angular accelerations. Let us summarise the procedure for solving the equations of motion for the planar kinetics of a rigid body, step-by-step:
- Sketch the rigid bodies, massless mechanisms, force vectors, moments, distances, dimensions and constraints.
- Choose and draw a suitable coordinate system (CS) and reference point \(P\) for moments.
- Determine the constraint equations.
- Determine the CoM \(\overrightarrow{\boldsymbol{r}}_{G}\) and moment of inertia \(I_{G}\) for each rigid body.
- Draw the free-body diagram (FBD) for each rigid body. Add both forces and moments \({ }^{1}\).
- Project force, moment and kinematic vectors onto the CS to obtain scalar components (like \(F_{x}, a_{G, x}, M_{z}\) and \(\alpha_{z}\) ).
- Determine the 3 equations of motion per rigid body. Two equations, for \(a_{G, x}\) and \(a_{G, y}\) from Euler’s first law, and one from Euler’s second law to determine \(\alpha_{z}\).
- Simplify the EoM by combining them and using constraint equations.
- Solve the simplified (differential) equations of motion (EoM), determining velocity and motion using integration or other methods.
This procedure follows to a large extent the methodologies for point masses that were discussed in Ch. 6. Nevertheless, there are some new aspects that need to be considered for rotating rigid bodies like:
- Selecting the easiest point of reference \(P\) for analysing the problem. In principle every point can be chosen, but the analysis can be facilitated if it allows simplifying moment or kinematic expressions (Sec. 9.5.3). The choice of the point \(P\) will not affect the dynamics, a smart choice can however simplify the calculation.
- For Euler’s second law to hold, the reference point \(P\) should be a fixed point in an IRF, any fixed point \(P\) in an IRF can be chosen. Alternatively the point \(G\) that moves along with the CoM of the rigid body, or system of point masses, can be chosen.
- It is important to use the same reference point \(P\) both for the moment and the angular momentum.
- Often, one first needs to use Euler’s first law and solve the corresponding EoMs to determine \(\overrightarrow{\boldsymbol{a}}_{G}\), before the term \(\overrightarrow{\boldsymbol{r}}_{G / P} \times\left(m \overrightarrow{\boldsymbol{a}}_{G}\right)\) in Euler’s second law Equation 10.22 can be evaluated to solve for the angular acceleration.
We are now going to solve the problem from Example 10.1 using Euler’s laws. Determine the angular acceleration of the dumbbell-shaped rigid body \(C\) in Figure 10.5.
Solution
We assume that the two masses are identical \(m_{A}=m_{B}=m\) and follow the steps introduced in this chapter.
- Step 1 and 2: First we use the sketch in Figure 10.5 and also indicate the Cartesian coordinate system and corresponding unit vectors in it.
- Step 3 and 4: The motion is unconstrained and the equation for the CoM is:
\[\overrightarrow{\boldsymbol{r}}_{G}=\frac{m_{A} \overrightarrow{\boldsymbol{r}}_{A}+m_{B} \overrightarrow{\boldsymbol{r}}_{B}}{m_{A}+m_{B}}=\frac{m \cdot 0+m L \hat{\boldsymbol{\imath}}}{2 m}=\frac{1}{2} L \hat{\boldsymbol{\imath}} \tag{10.23} \label{10.23}\]
The moment of inertia with respect to the CoM can be shown to be \(I_{G}=\frac{1}{2} m L^{2}\). We choose the CoM as the reference point for rotations, so \(P=G\).
- Step 5: Now we draw the FBD in Figure 10.6 according to the guidelines given in Sec. 6.8. It is important not to draw it in the ’trivial’ horizontal position, but at an angle \(\phi\).
- Step 6: We project the forces, moments and kinematic vectors: \(\overrightarrow{\boldsymbol{F}}_{B}=-F_{B} \hat{\boldsymbol{\jmath}}\), \(\overrightarrow{\boldsymbol{M}}_{B / G}=\overrightarrow{\boldsymbol{r}}_{B / G} \times \overrightarrow{\boldsymbol{F}}_{B}=-r_{B / G, x} F_{B} \hat{\boldsymbol{k}}=-\frac{L}{2} \cos (\phi) F_{B} \hat{\boldsymbol{k}}\). Furthermore we have \(\overrightarrow{\boldsymbol{a}}_{G}=a_{G, x} \hat{\boldsymbol{\imath}}+a_{G, y} \hat{\boldsymbol{\jmath}}\) and \(\overrightarrow{\boldsymbol{\alpha}}_{C}=\alpha_{C} \hat{\boldsymbol{k}}\).
- Step 7: We first apply Euler’s first law \(\sum \overrightarrow{\boldsymbol{F}}=m_{\text {tot }} \overrightarrow{\boldsymbol{a}}_{G}\).
\[\begin{align} \sum F_{x}=0 & =2 m a_{G, x} \tag{10.24} \label{10.24}\\[4pt] \sum F_{y}=-F_{B} & =2 m a_{G, y} \tag{10.25} \label{10.25}\end{align}\]
Now we apply Euler’s second law \(M_{G} \hat{\boldsymbol{k}}=\overrightarrow{\boldsymbol{r}}_{G / P} \times\left(m_{\text {tot }} \overrightarrow{\boldsymbol{a}}_{G}\right)+I_{G} \alpha \hat{\boldsymbol{k}}\). Since \(P=G\) the vector \(\overrightarrow{\boldsymbol{r}}_{G / G}=\overrightarrow{\mathbf{0}}\). Thus we have:
\[-\frac{L}{2} \cos (\phi) F_{B} \hat{\boldsymbol{k}}=I_{G} \alpha_{C} \hat{\boldsymbol{k}} \tag{10.26} \label{10.26}\]
- Step 8: By rewriting the last 3 equations, the simplified EoMs become:
\[\begin{align} a_{G, x}=\ddot{x}_{G} & =0 \tag{10.27} \label{10.27}\\[4pt] a_{G, y}=\ddot{y}_{G} & =-F_{B} /(2 m) \tag{10.28} \label{10.28}\\[4pt] \alpha_{C}=\ddot{\phi} & =\frac{-\frac{L}{2} \cos (\phi) F_{B}}{\frac{1}{2} m L^{2}}=\frac{-F_{B}}{m L} \tag{10.29} \label{10.29}\end{align}\]
where we used in the last step that \(I_{G}=\frac{1}{2} m L^{2}\) and \(\phi=0\) in our case. Note that the result for \(\alpha_{C}\) is identical to that in Equation 10.11, as expected.
- Step 9: In step 8 we have determined \(\alpha_{C}\) as requested. To obtain the full time dependence of \(x_{G}(t), y_{G}(t)\) and \(\phi_{C}(t)\) we would have to solve the EoMs differential equations.