10.1: Effects of a force on a rigid body
A rigid body is nothing else than a collection of many point masses that are rigidly connected and relatively constrained. So it can be analysed by the techniques we discussed earlier in Ch. 6. In this chapter we will introduce new techniques to facilitate such analysis, but before introducing these new techniques, let us give an example where we analyse the kinetics of a rigid body using the techniques from Ch. 6.
M Example 10.1 Problem: Consider the rigid body \(C\) in Figure 10.1 that consists of 2 point masses \(A\) and \(B\) connected by a rod of length \(L\). A force \(\overrightarrow{\boldsymbol{F}}_{B}=-F_{B} \hat{\boldsymbol{\jmath}}\) acts on mass \(B\).
- Determine the angular acceleration of \(C\) if the horizontal rigid body, with \(\phi=0\) is released from rest.
To solve this problem we first consider the constraint equation of the rod (constant length \(L)\) and take its time derivative twice:
\[\begin{align} \left|\overrightarrow{\boldsymbol{r}}_{B / A}\right|^{2} & =L^{2} \tag{10.1} \label{10.1}\\[4pt] x_{B / A}^{2}+y_{B / A}^{2} & =L^{2} \tag{10.2} \label{10.2}\\[4pt] 2 x_{B / A} \dot{x}_{B / A}+2 y_{B / A} \dot{y}_{B / A} & =0 \tag{10.3} \label{10.3}\\[4pt] \dot{x}_{B / A}^{2}+x_{B / A} \ddot{x}_{B / A}^{2}+\dot{y}_{B / A}^{2}+y_{B / A} \ddot{y}_{B / A}^{2} & =0 \tag{10.4} \label{10.4}\\[4pt] x_{B}\left(\ddot{x}_{B}-\ddot{x}_{A}\right)^{2} & =0 \tag{10.5} \label{10.5}\\[4pt] \ddot{x}_{B} & =\ddot{x}_{A}=0 \tag{10.6} \label{10.6}\end{align}\]
In the last steps we used that for the current position and at zero velocity the constants \(y_{B / A}=x_{A}=\dot{x}_{B / A}=\dot{y}_{B / A}=0\) are zero. The last equation shows that if the rod is horizontal and has zero velocity, then the constraint equalises accelerations of the point masses in the \(x\) direction, but not in the \(y\) direction. Since there are no forces acting in the \(x\)-direction, we conclude that the masses don’t accelerate in that direction.
Now we write down the equation of motion in the \(y\) direction for mass \(A\) and mass \(B\) :
\[\begin{align} \sum F_{A, y} & =0=m_{A} \ddot{y}_{A} \tag{10.7} \label{10.7}\\[4pt] \sum F_{B, y} & =-F_{B}=m_{B} \ddot{y}_{B} \tag{10.8} \label{10.8}\end{align}\]
Now the final step is to determine the angular acceleration \(\alpha_{C}\) using the kinematic equations from the previous chapter. Using that \(\overrightarrow{\boldsymbol{a}}_{A}=\overrightarrow{\mathbf{0}}, \omega_{C}=0\) and \(\hat{\boldsymbol{\phi}}_{A}=\hat{\boldsymbol{\jmath}}\) we get:
\[\begin{align} \overrightarrow{\boldsymbol{a}}_{B, 2 D} & =\overrightarrow{\boldsymbol{a}}_{A}+\alpha_{C}\left|\overrightarrow{\boldsymbol{r}}_{B / A}\right| \hat{\boldsymbol{\phi}}_{A}-\omega_{C}^{2} \overrightarrow{\boldsymbol{r}}_{B / A} \tag{10.9} \label{10.9}\\[4pt] -F_{B} / m_{B} \hat{\boldsymbol{\jmath}} & =\alpha_{C} L \hat{\boldsymbol{\jmath}} \tag{10.10} \label{10.10}\\[4pt] \alpha_{C} & =-F_{B} /\left(m_{B} L\right) \tag{10.11} \label{10.11}\end{align}\]
Although this example demonstrates that it is possible to analyse the the effect of forces on the dynamics of rigid bodies with Newton’s laws, both translation and rotation, it becomes very elaborate for rigid bodies consisting out of many point masses. In this chapter we will derive a simpler method to do this.