12.4: Using angular impulse and angular momentum
In contrast to a point mass, a rigid body has both translational and rotational degrees of freedom. By combining the methods from this chapter with those from Ch. 8, we can fully analyse the changes in both the velocities and angular velocities of rigid bodies. There are however two important limitations to these methods:
- Changes in position are not determined by this method, so the motion needs to be known in advance, or the impulse needs to be so short that the change in position can be neglected.
- Sufficient information about the forces and their points of action is needed to evaluate the angular impulse \(\overrightarrow{\boldsymbol{H}}_{\text {ang, } P, 12}\). This is substantially easier if the position of the point of action of the forces is approximately constant (e.g. in short collisions) or if only pure couples are acting for which the point of action is irrelevant.
Finally we note that in cases where (angular) impulse and momentum do not provide enough information, it can sometimes be combined with energy methods to solve a problem, e.g. by combining conservation of (angular) momentum with conservation of energy. Let us discuss an example.
Figure 12.1 shows a ball \(A\) (a point mass), with mass \(m\) that collides elastically with a dumbbell at rest \(B C\) that consists of two point masses \(B\) and \(C\) with identical mass \(m\). Determine the velocities and angular velocities of the objects after the collision.
Solution
To solve this problem we first note that because the collision is elastic, all forces are conservative forces. Also we note that there are no external moments or forces acting on the system (gravity is not indicated and can thus be neglected).
Secondly, we need to choose a reference point for analysing the rotations of the system. A logical choice is to use the point \(\overrightarrow{\boldsymbol{r}}_{B}\) as reference point. Since there are no external forces or moments acting on the system we have conservation of momentum and angular momentum, also we have conservation of energy. All forces and velocities are in the \(x\) direction, so we only need to consider their \(x\) components. This gives 3 equations, with three scalar unknowns after the collision: the velocity of mass A \(v_{A, 2, x}\), the velocity of the center of gravity of dumbbell \(B C v_{G, 2, x}\) and its angular velocity \(\omega_{2}\).
\[\begin{align} & m v_{A, 1, x}=m v_{A, 2, x}+2 m v_{G, 2, x} \tag{12.8} \label{12.8}\\[4pt] & \text { Conservation of momentum } \\[4pt] & \overrightarrow{\boldsymbol{r}}_{B / B} \times m \overrightarrow{\boldsymbol{v}}_{A, 1}=\overrightarrow{\boldsymbol{r}}_{B / B} \times m \overrightarrow{\boldsymbol{v}}_{A, 2}+\overrightarrow{\boldsymbol{r}}_{G / B} \times\left(m_{B C} \overrightarrow{\boldsymbol{v}}_{G, 2}\right)+I_{G} \omega_{2} \hat{\boldsymbol{k}} \\[4pt] & \text { Conservation of angular momentum } \\[4pt] & \overrightarrow{\boldsymbol{0}}=\overrightarrow{\boldsymbol{0}}-L / 2 \hat{\boldsymbol{\jmath}} \times\left(2 m v_{G, 2, x} \hat{\boldsymbol{\imath}}\right)+I_{G} \omega_{2} \hat{\boldsymbol{k}} \\[4pt] & 0=m L v_{G, 2, x}+I_{G} \omega_{2} \tag{12.9} \label{12.9}\\[4pt] & \frac{1}{2} m v_{A, 1, x}^{2}=\underset{\text { Conservation of energy }}{\frac{1}{2} m v_{A, 2, x}^{2}+m v_{G, 2, x}^{2}+\frac{1}{2} I_{G} \omega_{2}^{2}} \tag{12.10} \label{12.10}\\[4pt] & m v_{A, 1, x}^{2}=m\left(v_{A, 1, x}-2 v_{G, 2, x}\right)^{2}+2 m v_{G, 2, x}^{2}+\left(m L v_{G, 2, x}\right)^{2} /\left(\frac{1}{2} m L^{2}\right) \\[4pt] & 0=4 v_{G, 2, x}^{2}-4 v_{A, 1, x} v_{G, 2, x}+2 v_{G, 2, x}^{2}+2 v_{G, 2, x}^{2} \\[4pt] & v_{G, 2, x}=\frac{1}{2} v_{A, 1, x} \tag{12.11} \label{12.11}\\[4pt] & v_{A, 2, x}=\stackrel{0}{\text { Using Equation 12.8 }} \tag{12.12} \label{12.12}\\[4pt] & \omega_{2}=\left(-m L / I_{G)}\right) \frac{1}{2} v_{A, 1, x}=-v_{A, 1, x} / L \tag{12.13} \label{12.13}\end{align}\]
Here we simultaneously solved the equations of momentum conservation, angular momentum conservation and energy conservation, while using that for a dumbbell, \(I_{G}=\sum_{i} m_{i} \rho_{i / G}^{2}=2 m(L / 2)^{2}=\frac{1}{2} m L^{2}\).