D6. Examples of 2D dynamics
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This unit proposes a systematic method for solving problems in dynamics of systems consisting of any number of rigid bodies. If this number is high, defining the strategy to achieve the desired results is essential to avoid excessive development.
The examples focus on problems in plane dynamics (2D). Unit D7 analyses examples of 3D dynamics.
D6.1 2D kinematics and 2D dynamics
The condition for a system to exhibit 2D dynamics is not only that its kinematics be plane: it is also necessary that the angular momentum of each rigid body [math]\displaystyle{ \ss_\is }[/math] in the system relative to its center of inertia [math]\displaystyle{ \Gs_\is }[/math] be orthogonal to the plane of motion (secció D4.5). This is equivalent to saying that the direction orthogonal to the plane of motion is a principal direction of inertia for [math]\displaystyle{ \Gs_\is }[/math].
✏️ EXAMPLE D6.1: 2D kinematics and 3D dynamics
- The thin homogeneous bar is hinged at [math]\displaystyle{ \Os }[/math] to a support that rotates relative to the ground with constant vertical angular velocity [math]\displaystyle{ \Omega_0 }[/math]. The bar has a single-point contact with the smooth ground, so its motion is planar: all its points (except point [math]\displaystyle{ \Os }[/math]) describe horizontal circular trajectories.
- The fact that the kinematics of the bar is 2D may suggest that only the vertical component of the AMT is necessary to study its dynamics. But this is not the case.
- The angular momentum of the bar at point [math]\displaystyle{ \Os }[/math] is:
[math]\displaystyle{ \braq{\overline{\mathbf{H}}_{\text {RTO=E}}(\Os)}{}= \matriz{\Is}{0}{0}{0}{0}{0}{0}{0}{\Is}\vector{\Omega_0 \sin \beta}{\Omega_0 \cos \beta}{0}= \vector{\Is\Omega_0 \sin \beta}{0}{0} . }[/math]
- Only the horizontal component of this vector is variable (it has a variable direction because of [math]\displaystyle{ \Omega_0 }[/math]). Its geometrical time derivative yields:
[math]\displaystyle{ \dot{\overline{\mathbf{H}}}_{\text {RTO}} (\Os)=(\Uparrow \Omega_0) \times (\Rightarrow \Is \Omega_0 \sin \beta \cos \beta)= \otimes\Is \Omega_0^2 \sin\beta \cos\beta. }[/math]
This time derivative can also be obtained through the vector basis:
[math]\displaystyle{ \braq{\dot{\overline{\mathbf{H}}}_{\text {RTO}} (\Os)}{}= \vector{\Omega_0 \sin \beta}{\Omega_0 \cos \beta}{0} \times \vector{\Is \Omega_0 \sin \beta}{0}{0} =\vector{0}{0}{-\Is \Omega_0^2\sin\beta\cos\beta}. }[/math]
When applying the AMT at [math]\displaystyle{ \Os }[/math] to the bar, component 3, which is horizontal, cannot be ignored.
D6.2 Free-body diagram (FBD) and roadmap
When a system consisting of only one rigid body has 2D dynamics, the number of equations to solve the problem is 3: the two components of the LMT in the plane of motion and the component of the AMT perpendicular to this plane. Depending on what we want to calculate (equation of motion, constraint force...), it may not be necessary to apply both theorems.
Finding a quick way to achieve the desired result requires the development of a roadmap (which is a statement of the strategy to be followed). Exploring possible strategies and choosing an appropriate one requires a good understanding of the kinematics and dynamics of the rigid body. Therefore, it is always recommended to start by investigating its velocity distribution and the description of the interactions on it. The representation of these interactions constitutes the free-body diagram (FBD).