Planar rigid body kinematics: reference frame transformations
In planar kinematics, understanding how motion appears in different reference frames is a critical skill. Observations of a point’s position, velocity, and acceleration can change depending on the perspective, whether from a fixed or moving reference frame. These transformations are crucial for analyzing relative motion in both simple and complex systems.
To express the relationship between two reference frames, I use trigonometric relationships. For example, the unit vectors of a moving frame, expressed in terms of the fixed frame, are:
\begin{aligned} \mathbf{i}_2 &= \cos(\theta) \, \mathbf{i}_1 + \sin(\theta) \, \mathbf{j}_1 \\ \mathbf{j}_2 &= -\sin(\theta) \, \mathbf{i}_1 + \cos(\theta) \, \mathbf{j}_1 \end{aligned}
Differentiating these unit vectors with respect to time introduces the rotational effects of the moving frame. For instance, the time derivative of \mathbf{i}_2 in the fixed frame becomes:
\frac{\mathrm{d}\mathbf{i}_2}{\mathrm{d}t} = \dot{\theta} \, \mathbf{j}_2
where \dot{\theta} is the angular velocity of the moving frame. Similarly, the derivative of \mathbf{j}_2 is:
\frac{\mathrm{d}\mathbf{j}_2}{\mathrm{d}t} = -\dot{\theta} \, \mathbf{i}_2
Using these results, I can describe the motion of a vector \mathbf{V} in the moving frame, considering both its time-dependent components and the rotation of the frame. By applying the product rule for differentiation, the transformation of the vector’s derivative from the moving frame to the fixed frame becomes:
\frac{\mathrm{d}\mathbf{V}}{\mathrm{d}t} \bigg|_1 = \frac{\mathrm{d}\mathbf{V}}{\mathrm{d}t} \bigg|_2 + \dot{\theta} \, \mathbf{k} \times \mathbf{V}
where \mathbf{k} is the unit vector perpendicular to the plane of motion.
This formula simplifies the analysis of relative motion between frames and highlights the importance of rotational effects when dealing with moving reference frames.
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