Planar rigid body kinematics: acceleration
Understanding the motion of rigid bodies requires careful attention to both linear and angular dynamics. In this post, I focus on the kinematics and acceleration of a rigid body undergoing angular motion. By analyzing the relative motion between two points, P and Q, on the body, I derive key relationships that describe the system’s behavior.
The absolute velocity of a point Q can be expressed as:
\mathbf{v}_Q = \mathbf{v}_P + \omega \mathbf{k} \times \mathbf{r}_{PQ}
where \mathbf{v}_P is the absolute velocity of point P, \omega is the angular velocity, and \mathbf{r}_{PQ} is the position vector from P to Q. Differentiating this equation provides the acceleration of Q:
\mathbf{a}_Q = \mathbf{a}_P + (\alpha \mathbf{k}) \times \mathbf{r}_{PQ} - \omega^2 \mathbf{r}_{PQ}
Here, \mathbf{a}_P represents the acceleration of point P, \alpha \mathbf{k} \times \mathbf{r}_{PQ} corresponds to tangential acceleration, and -\omega^2 \mathbf{r}_{PQ} accounts for the normal (centripetal) acceleration.
The term -\omega^2 \mathbf{r}_{PQ} is particularly significant as it encapsulates the centripetal acceleration that ensures the point Q remains constrained to its rotational path. Additionally, the tangential acceleration (\alpha \mathbf{k}) \times \mathbf{r}_{PQ} reflects the contribution from angular acceleration.
To simplify computations involving the double cross product in the acceleration equation, I use the identity:
\mathbf{A} \times (\mathbf{B} \times \mathbf{C}) = \mathbf{B} (\mathbf{A} \cdot \mathbf{C}) - \mathbf{C} (\mathbf{A} \cdot \mathbf{B})
This aids in breaking down the components of motion effectively. By substituting the appropriate terms, I achieve a clear and concise representation of \mathbf{a}_Q in terms of angular variables.
For more insights into this topic, you can find the details here.