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Suppose we have two points $A, B \in \mathbb{R}^2$ where $A = (0 | y)$ and $B = (0|-y)$. Also we have a point $P$ which lies on the line between $A$ and $B$. If we can move the two points on the x-axis at velocity $v_x$ and the point $P$ on the line at velocity $v_{\lambda}$ then what is the kinematics equation?

I used $f(\lambda) = \lambda A + (1 - \lambda) B, \lambda \in [0,1]$ to model the line on which the point $P$ can be. I struggled at combining the three movements into one equation because $f(\lambda)$ depends on $A$ and $B$.

Erdogan Seref
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1 Answers1

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We have

$$\vec{v}_{P / \mathcal{I}} = \vec{v}_{A / \mathcal{I}} + \vec{v}_{P / \mathcal{B}} + \vec{\omega}_{\mathcal{B} / \mathcal{I}} \times \vec{r}_{A \to P}$$

where $\mathcal{I}$ is the inertial frame, and $\mathcal{B}$ is a frame attached to line segment $AB$. Note that $\mathcal{B}$ is not rotating in $\mathcal{I}$, so that the last term in the equation above can be ignored. The velocity of $A$ in the inertial frame is given; it is $v_x \hat{i}$ (where $\hat{i}$ is aligned with the positive $x$-axis). The velocity of $P$ in the frame $\mathcal{B}$ is also given; it is $-v_\lambda \hat{i}_\mathcal{B}$, where $\hat{i}_\mathcal{B}$ points in the direction from $B$ to $A$. Note that by coordinate transformation, $\hat{i}_\mathcal{B} = \cos \theta \hat{i} + \sin \theta \hat{j}$, where $\theta$ is the angle that $BA$ makes with the horizontal. Thus, in the inertial basis,

$$\vec{v}_{P / \mathcal{I}} = (v_x - v_\lambda \cos \theta) \hat{i} - v_\lambda \sin \theta \hat{j}$$

Since here, $\theta = \frac{\pi}{4}$, we obtain

$$\vec{v}_{P / \mathcal{I}} = \left( v_x - \frac{\sqrt{2}}{2} v_\lambda \right) \hat{i} - \frac{\sqrt{2}}{2} v_\lambda \hat{j}$$

K. Jiang
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  • What is $\hat{j}$? – Erdogan Seref Mar 19 '23 at 19:37
  • How can we use the final equation for the function $f(\Delta x_A, \Delta x_B, \Delta \lambda)$ which outputs the position of the point $P$ after moving the three points $A, B$ and $P$? – Erdogan Seref Mar 19 '23 at 19:43
  • I used $\hat{j}$ to refer to the unit vector pointing in the positive-$y$ direction. I think I may have misinterpreted your question. I thought that $A$ was constrained to move along the $s$-axis, and $B$ along the $y$-axis. Can you explain what $A = (0|y)$, $B = (0|-y)$ means? I'll update my answer when that is clear. – K. Jiang Mar 19 '23 at 23:39
  • $A = (0|y)$ and $B = (0|-y)$ are points who at the beginning lie on the y-axis and can both move along the x-axis. y is a free parameter which controls how long the line segment $AB$ is. – Erdogan Seref Mar 20 '23 at 11:54
  • graphic – Erdogan Seref Mar 20 '23 at 12:05