# Two Dimensional Kinematics

Attempting this kinematics question but not really understanding if what I'm doing is right or completely wrong. Should I be starting by finding Vp? Any help appreciated. Not an assignment, just practicing.

Let us work out the geometric relations first and then do the kinematics. I will follow a purely algorithmic approach to show the details. You can arrive at the same result from trigonometric relations.

Let us choose a fixed coordinate frame that has its origin at A, with the three unit vectors $$\mathbf{E}_1 = (1, 0, 0)$$, $$\mathbf{E}_2 = (0, 1, 0)$$, and $$\mathbf{E}_3 = (0, 0, 1)$$ where $$\mathbf{E}_1$$ is along AB, $$\mathbf{E}_2$$ is perpendicular to AB (in the plane ABP), and $$\mathbf{E}_3$$ is perpendicular to the plane ABP.

The position vectors of the three points are $$\mathbf{x}_A, \mathbf{x}_B, \mathbf{x}_P \,.$$ Then the direction vectors of interest are $$\mathbf{r}_{BA} = \mathbf{x}_B - \mathbf{x}_A ~,~~ \mathbf{r}_{PA} = \mathbf{x}_P - \mathbf{x}_A ~,~~ \mathbf{r}_{PB} = \mathbf{x}_P - \mathbf{x}_B \,.$$ From the figure, if $$\theta_A, \theta_B$$ are the angles made by PA and PB with AB, and $$r_A, r_B$$ are the lengths of PA and PB, we have \begin{align} \mathbf{r}_{BA} &= b \mathbf{E}_1 \\ \mathbf{r}_{PA} &= -r_A \cos\theta_A \mathbf{E}_1 + r_A \sin\theta_A \mathbf{E}_2 \\ \mathbf{r}_{PB} &= -r_B \cos\theta_B \mathbf{E}_1 + r_B \sin\theta_B \mathbf{E}_2 \end{align} Since $$\mathbf{r}_{PB} = -\mathbf{r}_{BA} + \mathbf{r}_{PA}$$, we have $$-r_B \cos\theta_B \mathbf{E}_1 + r_B \sin\theta_B \mathbf{E}_2 = -b \mathbf{E}_1 - r_A \cos\theta_A \mathbf{E}_1 + r_A \sin\theta_A \mathbf{E}_2$$ or \begin{align} -r_B \cos\theta_B & = -b - r_A \cos\theta_A \\ r_B \sin\theta_B & = r_A \sin\theta_A \end{align} In matrix form $$\begin{bmatrix} \cos\theta_A & -\cos\theta_B \\ \sin\theta_A & -\sin\theta_B \end{bmatrix} \begin{bmatrix} r_A \\ r_B \end{bmatrix} = \begin{bmatrix} - b \\ 0 \end{bmatrix}$$ The determinant of the $$2 \times 2$$ matrix is $$D := -\cos\theta_A \sin\theta_B + \sin\theta_A \cos\theta_B = \sin(\theta_A - \theta_B)$$ Therefore, inverting the matrix, we have $$\begin{bmatrix} r_A \\ r_B \end{bmatrix} = \frac{1}{\sin(\theta_A -\theta_B)} \begin{bmatrix} -\sin\theta_B & \cos\theta_B \\ -\sin\theta_A & \cos\theta_A \end{bmatrix}\begin{bmatrix} - b \\ 0 \end{bmatrix}$$ or \boxed{ \begin{align} r_A &= \frac{b\sin\theta_B}{\sin(\theta_A -\theta_B)} \\ r_B &= \frac{b\sin\theta_A}{\sin(\theta_A -\theta_B)} \end{align} } Plugging in $$b = 300$$ mm, $$\theta_A = 60^\circ$$, $$\theta_B = 20^\circ$$, we have $$r_A = 159.6$$ mm and $$r_B = 404.2$$ mm.

Now that we know the geometry, we can do the kinematics. Choose a rotating frame that has origin at point A, and the coordinate axes \begin{align} \mathbf{e}_1 &= \sin\theta_A \mathbf{E}_1 + \cos\theta_A \mathbf{E}_2 \\ \mathbf{e}_2 &= -\cos\theta_A \mathbf{E}_1 + \sin\theta_A \mathbf{E}_2 = \mathbf{r}_{PA}/||\mathbf{r}_{PA}||\\ \mathbf{e_3} &= \mathbf{E}_3. \end{align} or, inverting the relation, \begin{align} \mathbf{E}_1 &= \sin\theta_A \mathbf{e}_1 - \cos\theta_A \mathbf{e}_2 \\ \mathbf{E}_2 &= \cos\theta_A \mathbf{e}_1 + \sin\theta_A \mathbf{e}_2 \\ \mathbf{E_3} &= \mathbf{e}_3. \end{align}

Then the position vector of P in that coordinate system is $$\mathbf{r}_{PA} = r_A \mathbf{e}_2$$ and the velocity of P relative to the rotating frame is $$\mathbf{v}_{P/R} = \dot{r}_A \mathbf{e}_2 = \dot{r}_A \left[-\cos\theta_A \mathbf{E}_1 + \sin\theta_A \mathbf{E}_2\right]$$ The time derivative of $$r_A$$ is $$\dot{r}_A = \frac{b\cos\theta_B}{\sin(\theta_A -\theta_B)} \dot{\theta}_B + \frac{2b\sin\theta_B\cos(\theta_A-\theta_B)}{\cos[2(\theta_A -\theta_B)]-1} (\dot{\theta}_A - \dot{\theta}_B)$$ or, using $$\omega_A = \dot{\theta}_A$$ and $$\omega_B = \dot{\theta}_B$$, $$\boxed{ \dot{r}_A = \frac{2b\sin\theta_B\cos(\theta_A-\theta_B)}{\cos[2(\theta_A -\theta_B)]-1} \omega_A + \left[\frac{b\cos\theta_B}{\sin(\theta_A -\theta_B)} - \frac{2b\sin\theta_B\cos(\theta_A-\theta_B)}{\cos[2(\theta_A -\theta_B)]-1}\right] \omega_B }$$ where $$\boldsymbol{\Omega}_A = \omega_A \mathbf{E}_3 = \omega_A \mathbf{e}_3$$ and $$\boldsymbol{\Omega}_B = \omega_B \mathbf{E}_3 = \omega_B \mathbf{e}_3$$ are the angular velocities at A and B of AP and BP, respectively.

Plugging in the values that we know, we have $$\dot{r}_A = -190.24 \omega_A + 628.81 \omega_B \,.$$

At P, the velocity is $$\mathbf{v}_P = \boldsymbol{\Omega}_A \times \mathbf{r}_{PA} + \mathbf{v}_{P/R} = \boldsymbol{\Omega}_B \times \mathbf{r}_{PB}$$ Therefore, we have \begin{align} & (\omega_A \mathbf{e}_3) \times (r_A \mathbf{e}_2) + \\ & \left[\frac{2b\sin\theta_B\cos(\theta_A-\theta_B)}{\cos[2(\theta_A -\theta_B)]-1} \omega_A + \left[\frac{b\cos\theta_B}{\sin(\theta_A -\theta_B)} - \frac{2b\sin\theta_B\cos(\theta_A-\theta_B)}{\cos[2(\theta_A -\theta_B)]-1}\right] \omega_B\right]\mathbf{e}_2 \\ &= (\omega_B \mathbf{e}_3) \times (-r_B \cos\theta_B \mathbf{E}_1 + r_B \sin\theta_B \mathbf{E}_2) \end{align} Now \begin{align} \mathbf{e}_3 \times \mathbf{e}_1 &= \mathbf{e}_2 \\ \mathbf{e}_3 \times \mathbf{e}_2 &= -\mathbf{e}_1 \\ \mathbf{e}_3 \times \mathbf{E}_1 &= \sin\theta_A \mathbf{e}_3 \times \mathbf{e}_1 - \cos\theta_A \mathbf{e}_3 \times \mathbf{e}_2 = \sin\theta_A \mathbf{e}_2 + \cos\theta_A \mathbf{e}_1\\ \mathbf{e}_3 \times \mathbf{E}_2 &= \cos\theta_A \mathbf{e}_3 \times \mathbf{e}_1 + \sin\theta_A \mathbf{e}_3 \times \mathbf{e}_2 = \cos\theta_A \mathbf{e}_2 - \sin\theta_A \mathbf{e}_1 \end{align} Therefore, we have \begin{align} & -\omega_A r_A \mathbf{e}_1 + \\ & \left[\frac{2b\sin\theta_B\cos(\theta_A-\theta_B)}{\cos[2(\theta_A -\theta_B)]-1} \omega_A + \left[\frac{b\cos\theta_B}{\sin(\theta_A -\theta_B)} - \frac{2b\sin\theta_B\cos(\theta_A-\theta_B)}{\cos[2(\theta_A -\theta_B)]-1}\right] \omega_B\right]\mathbf{e}_2 \\ & = \omega_B r_B \left[-\cos\theta_B(\sin\theta_A \mathbf{e}_2 + \cos\theta_A \mathbf{e}_1) +\sin\theta_B(\cos\theta_A \mathbf{e}_2 - \sin\theta_A \mathbf{e}_1)\right]\\ & = -\omega_B r_B \left[\cos(\theta_A-\theta_B) \mathbf{e}_1 + \sin(\theta_A-\theta_B) \mathbf{e}_2 \right] \end{align} Comparing the components along $$\mathbf{e}_1$$ and $$\mathbf{e}_2$$, respectively, we have $$-\omega_A r_A = -\omega_B r_B \cos(\theta_A-\theta_B)$$ and $$\frac{2b\sin\theta_B\cos(\theta_A-\theta_B)}{\cos[2(\theta_A -\theta_B)]-1} \omega_A + \left[\frac{b\cos\theta_B}{\sin(\theta_A -\theta_B)} - \frac{2b\sin\theta_B\cos(\theta_A-\theta_B)}{\cos[2(\theta_A -\theta_B)]-1}\right] \omega_B = -\omega_B r_B \sin(\theta_A-\theta_B)$$

Plugging in $$\omega_A = 10$$ and the previously computed values of $$r_A$$ and $$r_B$$ into the first equation above, we have $$\omega_B =$$ 5.1555 rad/s.

Therefore, the relative velocity of the slider block is $$\dot{r}_A = -190.24 \omega_A + 628.81 \omega_B = 1339.4~\text{mm/s} \,.$$