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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.

Question Attempt

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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 $$ \dot{r}_A = -190.24 \omega_A + 628.81 \omega_B = 1339.4~\text{mm/s} \,. $$

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