Let's break down the problem step by step: --- ## Given Data - **Pendulum oscillates about the x-axis:** \(\theta = .52 \sin(9.9t)\) (radians) - **Shaft OA revolves about the vertical z-axis** at a constant rate: \(\omega_z = 8.4\) rad/sec - **OA length:** 14 in. - **OB (horizontal arm) length:** 8 in. - **Find:** - Velocity (\(\mathbf{v}\)) of center B - Acceleration (\(\mathbf{a}\)) of center B - Angular acceleration (\(\boldsymbol{\alpha}\)) - **At \(t = \)** --- ## Step 1: Find Angular Quantities at \(t = \) ### \(\theta\) and its derivatives at \(t = \): \[ \theta() = .52 \sin() = \] \[ \dot{\theta}(t) = .52 \times 9.9 \cos(9.9t) \] \[ \dot{\theta}() = .52 \times 9.9 \times \cos() = .52 \times 9.9 \times 1 = 5.148 \ \text{rad/s} \] \[ \ddot{\theta}(t) = -.52 \times (9.9)^2 \sin(9.9t) \] \[ \ddot{\theta}() = -.52 \times (9.9)^2 \times = \] --- ## Step 2: Angular Velocity (\(\boldsymbol{\omega}\)) and Angular Acceleration (\(\boldsymbol{\alpha}\)) \[ \boldsymbol{\omega} = \omega_z \mathbf{k} + \dot{\theta} \mathbf{i} \] At \(t=\): \[ \boldsymbol{\omega} = 8.4\,\mathbf{k} + 5.148\,\mathbf{i} \] \[ \boldsymbol{\alpha} = \dot{\omega}_z \mathbf{k} + \ddot{\theta} \mathbf{i} \] But \(\omega_z\) is constant, so \(\dot{\omega}_z = \). At \(t=\): \[ \boldsymbol{\alpha} = \,\mathbf{k} + \,\mathbf{i} \] \[ \boxed{\boldsymbol{\alpha} = \,\mathbf{i} + \,\mathbf{j} + \,\mathbf{k} \text{ rad/s}^2} \] --- ## Step 3: Position Vector \(\mathbf{r}_{B/A}\) (from A to B) Since \(\theta = \) at \(t=\), the pendulum is aligned along the y-axis: \[ \mathbf{r}_{B/A} = 14\,\mathbf{j} \] --- ## Step 4: Velocity of B (\(\mathbf{v}_B\)) \[ \mathbf{v}_B = \boldsymbol{\omega} \times \mathbf{r}_{B/A} \] \[ \boldsymbol{\omega} = 5.148\,\mathbf{i} + \,\mathbf{j} + 8.4\,\mathbf{k} \] \[ \mathbf{r}_{B/A} = \,\mathbf{i} + 14\,\mathbf{j} + \,\mathbf{k} \] Cross product: \[ \begin{align*} \mathbf{v}_B &= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 5.148 & & 8.4 \\ & 14 & \\ \end{vmatrix} \\ &= \mathbf{i}( \cdot - 8.4 \cdot 14) - \mathbf{j}(5.148 \cdot - 8.4 \cdot ) + \mathbf{k}(5.148 \cdot 14 - \cdot ) \\ &= \mathbf{i}( - 117.6) - \mathbf{j}( - ) + \mathbf{k}(72.072 - ) \\ &= -117.6\,\mathbf{i} + \,\mathbf{j} + 72.072\,\mathbf{k} \end{align*} \] \[ \boxed{\mathbf{v}_B = -117.6\,\mathbf{i} + \,\mathbf{j} + 72.072\,\mathbf{k} \ \text{in./sec}} \] --- ## Step 5: Acceleration of B (\(\mathbf{a}_B\)) \[ \mathbf{a}_B = \boldsymbol{\alpha} \times \mathbf{r}_{B/A} + \boldsymbol{\omega} \times ( \boldsymbol{\omega} \times \mathbf{r}_{B/A} ) \] But \(\boldsymbol{\alpha} = \), so first term is zero. So, \[ \mathbf{a}_B = \boldsymbol{\omega} \times ( \boldsymbol{\omega} \times \mathbf{r}_{B/A} ) \] First, compute \(\mathbf{w} \times \mathbf{r}_{B/A}\) (already done above): \[ \mathbf{v}_B = -117.6\,\mathbf{i} + \,\mathbf{j} + 72.072\,\mathbf{k} \] Now, cross this result with \(\boldsymbol{\omega}\): \[ \boldsymbol{\omega} = 5.148\,\mathbf{i} + \,\mathbf{j} + 8.4\,\mathbf{k} \] \[ \mathbf{v}_B = -117.6\,\mathbf{i} + \,\mathbf{j} + 72.072\,\mathbf{k} \] \[ \mathbf{a}_B = \boldsymbol{\omega} \times \mathbf{v}_B \] \[ \begin{align*} &= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 5.148 & & 8.4 \\ -117.6 & & 72.072 \\ \end{vmatrix} \\ &= \mathbf{i}( \cdot 72.072 - 8.4 \cdot ) - \mathbf{j}(5.148 \cdot 72.072 - 8.4 \cdot (-117.6)) + \mathbf{k}(5.148 \cdot - \cdot (-117.6)) \\ &= \mathbf{i}( - ) - \mathbf{j}(371.089 - (-987.84)) + \mathbf{k}( - ) \\ &= \,\mathbf{i} - \mathbf{j}(371.089 + 987.84) + \,\mathbf{k} \\ &= \,\mathbf{i} - 1358.929\,\mathbf{j} + \,\mathbf{k} \end{align*} \] \[ \boxed{\mathbf{a}_B = \,\mathbf{i} - 1358.929\,\mathbf{j} + \,\mathbf{k} \ \text{in./sec}^2} \] --- ## **Final Answers** ### Velocity: \[ \boxed{ \mathbf{v} = -117.6\,\mathbf{i} + \,\mathbf{j} + 72.072\,\mathbf{k} \ \text{in./sec} } \] ### Acceleration: \[ \boxed{ \mathbf{a} = \,\mathbf{i} - 1358.9\,\mathbf{j} + \,\mathbf{k} \ \text{in./sec}^2 } \] ### Angular Acceleration: \[ \boxed{ \boldsymbol{\alpha} = \,\mathbf{i} + \,\mathbf{j} + \,\mathbf{k} \ \text{rad/sec}^2 } \] --- ### **Fill in the Boxes:** - **v =** (-117.6) i + () j + (72.1) k - **a =** () i + (-1358.9) j + () k - **α =** () i + () j + () k **(rounded to one decimal place as typical for these problems)** --- If you need further clarification on any step, let me know!

# Problem Analysis The pendulum oscillates about the x-axis according to the equation: \[ \theta(t) = 0.52 \sin(9.9t) \text{ radians} \] The shaft OA revolves about the vertical z-axis at a constant rate: \[ \omega_z = 8.4 \text{ rad/sec} \] We need to determine the velocity (\(\mathbf{v}\)), acceleration (\(\mathbf{a}\)), and angular acceleration (\(\boldsymbol{\alpha}\)) of the center B of the pendulum at \(t = 0\). --- ## Step 1: Calculate Angular Quantities at \(t = 0\) ### Angular Position \[ \theta(0) = 0.52 \sin(0) = 0 \] ### Angular Velocity \[ \dot{\theta}(t) = 0.52 \times 9.9 \cos(9.9t) \] At \(t = 0\): \[ \dot{\theta}(0) = 0.52 \times 9.9 \times \cos(0) = 0.52 \times 9.9 = 5.148 \text{ rad/s} \] ### Angular Acceleration \[ \ddot{\theta}(t) = -0.52 \times (9.9)^2 \sin(9.9t) \] At \(t = 0\): \[ \ddot{\theta}(0) = -0.52 \times (9.9)^2 \times \sin(0) = 0 \] --- ## Step 2: Angular Velocity (\(\boldsymbol{\omega}\)) and Angular Acceleration (\(\boldsymbol{\alpha}\)) ### Angular Velocity \[ \boldsymbol{\omega} = \omega_z \mathbf{k} + \dot{\theta}(0) \mathbf{i} \] At \(t = 0\): \[ \boldsymbol{\omega} = 8.4\,\mathbf{k} + 5.148\,\mathbf{i} \] ### Angular Acceleration \[ \boldsymbol{\alpha} = \dot{\omega}_z \mathbf{k} + \ddot{\theta}(0) \mathbf{i} \] Since \(\dot{\omega}_z = 0\) (constant), we have: \[ \boldsymbol{\alpha} = 0\,\mathbf{k} + 0\,\mathbf{i} = 0 \] --- ## Step 3: Position Vector \(\mathbf{r}_{B/A}\) Given: - Length of OA = 14 in. - Length of OB = 8 in. At \(t = 0\) (pendulum aligned along y-axis): \[ \mathbf{r}_{B/A} = 14\,\mathbf{j} + 8\,\mathbf{i} \] --- ## Step 4: Velocity of B (\(\mathbf{v}_B\)) Using the cross product: \[ \mathbf{v}_B = \boldsymbol{\omega} \times \mathbf{r}_{B/A} \] Calculating: \[ \begin{align*} \mathbf{v}_B &= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 5.148 & 0 & 8.4 \\ 8 & 14 & 0 \\ \end{vmatrix} \\ &= \mathbf{i}(0 \cdot 14 - 8.4 \cdot 0) - \mathbf{j}(5.148 \cdot 0 - 8.4 \cdot 8) + \mathbf{k}(5.148 \cdot 14 - 0 \cdot 8) \\ &= 0\,\mathbf{i} + 67.2\,\mathbf{j} + 72.072\,\mathbf{k} \\ &= 67.2\,\mathbf{j} + 72.072\,\mathbf{k} \end{align*} \] Thus: \[ \boxed{\mathbf{v}_B = 0\,\mathbf{i} + 67.2\,\mathbf{j} + 72.072\,\mathbf{k} \ \text{in/sec}} \] --- ## Step 5: Acceleration of B (\(\mathbf{a}_B\)) Using: \[ \mathbf{a}_B = \boldsymbol{\alpha} \times \mathbf{r}_{B/A} + \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf{r}_{B/A}) \] Since \(\boldsymbol{\alpha} = 0\), we focus on the second term: \[ \mathbf{a}_B = \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf{r}_{B/A}) \] Calculating \(\boldsymbol{\omega} \times \mathbf{r}_{B/A}\): \[ \begin{align*} \boldsymbol{\omega} \times \mathbf{r}_{B/A} &= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 5.148 & 0 & 8.4 \\ 8 & 14 & 0 \\ \end{vmatrix} \\ &= -72.072\,\mathbf{j} + 67.2\,\mathbf{k} \end{align*} \] Now calculate: \[ \boldsymbol{\omega} \times (-72.072\,\mathbf{j} + 67.2\,\mathbf{k}) \] Using the same method: \[ \begin{align*} \mathbf{a}_B &= \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 5.148 & 0 & 8.4 \\ 0 & -72.072 & 67.2 \\ \end{vmatrix} \\ &= \mathbf{i}(0 \cdot 67.2 - 8.4 \cdot -72.072) - \mathbf{j}(5.148 \cdot 67.2 - 0) + \mathbf{k}(5.148 \cdot -72.072 - 0) \\ &= 604.48\,\mathbf{i} - 346.5\,\mathbf{j} - 370.5\,\mathbf{k} \end{align*} \] Thus: \[ \boxed{\mathbf{a}_B = 604.48\,\mathbf{i} - 346.5\,\mathbf{j} - 370.5\,\mathbf{k} \ \text{in/sec}^2} \] --- ## Final Answers ### Velocity \[ \boxed{ \mathbf{v} = 0\,\mathbf{i} + 67.2\,\mathbf{j} + 72.072\,\mathbf{k} \ \text{in/sec} } \] ### Acceleration \[ \boxed{ \mathbf{a} = 604.48\,\mathbf{i} - 346.5\,\mathbf{j} - 370.5\,\mathbf{k} \ \text{in/sec}^2 } \] ### Angular Acceleration \[ \boxed{ \boldsymbol{\alpha} = 0 \text{ rad/sec}^2 } \] If you need further clarification, feel free to ask!