# Problem 4: Kinematics of a Slider-Crank Mechanism **Given:** - Block \( B \): \( v_B = 4\, \text{m/s} \), \( a_B = 2\, \text{m/s}^2 \) - \( AC = 150\, \text{mm} = .15\, \text{m} \) - \( AB = 400\, \text{mm} = .4\, \text{m} \) --- ## (a) Type of Motion for Each Piece - **Block \( B \):** Pure **rectilinear (translational) motion** along the \( x \)-axis. - **Rod \( AB \):** **General plane motion** (rotation about \( A \) and translation). - **Wheel (Crank \( AC \)):** **Pure rotation** about fixed axis \( C \). --- ## (b) Angular Velocity (\(\omega_C\)) of the Wheel The velocity of point \( A \) (end of crank) is tangent to the wheel: \[ v_A = r_{AC} \omega_C \] But \( v_A \) is also the velocity of point \( B \) projected onto the direction of \( AB \): Let’s find \( v_A \) in terms of \( v_B \): - The velocity of \( B \) is entirely horizontal (\( x \)-direction): \( v_B = 4\, \text{m/s} \). - Triangle \( AB \): \( AB = .4\, \text{m} \), \( AC = .15\, \text{m} \). Let’s compute the **angle \( \theta \)** at this instant: \[ \sin \theta = \frac{AC}{AB} = \frac{.15}{.4} = .375 \] \[ \theta = \arcsin(.375) \approx 22^\circ \] The horizontal component of the velocity of \( A \) is: \[ v_{A_x} = v_B \] But \( v_A \) is perpendicular to \( AC \). So, the horizontal component of \( v_A \) is: \[ v_{A_x} = v_A \sin \theta \] Therefore, \[ v_A = \frac{v_B}{\sin \theta} = \frac{4}{.375} \approx 10.67\, \text{m/s} \] \[ \omega_C = \frac{v_A}{r_{AC}} = \frac{10.67}{.15} \approx 71.1\, \text{rad/s} \] --- ## (c) Angular Acceleration (\(\alpha_C\)) of the Wheel The acceleration at \( B \) is \( a_B = 2\, \text{m/s}^2 \) (horizontal). The acceleration of \( A \) in the \( x \)-direction is: \[ a_{A_x} = a_B \] But, \[ a_{A_x} = a_A \sin \theta \] So, \[ a_A = \frac{a_{A_x}}{\sin \theta} = \frac{2}{.375} \approx 5.33\, \text{m/s}^2 \] But \( a_A \) is due to both tangential and centripetal acceleration: \[ a_A = r_{AC} \alpha_C \] So, \[ \alpha_C = \frac{a_A}{r_{AC}} = \frac{5.33}{.15} \approx 35.6\, \text{rad/s}^2 \] --- ## (d) Angular Velocity (\(\omega_{AB}\)) of Rod AB Relative velocity at \( B \) with respect to \( A \): \[ \vec{v}_B = \vec{v}_A + \vec{\omega}_{AB} \times \vec{r}_{B/A} \] Since \( v_A \) is perpendicular to \( AC \), and \( v_B \) is horizontal: \[ v_B = v_A \cos \theta + \omega_{AB} \cdot AB \] But since \( v_A \) is almost vertical (since \( AC \) is nearly horizontal), \( v_A \cos \theta \) is the horizontal component. But more accurately: \[ v_B = v_A \cos \theta + \omega_{AB} \cdot AB \] \[ 4 = 10.67 \cdot \cos(22^\circ) + \omega_{AB} \cdot .4 \] \[ 4 = 10.67 \cdot .927 + \omega_{AB} \cdot .4 \] \[ 4 = 9.89 + .4\omega_{AB} \] \[ .4\omega_{AB} = 4 - 9.89 = -5.89 \] \[ \omega_{AB} = \frac{-5.89}{.4} \approx -14.7\, \text{rad/s} \] --- ## (e) Angular Acceleration (\(\alpha_{AB}\)) of Rod AB Using the acceleration relation: \[ \vec{a}_B = \vec{a}_A + \vec{\alpha}_{AB} \times \vec{r}_{B/A} - (\omega_{AB})^2 \vec{r}_{B/A} \] Horizontal component: \[ a_B = a_A \cos \theta + \alpha_{AB} \cdot AB - \omega_{AB}^2 \cdot AB \] \[ 2 = 5.33 \cdot .927 + \alpha_{AB} \cdot .4 - (14.7)^2 \cdot .4 \] \[ 2 = 4.94 + .4\alpha_{AB} - 86.35 \] \[ .4\alpha_{AB} = 2 - 4.94 + 86.35 = 83.41 \] \[ \alpha_{AB} = \frac{83.41}{.4} \approx 208.5\, \text{rad/s}^2 \] --- # **Summary of Answers** | Sub-Part | Quantity | Value | |----------|-------------------------------------------------------|----------| | (a) | **Type of Motion** | \( B \): rectilinear, \( AB \): general plane, Wheel: rotation | | (b) | **Angular velocity of wheel** (\(\omega_C\)) | \( 71.1\, \text{rad/s} \) | | (c) | **Angular acceleration of wheel** (\(\alpha_C\)) | \( 35.6\, \text{rad/s}^2 \) | | (d) | **Angular velocity of rod \( AB \)** (\(\omega_{AB}\))| \( -14.7\, \text{rad/s} \) | | (e) | **Angular acceleration of rod \( AB \)** (\(\alpha_{AB}\))| \( 208.5\, \text{rad/s}^2 \) | --- **Note:** Negative sign in angular velocity for rod \( AB \) indicates rotation direction (clockwise if positive is counterclockwise).