Let's break down the problem step by step: --- ## **Given Data** - **Slider velocity:** \( v_B = 4 \text{ m/s} \) (to the right) - **Linkage lengths:** - \( CA = 150 \text{ mm} = .15 \text{ m} \) - \( CB = 400 \text{ mm} = .4 \text{ m} \) - **Points on the mechanism:** - \( C \) is the wheel center, - \( A \) is a point on the rim, - \( B \) is the slider. --- ## **Step 1: Geometry and Instantaneous Center** The bar \( AB \) rotates about point \( C \). Since \( B \) is constrained to move horizontally, let's use relative velocity analysis. --- ## **Step 2: Velocity Analysis** ### **Relative Velocity Equation** The velocity of \( B \) with respect to \( C \): \[ \vec{v}_B = \vec{v}_C + \vec{\omega} \times \vec{r}_{B/C} \] But \( C \) is fixed (\( v_C = \)), so: \[ \vec{v}_B = \vec{\omega} \times \vec{r}_{B/C} \] - \( \vec{r}_{B/C} \) is the position vector from \( C \) to \( B \). Let’s find the direction of \( \vec{r}_{B/C} \): Since \( CA \) is vertical and \( CB \) is at an angle, let's assign coordinates: - \( C \) at \((, )\) - \( A \) at \((, .15)\) - \( B \) at \((x_B, )\), where \( CB = .4 \text{ m} \) By geometry: \[ CB^2 = x_B^2 + ( - )^2 \] \[ x_B = .4 \text{ m} \] But since the linkage is shown as a straight line, the direction from \( C \) to \( B \) is along the rod. The angle \( \theta \) between \( CB \) and the horizontal is: \[ \sin\theta = \frac{.15}{.4} \] \[ \theta = \arcsin\left(\frac{.15}{.4}\right) \] \[ \theta \approx 22^\circ \] --- ## **Step 3: Find Angular Velocity \(\omega\)** Since \( v_B \) is perpendicular to \( CB \): \[ v_B = \omega_{CB} \times CB \] \[ \omega_{CB} = \frac{v_B}{CB} \] \[ \omega_{CB} = \frac{4}{.4} = 10 \text{ rad/s} \] --- ## **Step 4: Velocity of Point A** \( A \) is at a distance \( r = .15 \text{ m} \) from \( C \). \[ v_A = \omega_{CB} \times r_{A/C} \] \[ v_A = 10 \times .15 = 1.5 \text{ m/s} \] Direction: Perpendicular to \( CA \) (i.e., **horizontal to the left**, since the wheel is rotating clockwise). --- ## **Step 5: Angular Acceleration \(\alpha\)** Use the acceleration equation for point \( B \): \[ a_B = a_C + \alpha_{CB} \times r_{B/C} - \omega_{CB}^2 r_{B/C} \] Since \( C \) is fixed (\( a_C = \)), and \( a_B \) is horizontal (slider): \[ a_B = \alpha_{CB} \times r_{B/C} - \omega_{CB}^2 r_{B/C} \] But to find \( \alpha_{CB} \), we need \( a_B \). Since the slider is moving at constant velocity (\( v_B \)), \( a_B = \): \[ = \alpha_{CB} \times .4 - (10)^2 \times .4 \] \[ = .4\alpha_{CB} - 40 \] \[ .4\alpha_{CB} = 40 \] \[ \alpha_{CB} = 100 \text{ rad/s}^2 \] This is **positive**, so the angular acceleration is **counter-clockwise** (slowing down the clockwise rotation). --- ## **Final Answers** ### **1. Velocity of Point A** \[ \boxed{v_A = 1.5 \text{ m/s} \text{ (to the left)}} \] ### **2. Angular Velocity of the Wheel** \[ \boxed{\omega_{CB} = 10 \text{ rad/s} \text{ (clockwise)}} \] ### **3. Angular Acceleration of the Wheel** \[ \boxed{\alpha_{CB} = 100 \text{ rad/s}^2 \text{ (counter-clockwise)}} \] --- **Summary Table** | Quantity | Value | Direction | |----------------------------------|--------------------------|----------------------| | Velocity of A (\(v_A\)) | 1.5 m/s | Left | | Angular velocity (\(\omega\)) | 10 rad/s | Clockwise | | Angular acceleration (\(\alpha\))| 100 rad/s² | Counter-clockwise |