### Step 1: Conceptual Introduction This problem involves determining the **angular acceleration** of a rigid link (EC) in a mechanical system with a pin and a slot. The link EC rotates about point A, which moves in a circular slot of link DO. The slot has a fixed radius, and both the angular velocity and angular acceleration of the slot are given. We are to find the angular acceleration of EC at a specific instant. **Explanation Block:** The situation describes a planar mechanism with two rotating links connected via a pin. The angular acceleration of one link (DO) is known, and we need to relate the motion of the pin (A) to the rotation of the other link (EC). This typically involves kinematic relationships between angular velocities and accelerations, using the geometry and constraints of the mechanism. --- ### Step 2: Formulae Used - **Relative Acceleration Equation:** \[ \vec{a}_A = \vec{a}_C + \vec{\alpha}_{EC} \times \vec{r}_{CA} - \omega_{EC}^2 \vec{r}_{CA} \] For the slot (A on the circular path of DO): \[ \vec{a}_A = \vec{a}_O + \vec{\alpha}_{DO} \times \vec{r}_{OA} - \omega_{DO}^2 \vec{r}_{OA} \] Where: - \(\omega\) and \(\alpha\) are angular velocity and angular acceleration. - \(\vec{r}\) is the position vector between points. **Explanation Block:** The fundamental approach is to use the relative acceleration equation for points on a rigid body. For a pin moving in a slot, its acceleration can be described both from the link EC and DO. Equating both expressions for \(\vec{a}_A\) allows us to solve for the unknown angular acceleration (\(\alpha_{EC}\)), using the known geometric and kinematic properties. --- ### Step 3: Step-by-Step Calculation #### 1. **Assign Known Values** - \(\omega_{DO} = \dot{\beta} = 5.0 \text{ rad/s}\) - \(\alpha_{DO} = \ddot{\beta} = 2.9 \text{ rad/s}^2\) - \(r_{OA} = r_{CA} = 5\,\text{in}\) (since both are radius of curvature) - \(\theta = \beta = 60^\circ\) #### 2. **Express Velocity of A via both links:** - For link DO: \[ \vec{v}_A = \vec{\omega}_{DO} \times \vec{r}_{OA} \] - For link EC: \[ \vec{v}_A = \vec{\omega}_{EC} \times \vec{r}_{CA} \] For position shown, \(\vec{v}_A\) is tangential to the slot (perpendicular to OA). #### 3. **Express Acceleration of A:** - For link DO: \[ \vec{a}_A = \vec{\alpha}_{DO} \times \vec{r}_{OA} - \omega_{DO}^2 \vec{r}_{OA} \] - For link EC: \[ \vec{a}_A = \vec{\alpha}_{EC} \times \vec{r}_{CA} - \omega_{EC}^2 \vec{r}_{CA} \] #### 4. **Write Components:** Let’s take CCW as positive. Both position vectors have same length, so: - \(\vec{r}_{OA}\) at \(60^\circ\) to horizontal. - \(\vec{r}_{CA}\) at \(60^\circ\) to horizontal. Tangential direction to slot is perpendicular to \(\vec{r}_{OA}\). #### 5. **Set Up Equation in Tangential Direction:** Equate tangential accelerations: \[ \alpha_{DO} r_{OA} = \alpha_{EC} r_{CA} \] Because both \(\omega_{EC}\) and \(\omega_{DO}\) are equal (\(\dot{\beta}\)), and with the same radius, the centripetal terms cancel out. \[ \alpha_{EC} = \alpha_{DO} = 2.9 \text{ rad/s}^2 \] --- **Explanation Block:** In this mechanism, the slot and the link EC rotate at the same angular velocity and with the same angular acceleration at the instant considered, because the pin’s velocity and acceleration must match in both links at that point. Thus, the angular acceleration of EC is simply equal to that of DO. --- ### Final Answer \[ \boxed{\text{The angular acceleration of link EC is } 2.9\ \text{rad/s}^2} \]

Let's reverify the solution step-by-step to ensure correctness. **Given Data:** - Angular velocity of link DO: \(\omega_{DO} = 5.0\, \text{rad/sec}\) - Angular acceleration of link DO: \(\alpha_{DO} = 2.9\, \text{rad/sec}^2\) - Radius of curvature of slot in link DO: \(r_{OA} = 5\, \text{in}\) - Position angle: \(\theta = 60^\circ\) --- ### Step-by-step verification: **1. Understanding the mechanism:** - The link EC rotates about point A. - The pin A contacts a slot in link DO, which moves along a circular path of radius 5 in. - The angular velocity and acceleration of link DO are given. - The goal is to find the angular acceleration \(\alpha_{EC}\). --- **2. Key kinematic relationships:** - The velocity of point A (on link EC) must match the velocity of the contact point in the slot, which moves along the circular path of radius 5 in. - The acceleration of point A can be expressed from both links, and equating these provides relations to find \(\alpha_{EC}\). --- **3. Velocity considerations:** - Velocity of A due to link DO: \[ \vec{v}_A = \vec{\omega}_{DO} \times \vec{r}_{OA} \] Since \(\vec{r}_{OA}\) is at \(60^\circ\), the magnitude of \(\vec{v}_A\): \[ v_A = \omega_{DO} \times r_{OA} = 5.0 \times 5 = 25\, \text{in/sec} \] - Velocity of A due to link EC: \[ v_A = \omega_{EC} \times r_{CA} \] But since the problem states the tangent to the slot at contact is parallel to AO, the velocity directions are aligned, implying: \[ \omega_{EC} = \omega_{DO} = 5\, \text{rad/sec} \] **Note:** The problem directly states \(\omega_{i} = 5.0\, \text{rad/sec}\) and \(\dot{\theta} = 60^\circ\). It also indicates that the tangents are parallel, which suggests the angular velocities of the links are equal at this position. --- **4. Acceleration considerations:** - Acceleration of A on link DO: \[ a_A^{(DO)} = \alpha_{DO} \times r_{OA} - \omega_{DO}^2 \times r_{OA} \] \[ a_A^{(DO)} = 2.9 \times 5 - (5)^2 \times 5 \] \[ a_A^{(DO)} = 14.5 - 125 = -110.5\, \text{in/sec}^2 \] The negative sign indicates the direction is opposite to the assumed positive. - Acceleration of A on link EC: \[ a_A^{(EC)} = \alpha_{EC} \times r_{CA} - \omega_{EC}^2 \times r_{CA} \] Since \(\omega_{EC} = 5\, \text{rad/sec}\), same as \(\omega_{DO}\): \[ a_A^{(EC)} = \alpha_{EC} \times 5 - 25 \times 5 = 5 \alpha_{EC} - 125 \] **5. Equating accelerations:** Because the contact point A must have the same acceleration from both perspectives: \[ a_A^{(DO)} = a_A^{(EC)} \] \[ -110.5 = 5 \alpha_{EC} - 125 \] \[ 5 \alpha_{EC} = -110.5 + 125 = 14.5 \] \[ \alpha_{EC} = \frac{14.5}{5} = 2.9\, \text{rad/sec}^2 \] --- ### **Final conclusion:** The reverified calculation confirms that the angular acceleration of link EC is **2.9 rad/sec\(^2\)**, matching the initial solution. --- ### **Final answer:** \[ \boxed{\text{The angular acceleration of link EC is } \mathbf{2.9\, \text{rad/sec}^2}} \]