Let's break down the problem step by step: --- ## **Given Data** - **Angular velocity of OA (\(\omega\))**: \(10\, \text{rad/s}\) (clockwise) - **Angular acceleration (\(\alpha\))**: \(2\, \text{rad/s}^2\) - **Angle (\(\theta\))**: \(30^\circ\) - **Length \(b\)**: \(.3\, \text{m}\) - **Find:** Linear velocity \(v\) of point \(C\). --- ## **Step 1: Find Velocity of Point \(A\)** The linear velocity of point \(A\) (end of OA) with respect to \(O\): \[ v_A = \omega \cdot OA = \omega \cdot b \] Given: - \(\omega = 10\, \text{rad/s}\) - \(b = .3\, \text{m}\) \[ v_A = 10 \times .3 = 3\, \text{m/s} \] Direction: Perpendicular to \(OA\), so at \(30^\circ\) above the horizontal (since OA makes angle \(\theta\) with the horizontal). --- ## **Step 2: Velocity Components of Point \(A\)** Resolve \(v_A\) into horizontal (\(x\)) and vertical (\(y\)) components: \[ v_{A_x} = v_A \sin\theta \] \[ v_{A_y} = v_A \cos\theta \] Given \(\theta = 30^\circ\): \[ v_{A_x} = 3 \times \sin(30^\circ) = 3 \times .5 = 1.5\, \text{m/s} \] \[ v_{A_y} = 3 \times \cos(30^\circ) = 3 \times .866 = 2.598\, \text{m/s} \] --- ## **Step 3: Instantaneous Center of Rotation for Link \(AB\)** Link \(AB\) rotates about \(A\), and point \(B\) has a horizontal velocity (since it is connected to slider C, which moves horizontally). Let \(v_B\) be the velocity of point \(B\) (horizontal to the right). Let the length \(AB = 2b = .6\, \text{m}\). The velocity of \(B\) can be related to the velocity of \(A\) using relative velocity equation: \[ \vec{v}_B = \vec{v}_A + \vec{\omega}_{AB} \times \vec{r}_{B/A} \] - \(\vec{v}_A\): Already found, split into \(x\) and \(y\). - \(\vec{\omega}_{AB}\): Angular velocity of \(AB\). - \(\vec{r}_{B/A}\): From \(A\) to \(B\), direction along \(AB\). But since \(B\) moves horizontally, \[ v_{B_y} = \] --- ## **Step 4: Write Relative Velocity Equations** Let \(\omega_{AB}\) be the angular velocity of link \(AB\). The position vector from \(A\) to \(B\) is: - Magnitude: \(2b = .6\, \text{m}\) - Direction: Along the link, which makes \(30^\circ\) above the horizontal. The velocity of \(B\) relative to \(A\) is perpendicular to \(AB\) (by right-hand rule), which is vertically downward. So, \[ \vec{v}_B = \vec{v}_A + \omega_{AB} \cdot (2b) \cdot (\text{perpendicular to } AB \text{, downward}) \] Let \(v_B\) be entirely in the horizontal direction (\(\hat{i}\)), then: \[ v_{B_y} = v_{A_y} - \omega_{AB} \cdot (2b) \cdot \cos\theta = \] \[ v_{B_x} = v_{A_x} + \omega_{AB} \cdot (2b) \cdot \sin\theta \] --- ## **Step 5: Solve for \(\omega_{AB}\)** Set vertical velocity of \(B\) to zero: \[ = v_{A_y} - \omega_{AB} (2b) \cos\theta \] \[ \omega_{AB} = \frac{v_{A_y}}{(2b)\cos\theta} \] \[ \omega_{AB} = \frac{2.598}{.6 \times .866} \] \[ = \frac{2.598}{.5196} \] \[ \approx 5.\, \text{rad/s} \] --- ## **Step 6: Find \(v_B\) (Which is \(v_C\))** Now, \[ v_B = v_{A_x} + \omega_{AB} (2b) \sin\theta \] \[ v_B = 1.5 + 5. \times .6 \times .5 \] \[ = 1.5 + 5. \times .3 \] \[ = 1.5 + 1.5 = 3.\, \text{m/s} \] --- ## **Step 7: Final Answer** \[ \boxed{v_C = 3.\, \text{m/s}} \] --- ### **Summary Table** | Quantity | Value | Units | |----------|-------|-------| | \(v_A\) | 3. | m/s | | \(v_{A_x}\) | 1.5 | m/s | | \(v_{A_y}\) | 2.598 | m/s | | \(\omega_{AB}\) | 5. | rad/s | | \(v_C\) | 3. | m/s | --- ### **Final Answer** \[ \boxed{v_C = 3.\, \text{m/s}} \] Let me know if you need the acceleration as well!