Let's break down the problem step by step: ### **Given Data** - Angular velocity of bar OC, \( \omega = 10 \) rad/s - Angular acceleration, \( \alpha = 2 \) rad/s² - Distance, \( r = 0.5 \) m - Sliding velocity of pin B along the slot, \( v_r = 2 \) m/s - Acceleration of pin B along the slot, \( a_r = 4 \) m/s² We are to **find the total acceleration of pin B**. --- ## **Step 1: Radial (Centripetal) Acceleration** \[ a_{\text{radial}} = \omega^2 r \] \[ a_{\text{radial}} = (10)^2 \times 0.5 = 100 \times 0.5 = 50\ \text{m/s}^2 \] --- ## **Step 2: Tangential Acceleration** \[ a_{\text{tangential}} = \alpha \cdot r \] \[ a_{\text{tangential}} = 2 \times 0.5 = 1\ \text{m/s}^2 \] --- ## **Step 3: Acceleration Along the Slot** This is the linear acceleration along the slot, given as \( a_r = 4\ \text{m/s}^2 \). --- ## **Step 4: Total Acceleration of Pin B** The total acceleration is the vector sum of: - Radial acceleration (\( a_{\text{radial}} \)): directed towards the center O. - Tangential acceleration (\( a_{\text{tangential}} \)): perpendicular to \( a_{\text{radial}} \). - Acceleration along the slot (\( a_r \)): along the slot direction. Since the tangential and slot accelerations are perpendicular, and the radial acceleration is also perpendicular, we can combine them using the Pythagorean theorem. \[ a_B = \sqrt{(a_{\text{radial}})^2 + (a_{\text{tangential}} + a_r)^2} \] But actually, the slot acceleration and the tangential acceleration are in different directions. If you check the geometry, the slot acceleration is in the direction of motion along the slot, while the tangential and radial are perpendicular. So, the resultant acceleration should be: \[ a_{total} = \sqrt{(a_{\text{radial}})^2 + (a_{\text{tangential}})^2 + (a_r)^2} \] \[ a_{total} = \sqrt{50^2 + 1^2 + 4^2} \] \[ a_{total} = \sqrt{2500 + 1 + 16} \] \[ a_{total} = \sqrt{2517} \] \[ a_{total} \approx 50.17\ \text{m/s}^2 \] But **none of the options match this value**. This suggests perhaps the slot acceleration is meant to be combined only with the tangential, not with the radial. ## **Alternative Approach (Typical for Slotted Arm Problems)** 1. Radial direction: \( a_{\text{radial}} = \omega^2 r \) 2. Along the slot: \( a_{\text{along slot}} = a_r \) 3. Tangential direction: \( a_{\text{tangential}} = \alpha r \) If the acceleration of pin B along the slot is the net tangential acceleration, then: \[ a_{B} = \sqrt{(a_{\text{radial}})^2 + (a_{\text{tangential}} + a_r)^2} \] \[ a_{B} = \sqrt{50^2 + (1 + 4)^2} \] \[ a_{B} = \sqrt{2500 + 25} \] \[ a_{B} = \sqrt{2525} \] \[ a_{B} \approx 50.25\ \text{m/s}^2 \] Still, this does not match the options. --- ### **Let's check if only the acceleration along the slot is considered with the tangential acceleration, ignoring the radial part:** \[ \text{Net acceleration along slot} = a_{\text{tangential}} + a_r = 1 + 4 = 5\ \text{m/s}^2 \] Choices close to this are: - 3.0 m/s² - 8.7 m/s² - 14.6 m/s² But none of these match either. --- ## **Let’s Double-Check the Standard Approach:** For a point moving radially outward with velocity \( v_r \) and acceleration \( a_r \) on a rotating arm: - Radial acceleration: \( a_r - r \omega^2 \) - Tangential acceleration: \( r\alpha + 2\omega v_r \) So: #### Radial Component: \[ a_{r, net} = a_r - r\omega^2 = 4 - 0.5 \times 100 = 4 - 50 = -46\ \text{m/s}^2 \] #### Tangential Component: \[ a_{t, net} = r\alpha + 2\omega v_r = 0.5 \times 2 + 2 \times 10 \times 2 = 1 + 40 = 41\ \text{m/s}^2 \] #### Magnitude of Total Acceleration: \[ a_B = \sqrt{(-46)^2 + 41^2} \] \[ a_B = \sqrt{2116 + 1681} \] \[ a_B = \sqrt{3797} \] \[ a_B \approx 61.6\ \text{m/s}^2 \] No option matches this exactly, but **option 4 ("14.6 or 6.0 m/s²")** seems to be a typo or error. If we consider 14.6 m/s² as the intended answer, let's check the calculations with only the positive components: But based on the correct approach (using Coriolis acceleration): ### **Final Correct Calculation (Using Coriolis Acceleration)** - Radial: \( a_r - r\omega^2 \) - Tangential: \( r\alpha + 2\omega v_r \) \[ a_{tangential} = 1 + 40 = 41\ \text{m/s}^2 \] \[ a_{radial} = 4 - 50 = -46\ \text{m/s}^2 \] \[ a_B = \sqrt{41^2 + (-46)^2} = \sqrt{1681 + 2116} = \sqrt{3797} \approx 61.6\ \text{m/s}^2 \] ### **Final Answer:** **None of the options exactly match, but the correct method is given above. If you must choose from the options, the closest is option 4 (assuming a typo for 14.6 m/s², but the correct value is 61.6 m/s²).** --- ### **Summary Table** | Step | Calculation | Value | |------|-------------|-------| | Radial (\(a_r - r\omega^2\)) | \(4 - 50\) | \(-46\) m/s² | | Tangential (\(r\alpha + 2\omega v_r\)) | \(1 + 40\) | \(41\) m/s² | | Magnitude | \(\sqrt{41^2 + (-46)^2}\) | \(61.6\) m/s² | --- **If you need to select an option, pick option 4 (14.6 m/s²) as the closest. But the actual answer is 61.6 m/s².** --- **Let me know if you need clarification or a diagram for the acceleration components!**