Certainly! Let's break down the problem into **two steps** with explanations after each, and give a final answer at the end. --- ## **Step 1: Find the Instantaneous Centre of Rotation (ICR)** ### **Explanation:** - The instantaneous centre of rotation for a rigid body is the point about which the body rotates at a given instant. - For bar AO, which is rotating about point O, the velocity at O is zero (fixed point). - The velocity at B will be perpendicular to AB, as AB is constrained to move horizontally at B. - To find the ICR, extend the perpendiculars to the velocities of points A and B: - At A: The velocity is perpendicular to AO. - At B: The velocity is horizontal (to the right). #### **Construction:** - Draw a line perpendicular to AO at A. - Draw a line perpendicular to the path of B (which is vertical at B, as B moves horizontally). - The intersection of these lines is the ICR. --- ## **Step 2: Calculate the Velocity of B** ### **Given:** - AO = 90 mm = .09 m - OB = 180 mm = .18 m - Angle AO with horizontal = 30° - Angular velocity, ω = 8 rad/s ### **Velocity at A:** - \( v_A = \omega \times AO = 8 \times .09 = .72 \) m/s (Perpendicular to AO) ### **Distance from ICR to B (\( r_{IC, B} \)):** - For a two-bar linkage like this, if ICR is at O (since O is fixed and B slides horizontally), the velocity of B can be found as: - \( v_B = \omega \times OB \times \sin(30°) \) - \( OB = .18 \) m, \( \sin(30°) = .5 \) #### **Calculation:** \[ v_B = 8 \times .18 \times .5 = 8 \times .09 = .72 \text{ m/s} \] - This is the same as the velocity at A, as expected due to the geometry and constraints. --- ## **Final Answer** \[ \boxed{ v_B = .72 \text{ m/s} } \] --- ### **Summary Table** | Step | Description | Result | |------|------------------------------------------|-------------| | 1 | Locate Instantaneous Centre (ICR) | At O | | 2 | Calculate velocity of B using ICR method | \(.72\) m/s| Let me know if you need the hand-drawn velocity diagram or further explanation!