To find the angular velocity \(\omega\) of member ABC, we need to analyze the motion of point C given its acceleration. ### Given: - Total acceleration of point C: \( a_C = 240 \, \text{in/s}^2 \) - Angle of acceleration: \( 50^\circ \) North of East - Geometry of the rod: - \( AC = 12 \, \text{in} \) - \( CB = 2 \, \text{in} \) - \( BC = 1.5 \, \text{in} \) ### Steps to Solve: 1. **Resolve the Acceleration into Components:** - \( a_{C_x} = a_C \cos(50^\circ) \) - \( a_{C_y} = a_C \sin(50^\circ) \) 2. **Use Acceleration Components:** - The acceleration of point C can be written as: \[ a_C = \sqrt{a_{C_x}^2 + a_{C_y}^2} \] - We need to find the relation between linear acceleration and angular velocity. 3. **Relation Between Linear and Angular Quantities:** - The tangential acceleration \( a_t \) of point C is due to change in angular velocity: \[ a_t = r \alpha \] - The centripetal (radial) acceleration \( a_n \) is: \[ a_n = r \omega^2 \] 4. **Calculate Radius \( r \):** - The effective radius \( r \) for point C from point B is the length of CB: \[ r = 1.5 \, \text{in} \] 5. **Use Components to Find \(\omega\):** - Assume \( a_n \approx a_{C_x} \) since it's pointing towards the center. - \( a_{C_x} = r \omega^2 \) - Rearrange to solve for \(\omega\): \[ \omega = \sqrt{\frac{a_{C_x}}{r}} \] ### Calculation: - Calculate \( a_{C_x} \): \[ a_{C_x} = 240 \cos(50^\circ) \approx 154.64 \, \text{in/s}^2 \] - Insert into the equation for \(\omega\): \[ \omega = \sqrt{\frac{154.64}{1.5}} \approx 10.1567 \, \text{rad/s} \] Therefore, the angular velocity \(\omega\) is approximately \(10.16 \, \text{rad/s}\) (rounded to four significant figures).