Collar B moves up with constant velocity of \(V_B = 2\) m/s. Rod AB is pinned at B. Find the angular velocity of rod AB and velocity of A.[Figure: collar B slides on a fixed vertical rod. Rod AB of length 1.5 m runs from B down to the left to end A;…

Given: \(v_B = 2\) m/s ↑, AB = 1.5 m at 50° to horizontal, A moves along a 30° incline.

Relative velocity: \(\vec v_B = \vec v_A + \vec\omega \times \vec r_{B/A}\), with \(\vec r_{B/A} = 1.5(\cos50^\circ\,i + \sin50^\circ\,j) = 0.9642\,i + 1.1491\,j\) m and \(\vec v_A = v_A(\cos30^\circ\,i + \sin30^\circ\,j)\).

\(\vec\omega \times \vec r_{B/A} = \omega k \times (0.9642 i + 1.1491 j) = -1.1491\omega\, i + 0.9642\omega\, j\)

i-component: \(0 = 0.8660 v_A - 1.1491\omega \Rightarrow v_A = 1.3269\omega\)

j-component: \(2 = 0.5 v_A + 0.9642\omega = 0.6634\omega + 0.9642\omega = 1.6276\omega\)

\(\omega_{AB} = 1.229\) rad/s (anticlockwise), \(v_A = 1.3269 \times 1.229 = 1.630\) m/s

Check (components along the rod must be equal): \(v_A\cos(50^\circ - 30^\circ) = 1.630 \times 0.9397 = 1.532\); \(v_B\cos(90^\circ - 50^\circ) = 2 \times 0.766 = 1.532\). ✔ (The ICR method gives the same result.)

Answer: \(\omega_{AB} = 1.23\) rad/s anticlockwise; \(v_A = 1.63\) m/s directed up the 30° incline.