Statics and Dynamics Quiz
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In a four-bar linkage, the fixed ground link is O₁O₂. Crank O₁A rotates with constant angular velocity ω1=10 rad/s\omega_1 = 10\,\text{rad/s} CCW and has length 0.2 m. The coupler AB has length 0.5 m. The follower O₂B has length 0.3 m. At the instant of interest, O₁A is horizontal (pointing right), AB makes 30° above horizontal (pointing up-right), and O₂B is vertical (pointing upward). The velocity of A is therefore vA=(0)(i^)+(2)j^=2j^ m/s\mathbf{v}_A = (0)(\hat{i}) + (2)\hat{j} = 2\hat{j}\,\text{m/s} (since vA=ω1⋅∣O1A∣=2 m/sv_A = \omega_1 \cdot |O_1A| = 2\,\text{m/s} upward).
Using the velocity loop equation vB=vA+ωAB×rB/A\mathbf{v}_B = \mathbf{v}_A + \boldsymbol{\omega}_{AB}\times\mathbf{r}_{B/A}, combined with the constraint that B moves perpendicular to O₂B (i.e., B's velocity is horizontal at this instant), find ωAB\omega_{AB} (angular velocity of the coupler AB).
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