Project ForgeThe input link O2P of a four bar linkage is rotated at 2 rad/s in counter clockwise direction, with PQ=O4Q=√2·a and O2P=O2O4=a. The angular velocity of the coupler PQ in rad/s, at an instant when ∠O4O2P=180°, is
- A4
- B2√2
- 1
- D1/√2
Explanation
With ∠O4O2P=180°, points P, O2, O4 are colinear (P and O4 on opposite sides of O2), separated by 2a. Since PQ=O4Q=√2a and PO4=2a exactly satisfy Pythagoras (PQ²+O4Q²=PO4²), triangle PQO4 has a right angle at Q. Placing O2=(0,0), P=(−a,0), O4=(a,0) gives Q=(0,a). Velocity of P (ω2=2 rad/s CCW) is (0,−2a). Writing V_Q=V_P+ω_PQ ẑ×(Q−P) and requiring V_Q ⊥ O4Q (since Q rotates about fixed O4) gives ω_PQ=1 rad/s.
Source: GATE ME 2007 (2007)
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