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A particle of mass m is moving along a trajectory given byx = x₀ + a cos ω₁ty = y₀ + b sin ω₂tThe torque, acing on the particle about the origin, at t = 0 is:(A) my₀aω₁² kˆ(B) m(-x₀b + y₀a)ω₁² kˆ(C) -m(-x₀bω₂² - y₀aω₁²)kˆ(D) Zero |
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Answer» D ) ZERO Explanation: is = o x.y |
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