PhysicsEasy49×since 2002Q8108A particle of mass m is moving along a trajectory given by x = x₀ + a cosω\omegaω₁t y = y₀ + b sinω\omegaω₂t The torque, acting on the particle about the origin, at t = 0 is :AZeroB+my₀a ω12\omega _1^2ω12k^\widehat kkC−m(x0bω22−y0aω12)k^- m\left( {{x_0}b\omega _2^2 - {y_0}a\omega _1^2} \right)\widehat k−m(x0bω22−y0aω12)kDm (–x₀b + y₀a) ω12\omega _1^2ω12k^\widehat kkCheck answerSkip