• # question_answer A particle of mass m is moving along a trajectory given by $x={{x}_{0}}+a\cos {{\omega }_{1}}t$ $y={{y}_{0}}+b\sin {{\omega }_{2}}t$ The torque, acting on the particle about the origin, at t = 0 is : [JEE Main 10-4-2019 Morning] A) $m(-{{x}_{0}}b+{{y}_{0}}a)\omega _{1}^{2}\hat{k}$B) $+m{{y}_{0}}a\omega _{1}^{2}\hat{k}$C) $-m({{x}_{0}}b\omega _{2}^{2}-{{y}_{0}}a\omega _{1}^{2})\hat{k}$D) Zero

$F=-m(a\omega _{1}^{2}\cos {{\omega }_{1}}t\hat{i}+b\omega _{2}^{2}sin{{\omega }_{2}}t\hat{j})$           $\vec{r}=({{x}_{0}}+a\cos {{\omega }_{1}}t)\hat{i}+({{y}_{0}}+b\sin {{\omega }_{2}}t)\hat{j}$           $\vec{T}=\vec{r}\times \vec{F}=-m({{x}_{0}}+a\cos {{\omega }_{1}}t)b\omega _{2}^{2}sin{{\omega }_{2}}t\hat{k}$           $+m({{y}_{0}}+b\sin {{\omega }_{2}}t)a\omega _{1}^{2}\cos {{\omega }_{1}}t\hat{k}$ $=ma\omega _{1}^{2}{{y}_{0}}\hat{k}$