A particle of mass $m$ is moving along a trajectory given by
$x = x_0 + a\, cos\,\omega_1 t$
$y = y_0 + b\, sin\,\omega_2t$
The torque, acing on the particle about the origin, at $t = 0$ is
$x = x_0 + a\, cos\,\omega_1 t$
$y = y_0 + b\, sin\,\omega_2t$
The torque, acing on the particle about the origin, at $t = 0$ is
JEE MAIN 2019, Diffcult
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$\overrightarrow{\mathrm{F}}=\mathrm{m} \overrightarrow{\mathrm{a}}=\mathrm{m}\left[-\mathrm{a} \omega_{1}^{2} \cos \omega, \mathrm{t} \hat{\mathrm{i}}-\mathrm{b} \omega_{2}^{2} \sin \omega_{2} \mathrm{t} \hat{\mathrm{j}}\right.$
$\overrightarrow{\mathrm{f}}_{\mathrm{t}-0}=-\mathrm{ma} \omega_{1}^{2} \hat{\mathrm{i}}$
$\overrightarrow{\mathrm{r}}_{\mathrm{t}-0}=\left(\mathrm{x}_{0}+\mathrm{a}\right) \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}$
$\vec{\tau}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}=\mathrm{my}_{0} \mathrm{a} \omega_{1}^{2} \hat{\mathrm{k}}$
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