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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2019
  • Hard
  • MCQ

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

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Answer & Solution

Correct Answer

(A) \(m{y_0}a\omega _1^2\hat k\)

Step-by-step Solution

Detailed explanation

\(\overrightarrow{\mathrm{F}}=\mathrm{m} \overrightarrow{\mathrm{a}}=\mathrm{m}\left[-\mathrm{a} \omega_{1}^{2} \cos \omega, \mathrm{t} \hat{\mathrm{i}}-\mathrm{b} \omega_{2}^{2} \sin \omega_{2} \mathrm{t} \hat{\mathrm{j}}\right.\)

\(\overrightarrow{\mathrm{f}}_{\mathrm{t}-0}=-\mathrm{ma} \omega_{1}^{2} \hat{\mathrm{i}}\)

\(\overrightarrow{\mathrm{r}}_{\mathrm{t}-0}=\left(\mathrm{x}_{0}+\mathrm{a}\right) \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}\)

\(\vec{\tau}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}=\mathrm{my}_{0} \mathrm{a} \omega_{1}^{2} \hat{\mathrm{k}}\)

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