Q 11-06-207JEE MainJEE Main 2019 (10 Apr, Shift 1)Medium
A particle of mass $m$ is moving along a trajectory given by
$$x = x_0 + a\cos\omega_1t,\qquad y = y_0 + b\sin\omega_2t$$
The torque, acting on the particle about the origin, at $t = 0$ is:
Answer: (A) $+my_0a\omega_1^2\,\hat k$
At $t = 0$ the position is $(x_0 + a,\ y_0)$. The accelerations are
$$a_x = -a\omega_1^2\cos\omega_1t = -a\omega_1^2,\qquad a_y = -b\omega_2^2\sin\omega_2t = 0$$
so $\vec F = -ma\omega_1^2\,\hat i$.
$$\vec\tau = \vec r\times\vec F = \left[(x_0+a)\hat i + y_0\hat j\right]\times\left(-ma\omega_1^2\hat i\right) = my_0a\omega_1^2\,\hat k$$
Solution by Sreeraj P, M.Sc Physics