A particle of mass m is moving along a trajectory given by x = x0 + a cos \omega 1t y = y0 + b sin \omega 2t The torque, acting on the particle about the origin, at t = 0 is :

- m( x_0b\omega _2^2 - y_0a\omega _1^2 )\widehat k

m (–x0b + y0a) \omega _1^2\widehat k

Rotational Motion PYQ — JEE Main 2019 · 10 Apr · Shift 1 · Q53 with Solution

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Correct answer: B

The coordinates of the particle are $x = x_0 + a\cos(\omega_1 t), \qquad y = y_0 + b\sin(\omega_2 t).$

We need the torque about the origin at $t=0$ .

Torque about the origin is $\vec\tau = \vec r \times \vec F = \vec r \times m\vec a.$

Since motion is in the $xy$ -plane, $\vec r = x\hat i + y\hat j, \qquad \vec a = \ddot x\hat i + \ddot y\hat j.$

Thus,

= m(x\ddot y - y\ddot x)\hat k.$$ 3. **Compute accelerations** From $$x = x_0 + a\cos(\omega_1 t),$$ we get $$\dot x = -a\omega_1 \sin(\omega_1 t),$$ $$\ddot x = -a\omega_1^2 \cos(\omega_1 t).$$ From $$y = y_0 + b\sin(\omega_2 t),$$ we get $$\dot y = b\omega_2 \cos(\omega_2 t),$$ $$\ddot y = -b\omega_2^2 \sin(\omega_2 t).$$ 4. **Evaluate at $t=0$** At $t=0$, $$x(0) = x_0 + a,$$ $$y(0) = y_0,$$ $$\ddot x(0) = -a\omega_1^2,$$ $$\ddot y(0) = 0.$$ 5. **Substitute into torque expression** $$\vec\tau(0) = m\big(x(0)\ddot y(0) - y(0)\ddot x(0)\big)\hat k.$$ So, $$\vec\tau(0) = m\big((x_0+a)(0) - y_0(-a\omega_1^2)\big)\hat k$$ $$\vec\tau(0) = my_0 a\omega_1^2\hat k.$$ 6. **Match with options** This matches: $$\boxed{\text{B: } +my_0 a\omega_1^2\hat k}$$ 7. **Comparison with stored answer** Stored correct answer is **B**, which agrees with the derived result.