A particle of mass m is moving along a trajectory given by x = x0 + a cos ω 1t y = y0 + b sin ω 2t The torque, acting on the particle about the origin, at t = 0 is: (a) Zero (b) +my0a ω₁²k (c) -m(x₀ bω₂² - y₀ aω₁²)k (d) m (–x0b + y0a) ω₁²k

Answer: (b). At t = 0: r = (x₀ + a) i + y₀ j. ax = -aω₁²cos ω₁ t = -aω₁², ay = -bω₂²sin ω₂ t = 0, so F = -maω₁² i. τ = r× F = (xFy - yFx) k = -y₀(-maω₁²) k So the answer is (b) +my₀aω₁² k.

A particle of mass m is moving along a trajectory given by x = x0 + a… | JEE Main 2019 (10 Apr, Shift 1)

Q55JEE Main 2019 · 10 APR · SHIFT 1Medium

Rotational Kinematics & Dynamics

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:

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Answer: (b)

At $t = 0$ : $\vec r = (x_0 + a)\hat i + y_0\hat j$ .
$a_x = -a\omega_1^2\cos\omega_1 t = -a\omega_1^2$ , $a_y = -b\omega_2^2\sin\omega_2 t = 0$ , so $\vec F = -ma\omega_1^2\hat i$ .
$\vec\tau = \vec r\times\vec F = (xF_y - yF_x)\hat k = -y_0(-ma\omega_1^2)\hat k$
So the answer is (b) $+my_0a\omega_1^2\hat k$ .

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