A particle of mass $m$ is moving along a trajectory given by $$ \begin{aligned} & x=x_{0}+a \cos \omega_{1} t \\ & y=y_{0}+b \sin \omega_{2} t \end{aligned} $$ The torque, acting on the particle about the origin, at $t=0$ is:

(A) $my_{0} a \omega_{1}^{2} \hat{k}$

(B) $m\left(-x_{0} b+y_{0} a\right) \omega_{1}^{2} \hat{k}$

(C) $-m\left(-x_{0} b \omega_{2}^{2}-y_{0} a \omega_{1}^{2}\right) \hat{k}$

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Rotational Mechanics

A particle of mass

m

is moving along a trajectory given by

x = x_{0} + a cos ω_{1} t y = y_{0} + b sin ω_{2} t

The torque, acting on the particle about the origin, at

t = 0

is:

Answer: (A)

The force on the particle is

F = m a

, and differentiating the position twice gives

a = - a ω_{1}^{2} cos (ω_{1} t) \hat{i} - b ω_{2}^{2} sin (ω_{2} t) \hat{j}

, so at

t = 0

only the

x

-part survives:

F = - ma ω_{1}^{2} \hat{i}

. The position at

t = 0

r = (x_{0} + a) \hat{i} + y_{0} \hat{j}

. The torque is

τ = r \times F = [(x_{0} + a) \hat{i} + y_{0} \hat{j}] \times [- ma ω_{1}^{2} \hat{i}] = - ma ω_{1}^{2} y_{0} (\hat{j} \times \hat{i}) = m y_{0} a ω_{1}^{2} \hat{k}

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Single correct option