Question

A particle of mass $m$ is moving along a trajectory given by $x = x_0 + a \; \cos \omega_1 t$
$y = y_0 + b \; \sin \omega_2 t$
The torque, acting on the particle about the origin, at $t = 0$ is :

• A. $m (- x_0 b + y_0 a) \omega_1^2 \hat{k}$
• B. $ + my_0 a \omega^2_1 \hat{k}$
• C. $ - m (x_0 b \omega^2_2 - y_0 a \omega_1^2 ) \hat{k}$
• D. Zero

Answer 1

Answer: (B)

$F =-m \left(a\omega^{2}_{1} \cos\omega_{1}t \hat{i} + b\omega^{2}_{2} \sin \omega_{2} t \hat{j}\right)$
$ \vec{r} = \left(x_{0} + a \cos \omega_{1} t\right) \hat{i} + \left(y_{0} +b \sin\omega_{2} t\right) \hat{j} $
$ \vec{T} = \vec{r} \times\vec{F} = - m\left(x_{0} +a \cos\omega_{1} t\right) b \omega^{2}_{2} \sin\omega_{2} t \hat{k} + m \left(y_{0} + b\sin\omega_{2} t\right) a\omega^{2}_{1} \cos\omega_{1} t \hat{k} $
$ = ma \omega_{1}^{2} y_{0 } \hat{k} $