A particle of mass m is moving along a trajec | vedclass

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(A) The position vector of the particle is $\vec{r} = (x_0 + a \cos \omega_1 t) \hat{i} + (y_0 + b \sin \omega_2 t) \hat{j}$.At $t = 0$,$\vec{r} = (x_

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$m y_0 a \omega_1^2 \hat{k}$

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(A) The position vector of the particle is $\vec{r} = (x_0 + a \cos \omega_1 t) \hat{i} + (y_0 + b \sin \omega_2 t) \hat{j}$.At $t = 0$,$\vec{r} = (x_0 + a) \hat{i} + y_0 \hat{j}$.The acceleration is $\vec{a} = \frac{d^2\vec{r}}{dt^2} = (-a \omega_1^2 \cos \omega_1 t) \hat{i} + (-b \omega_2^2 \sin \omega_2 t) \hat{j}$.At $t = 0$,$\vec{a} = -a \omega_1^2 \hat{i}$.The force acting on the particle is $\vec{F} = m \vec{a} = -m a \omega_1^2 \hat{i}$.The torque about the origin is $\vec{	au} = \vec{r} 	imes \vec{F}$.$\vec{	au} = [(x_0 + a) \hat{i} + y_0 \hat{j}] 	imes [-m a \omega_1^2 \hat{i}]$.Using the cross product rules $\hat{i} 	imes \hat{i} = 0$ and $\hat{j} 	imes \hat{i} = -\hat{k}$,we get:$\vec{	au} = y_0 (-m a \omega_1^2) (\hat{j} 	imes \hat{i}) = y_0 (-m a \omega_1^2) (-\hat{k}) = m y_0 a \omega_1^2 \hat{k}$.

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

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