A particle of mass 1 kg is subjected to a fo | vedclass

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(A) The force $\vec{F} = -k(x \hat{i} + y \hat{j})$ is a central force directed towards the $z$-axis. The torque $\vec{	au} = \vec{r} 	imes \vec{F}$

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$3$

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(A) The force $\vec{F} = -k(x \hat{i} + y \hat{j})$ is a central force directed towards the $z$-axis. The torque $\vec{	au} = \vec{r} 	imes \vec{F}$ about the origin is zero because the force vector is always in the $xy$-plane and the position vector $\vec{r}$ has components $(x, y, z)$. Specifically,$\vec{	au} = (x \hat{i} + y \hat{j} + z \hat{k}) 	imes (-kx \hat{i} - ky \hat{j}) = -k(x \hat{i} + y \hat{j} + z \hat{k}) 	imes (x \hat{i} + y \hat{j}) = -k(z y \hat{i} - z x \hat{j} + 0 \hat{k})$. However,since the force is purely radial in the $xy$-plane,the angular momentum component $L_z = m(x v_y - y v_x)$ is conserved.At $t=0$,$\vec{r}_0 = (\frac{1}{\sqrt{2}}, \sqrt{2}, 0)$ and $\vec{v}_0 = (-\sqrt{2}, \sqrt{2}, \frac{2}{\pi})$.The $z$-component of angular momentum $L_z = m(x_0 v_{y0} - y_0 v_{x0}) = 1 	imes [(\frac{1}{\sqrt{2}})(\sqrt{2}) - (\sqrt{2})(-\sqrt{2})] = 1 + 2 = 3 \ kg \ m^2 \ s^{-1}$.Since $L_z$ is conserved,at any time,$x v_y - y v_x = 3 \ m^2 \ s^{-1}$.

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