The velocities of two particles A and B,separ | vedclass

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(D) The angular velocity of $B$ with respect to $A$ is given by $\omega_{BA} = \frac{v_{\perp}}{r}$,where $v_{\perp}$ is the relative velocity of $B$

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None of these

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(D) The angular velocity of $B$ with respect to $A$ is given by $\omega_{BA} = \frac{v_{\perp}}{r}$,where $v_{\perp}$ is the relative velocity of $B$ with respect to $A$ perpendicular to the line joining $A$ and $B$.Let the line $AB$ be along the $x$-axis.Velocity of $B$ is $\vec{v}_B = (-100\sqrt{3} \cos 60^{\circ}) \hat{i} + (100\sqrt{3} \sin 60^{\circ}) \hat{j} = -150 \hat{i} + 150 \sqrt{3} \hat{j} \ m/s$.Velocity of $A$ is $\vec{v}_A = (-50 \cos 30^{\circ}) \hat{i} - (50 \sin 30^{\circ}) \hat{j} = -25\sqrt{3} \hat{i} - 25 \hat{j} \ m/s$.Relative velocity $\vec{v}_{BA} = \vec{v}_B - \vec{v}_A = (-150 + 25\sqrt{3}) \hat{i} + (150\sqrt{3} + 25) \hat{j} \ m/s$.The component of $\vec{v}_{BA}$ perpendicular to $AB$ (which is along the $y$-axis) is $v_{\perp} = 150\sqrt{3} + 25 \ m/s$.$\omega_{BA} = \frac{150\sqrt{3} + 25}{20} = \frac{150(1.732) + 25}{20} = \frac{259.8 + 25}{20} = \frac{284.8}{20} = 14.24 \ rad/s$.Since $14.24 \ rad/s$ is not among the options,the correct choice is $D$.

The velocities of two particles $A$ and $B$,separated by a distance of $20 \ m$,are as shown in the figure. Find the angular velocity of $B$ with respect to $A$ in $rad/s$.

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