Two particles A and B are situated on a rigid body at a distance d = 2m apart. Particle A has a velocity of 10/ms at an angle of 60° and particle B has velocity V at an angle 30° as shown in figure. The distance d between A and B is constant. The angular velocity of A with respect to B is

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$10 \sqrt{3} rad / s$

$\frac{5}{\sqrt{3}} rad / s$

$5 \sqrt{3} rad / s$

$\frac{10}{\sqrt{3}} rad / s$

answer is B.

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Detailed Solution

Distance d is constant , i.e.,
$Vcos 30^{\circ} = ucos 60^{\circ}$
Or $\frac{V \sqrt{3}}{2} = \frac{10}{2} or V = \frac{10}{\sqrt{3}} m / s$
Magnitude of angular velocity of B with respect to A is
$ω = \frac{usin 60^{\circ} - vsin 30^{\circ}}{d}$
$\begin{array}{l} = \frac{(10) \frac{\sqrt{3}}{2} - \frac{10}{\sqrt{3}} \cdot \frac{1}{2}}{2} \\ = \frac{5 \sqrt{3} - \frac{5}{\sqrt{3}}}{2} = \frac{5}{\sqrt{3}} rad / s \end{array}$