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Two points A and B on a disc have velocities $v_{1} and v_{2}$ , respectively, at some moment. Their directions make angles $60^{0} and 30^{0}$ , respectively, with the line of separation as shown in figure. The angular velocity of disc is:

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$\frac{v_{2} - v_{1}}{d}$

$\frac{\sqrt{3} v_{1}}{d}$

$\frac{v_{2}}{d}$

$\frac{v_{2}}{\sqrt{3} d}$

answer is D.

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Detailed Solution

For rigid body, separation between two points remains same.

$v_{1} \cos 60^{0} = v_{2} \cos 30^{0}$

$\frac{v_{1}}{2} = \frac{\sqrt{3} v_{2}}{2} \Rightarrow v_{1} = \sqrt{3} v_{2}$

$ω_{disc} | \frac{v_{2} \sin 30^{0} - v_{1} \sin 60^{0}}{d} | = \frac{| \frac{v}{2} - \frac{\sqrt{3} v_{1}}{2} |}{d}$

= $\frac{| v_{2} - \sqrt{3} \times \sqrt{3} v_{2} |}{2 d} = \frac{2 v_{2}}{2 d} = \frac{v_{2}}{d} \Rightarrow ω_{disc} = \frac{v_{2}}{d}$

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