question_answer

Two points A & B on a disc have velocities \[{{v}_{1}}\] & \[{{v}_{2}}\] at some moment. Their directions make angles \[60{}^\circ \] and \[30{}^\circ \] respectively with the line of separation as shown in figure. The angular velocity of disc is:

A) \[\frac{\sqrt{3}{{v}_{1}}}{d}\]             

B) \[\frac{{{v}_{2}}}{\sqrt{3}\,d}\]

C) \[\frac{{{v}_{2}}-{{v}_{1}}}{d}\]

D) \[\frac{{{v}_{2}}}{d}\]

Correct Answer: D

Solution :

For rigid body separation between two point remains same.

\[{{v}_{1}}\cos 60{}^\circ ={{v}_{2}}\cos 30{}^\circ \]

\[\frac{{{v}_{1}}}{2}=\frac{\sqrt{3}\,\,{{v}_{2}}}{2}\] \[\Rightarrow \]   \[{{V}_{1}}=\sqrt{3}\,\,{{v}_{2}}\]

\[{{\omega }_{\text{disc}}}=\left| \frac{{{v}_{2}}\sin 30{}^\circ -{{v}_{1}}\sin 60{}^\circ }{d} \right|=\left| \frac{\frac{{{v}_{2}}}{2}-\frac{\sqrt{3}\,{{v}_{1}}}{2}}{d} \right|\]

\[=\left| \frac{{{v}_{2}}-\sqrt{3}\times \sqrt{3}\,{{v}_{2}}}{2\,d} \right|=\frac{2{{v}_{2}}}{2d}=\frac{{{v}_{2}}}{d}\]\[\Rightarrow \]      \[{{\omega }_{\text{disc}}}=\frac{{{v}_{2}}}{d}\]