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