| Two particles A and B are situated at a distance d = 2m apart. Particle A has a velocity of 10 m/s at an angle of \[60{}^\circ \] and particle B has a velocity \[v\] at an angle \[30{}^\circ \] as shown in figure. The distance d between A and B the instant shown in figure is constant. The angular velocity of B with respect to A is - |
 |
A) \[5\sqrt{3}\,\,rad/s\]
B) \[\frac{5}{\sqrt{3}}\,\,rad/s\]
C) \[10\sqrt{3}\,\,rad/s\]
D) \[\frac{10}{\sqrt{3}}\,\,rad/s\]
Correct Answer: D
Solution :
| Distance d is constant, i.e., \[v\cos 30{}^\circ =u\cos 60{}^\circ \] 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: |
| \[\omega =\frac{u\sin 60{}^\circ -v\sin 30{}^\circ }{d}\]\[=\frac{(10)\frac{\sqrt{3}}{2}-\frac{10}{\sqrt{3}}.\frac{1}{2}}{2}=\frac{5\sqrt{3}-\frac{5}{\sqrt{3}}}{2}=\frac{5}{\sqrt{3}}rad/s\] |
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