Two particles A, B are moving on two concentric circles of radii \[{{R}_{1}}\]and \[{{R}_{2}}\]with equal angular speed\[\omega \]. At t = 0, their positions and direction of motion are shown in the figure. [JEE Main 12-Jan-2019 Evening] |
The relative velocity \[{{\vec{v}}_{A}}-{{\vec{v}}_{B}}\]at \[t=\frac{\pi }{2\omega }\]is given by |
A) (a)\[-\omega ({{R}_{1}}+{{R}_{2}})\hat{i}\]
B) (b)\[\omega ({{R}_{1}}-{{R}_{2}})\hat{i}\]
C) (c)\[\omega ({{R}_{2}}-{{R}_{1}})\hat{i}\]
D) (d)\[-\omega ({{R}_{1}}+{{R}_{2}})\hat{i}\]
Correct Answer: C
Solution :
[c] The angle transversed in time \[\frac{\pi }{2\omega }\]is\[\theta =\omega t=\frac{\omega \pi }{2\omega }=\frac{\pi }{2}\]i.e., at\[t=\frac{\pi }{2\omega },\]the position of two particles is shown in the figure |
\[\therefore \] The relative velocity\[{{\vec{v}}_{A}}-{{\vec{v}}_{B}}\]is |
\[=-{{R}_{1}}\omega \hat{i}-(-{{R}_{2}}\omega \hat{i})\] |
\[=\omega ({{R}_{2}}-{{R}_{1}})\hat{i}\] |
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