In the figure given below, the end B of the rod AB which makes angle \[\theta \] with the floor is pulled with a constant velocity \[{{v}_{0}}\] as shown. The length of rod is \[\ell .\] At an instant when \[\theta =37{}^\circ \] |
A) Velocity of end A is \[\frac{4{{v}_{0}}}{3}\]
B) Angular velocity of rod is \[\frac{5{{v}_{0}}}{6\ell }\]
C) Angular velocity of rod is constant
D) Velocity of end A is constant
Correct Answer: A
Solution :
\[{{x}^{2}}+{{y}^{2}}={{\ell }^{2}}\]\[\Rightarrow \frac{dy}{dt}=-\,\left( \frac{x}{y} \right)\,\,\frac{dx}{dt}\] |
\[\therefore \,\,\,{{v}_{A}}=-\,\frac{4}{3}\,\,{{v}_{0}}\] |
Now, \[x=\ell \cos \theta \] |
\[\frac{dx}{dt}=-\,\ell \,\,\sin \theta \,\,\frac{d\theta }{dt}\]\[\Rightarrow \omega =-\frac{5}{3}\left( \frac{{{v}_{0}}}{\ell } \right)\] |
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