question_answer

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, \[x=\ell \cos \theta \]

\[\frac{dx}{dt}=-\,\ell \,\,\sin \theta \,\,\frac{d\theta }{dt}\]\[\Rightarrow \omega =-\frac{5}{3}\left( \frac{{{v}_{0}}}{\ell } \right)\]