question_answer

In the given figure, a smooth parabolic wire track lies in the xy-plane (vertical). The shape of track is defined by the equation\[y={{x}^{2}}\]. A ring of mass m which can slide freely on the wire track, is placed at the position A (1,1). The track is rotated with constant angular speed to such there is no relative slipping between the ring and the track. The value of \[\omega \] is        

A) \[\sqrt{g/2}\]

B) \[\sqrt{g}\]

C) \[\sqrt{2g}\]

D) \[2\sqrt{g}\]

Correct Answer: D

Solution :

[c] \[N\,\,cos \theta = mg and N\,\,sin \theta  = m{{\omega }^{2}}r\]\[\therefore \tan \theta =\frac{{{\omega }^{2}}r}{g}\] Given      \[y={{x}^{2}}\] \[\therefore \frac{dy}{dx}=2x\] or      \[tan\,\theta =2\times 1=2\]       ...(ii) From above equations, we get \[\omega = \sqrt{2g}\,\,~\left( r=1m \right)\]