Question
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 $$A\left( {1,1} \right).$$ The track is rotated with constant angular speed $$\omega $$ such there is no relative slipping between the ring and the track. The value of $$\omega $$ is
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 $$A\left( {1,1} \right).$$ The track is rotated with constant angular speed $$\omega $$ such there is no relative slipping between the ring and the track. The value of $$\omega $$ is
A.
$$\sqrt {\frac{g}{2}} $$
B.
$$\sqrt g $$
C.
$$\sqrt {2g} $$
D.
$$2\sqrt g $$
Answer :
$$\sqrt {2g} $$
Solution :
$$\eqalign{ & N\cos \theta = mg\,\,{\text{and}}\,\,N\sin \theta = m{\omega ^2}r \cr & \therefore \tan \theta = \frac{{{\omega ^2}r}}{g}\,......\left( {\text{i}} \right) \cr & {\text{Given}}\,y = {x^2} \cr & \therefore \frac{{dy}}{{dx}} = 2x \cr & {\text{or}}\,\,\tan \theta = 2 \times 1 = 2\,......\left( {{\text{ii}}} \right) \cr} $$
From above equations, we get
$$\omega = \sqrt {2g} \,\,\left( {r = 1\;m} \right)$$
$$\eqalign{ & N\cos \theta = mg\,\,{\text{and}}\,\,N\sin \theta = m{\omega ^2}r \cr & \therefore \tan \theta = \frac{{{\omega ^2}r}}{g}\,......\left( {\text{i}} \right) \cr & {\text{Given}}\,y = {x^2} \cr & \therefore \frac{{dy}}{{dx}} = 2x \cr & {\text{or}}\,\,\tan \theta = 2 \times 1 = 2\,......\left( {{\text{ii}}} \right) \cr} $$
From above equations, we get
$$\omega = \sqrt {2g} \,\,\left( {r = 1\;m} \right)$$
