A smooth wire frasme is in the shape of a parabola `y = 5x^(2)` . It is being rotated about y-axis with an angular velocity `omega`. A small bead of mass `(m = 1 kg)` is at point `P` and is in equilibrium with respect to the frame. Then the angular velocity `omega` is `(g = 10 m//s^(2))`

Options

A. `omega = 5 rad//sec`
B. `omega = 10 rad//sec`
C. `omega = 20 rad//sec`
D. Depends of x-coordinate of particle

Answer

B

Explanation

A smooth wire …………

`(1)m omega^(2)x cos theta = mg sin theta implies tan theta = (omega^(2)x)/(g)`
`tan theta = (dy)/(dx) = 10x`
`omega = 10 rad//sec`