A block of mass m is at rest on an incline having an angle of inclination `theta` with the horizontal. The incline is now being rotated with constant angular velocity `omega` about an axis passing through the perpendicular side. If the surface of the incline is rough having coefficient of friction as `mu`, then find the distance 'x' aong the incline so that block just remains stationary with restpect to incline and is on the verge of slipping. [ Take 'x' from the top of the incline as shown in the figure.]

Options

A. `(g)/(omega^(2)) ((mu + tan theta))/((mu tan theta - 1))`
B. `(g)/(omega^(2)) ((mu - tan theta))/((mu tan theta + 1))`
C. `(g)/(omega^(2) cos theta) ((mu - tan theta))/((mu tan theta + 1))`
D. `(g)/(omega^(2)) ((mu - tan theta))/((mu tan theta - 1))`

Answer

C

Explanation

`N + mr omega^(2) sin theta = mg cos theta`
`mg sin theta + mr omega^(2) cos theta = F_("down")`

`mu (mg cos theta - mr omega^(2) sin theta) = F_("up")`
`F_("down") = F_("up")`
`r = (g (mu cos theta - sin theta))/(omega^(2) (mu sin theta + cos theta))`
`x cos theta = r`
`x = (r)/(cos theta) = (g)/(omega^(2) cos theta) ((mu cos theta - sin theta))/((mu sin theta + cos theta))`
`= (g)/(omega^(2) cos theta) ((mu - tan theta))/((mu tan theta + 1))`