A disc of radius `R` has a light pole fixed perpendicular to the disc at its periphery which in turn has a pendulum of length `R` attached to its other end as shown in figure. The disc is rotated with a constant angular velocity `omega` The string is making an angle `45^(@)` with the rod. Then the angular velocity `omega` of disc is

Options

A. `((sqrt(3)g)/(R))^(1//2)`
B. `((sqrt(3)g)/(2R))^(1//2)`
C. `((g)/(sqrt(3)R))^(1//2)`
D. `((2g)/(3sqrt(3)R))^(1//2)`

Answer

D

Explanation

`(D)` The bob of the pendulum moves in a circle of radius `(R+R sin 30^(@))=(3R)/(2)`

Force equations `:T sin 30^(@)=m((3R)/(2))omega^(2)`
`T cos 30^(@)=(3)/(2) (omega^(2)R)/(g)=(1)/(sqrt(3))`
`rArr omega=sqrt((2g)/(3sqrt(3R)))` Ans.