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. `((sqrt3g)/(R))^(1//2)`
B. `((sqrt3g)/(2R))^(1//2)`
C. `((g)/(sqrt3R))^(1//2)`
D. `((sqrt2 g)/((sqrt2 + 1)R))^(1//2)`

Answer

D

Explanation

The bob of the pendulum moves in a circle of radius
`(R + R "sin" 45^(@)) = ((sqrt2 + 1)/(sqrt2))R`
T sin `45^(@) m ((sqrt2 + 1)/(sqrt2)) R omega^(2)`
T cos `45^(@) = "mg" " " omega = sqrt((sqrt2 g)/((sqrt2 + 1) R))`