Question

A disc of radius `R` has a light pole fixed perpendicular to the disc at its periphery whish in turn has a pendulum of legth `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

A. `(sqrt(3g)/R)^(1//2)`
B. `(sqrt(3g)/(2R))^(1//2)`
C. `(g/sqrt(3R))^(1//2)`
D. `(sqrt(2g)/((sqrt2+1)R))^(1//2)`

Answer 1

Correct Answer - D
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)Romega^(2)`
`T cos 45= mg, omega=sqrt((sqrt2g)/((sqrt2+1)R))`