A ball is placed on a smooth inclined plane of inclination `theta=30^(@)` to the horizontal, which is rotating at frequency `0.5` Hz about a vertical axiz passing through its lower end. At what distance from the lower end does the ball remain at rest?

Options

A. `0.87m`
B. `0.33m`
C. `0.5m`
D. `0.67m`

Answer

D

Explanation

`omega=2pif=2pi(0.5)=pi"rad s"^(-1)`

From the figure,
`Nsintheta=mRomega^(2)" and "Ncostheta=mg`
or `tan theta=(Romega^(2))/(g)`
`therefore" "R=(g tantheta)/(omega^(2))`
`rArr" "R=(10.tan 30^(@))/((pi^(2)))~~(1)/(sqrt3)m`
Now, `d=(R)/(costheta)=(1//sqrt3)/(costheta)=(1//sqrt3)/(sqrt3//2)=(2)/(3)=0.67m`