A block of mass `m` is revolving in a smooth horizontal plane with a constant speed v.If the radius of the circle path is R find the total contact force received by the block

Options

A. `(mv^(2))/(R )`
B. `mg`
C. `m sqrt((v^(4))/(R^(2)) + 4g^(2))`
D. `m sqrt((v^(4))/(R^(2)) + g^(2))`

Answer

D

Explanation

FBD Let `N_(1) N_(2)uarr` be the normal reaction offered by the vetrical and horizontal surface respectively `N_(1)` pushes the block toward the center of circle and `N_(2)` equilibrium the block in vertical by millifying the weight `mg^(-2`) force equation

`N_(1) = ma_(1) = (mv^(2))/(R )`...(i)
`N_(2) = mg` ...(ii)
Hence the net reaction (contact) force is `vec N= vec N_(1) + vecN_(2)`
`|vecN|= | vecN_(2)| = sqrt(N_(2)^(2) + N_(1)^(2)`
Substinuting `N_(1) and N_(2)` we have
`N = msqrt((v^(4))/(R^(2)) + g^(2))`