A rotor is rotating with a constant angular velocity `omega` about its own axis. At the centre of the rotor a spring is fixed of nature length `(l_(0))` whose other end is connected to a block of mass m. Find the minimum value of coefficient of friction between block and rotor wall for which spring remains horizontal. The radius of rotor is `R (R gt l_(0))`.

Options

A. `(mg)/(m omega^(2) R + K (R - l_(0)))`
B. `(mg)/(m omega^(2) R - K (R - l_(0)))`
C. `(mg)/((m omega^(2) + K)R)`
D. `(mg)/((m omega^(2) - K)R)`

Answer

B

Explanation

`N = m omega^(2)R - K (R - l_(0))` ….(1)

For the block to remains in the vertical equilibrium,
f = mg
`rArr N = mg`
`rArr mu (m omega^(2) R - K (R - l_(0)) = mg`
[from equation (1)]
`:. mu = (mg)/(m omega^(2) R - K (R - l_(0)))`