A particle of `m_(1)` moves in a circular path of radius R on a rotating .A string connecting the particles `m_(1)` and `m_(2)` passes over a smooth bole made on the table as shown in figure If mass `m_(1)` does not slide relative to the rotating table , mark the correct option as applicable.

Options

A. The friction force acting on the block `m_(1) is (m_(1) omega^(2)R - m_(2)g)` along radial rotation towards center of rotation
B. The friction force acting on the block `m_(1) is (m_(1) omega^(2)R - m_(2)g)` along tangle direction in the direction oppesite to`vecv`
C. The maximum angular velocity of the particle is `sqrt(((m_(2) + mu m_(1))/(m_(1))) (g)/(R ))`
D. The minimum angular velocity of the particle is `sqrt(((m_(2) + mu m_(1))/(m_(1))) (g)/(R ))`

Answer

a,c,d

Explanation

FBD Let the friction radially inward .The force on `m_(1)` are `f larr`(assumed),`T larr N uarr m_(t)g darr` The force on `m_(2)` are `m_(2)g darrand T uarr` as shown in figure
Force equation:
For `m_(1)`:
`sum F_(y) = N -m_(1)g = m_(a)a_(1)y` (i)
`sum F_(r) = f +T = m_(1) a_(r)` (ii)
Fro `m_(2)`:
`sum F_(y) = T -m_(2)g = m_(2)a_(2)y` (iii)
Law of static friction:
`f le mu N`
Kinematics: `a_(1)y = 0`
`a_(r) = (v^(2))/(R)`
`a_(2)y = 0` (vii)
Solving the above equations, we have
`f = (m_(1)v^(2))/(R) - m_(2)g` or `m_(1)omega^(2)R - mg`
Now substituinh the above value of f and `N` in Eq (iv).
we have `((m_(1)v^(2))/(R)-m_(2)g) le mu m_(1)g`
This gives `v_(max) = sqrt(((m_(2)+mum_(1))/(m_(1)))gR)`