Two block of masses `m` and `M` are connected by a chord passing around a frictionless pulley which is attached to a rotating frame , which rotates about a vertical axis with an angular velocity `omega` If the coefficient of friction between the two masses and the surface are `mu_(1)` and `mu_(2)` respectively determine the value of `omega` at which the block starts sliding radially `(M gt m)`

Answer

`omega = [(g[mu_(1) m+mu_(2) M])/(Mr_(2) - mr_(1))]^(1//2)`

Explanation

Evidently the larger block of mass experience move centrifugal force redially outwards compared to the block of smaller block m
`[M gt m and r_(2) gt r_(1)]`
figure shown than FBD

Owing to the large force exerienced by block of mass `M` is tends in fly off redally
In the situation of limiting equilibrium we have
`T = m omega^(2) r_(1) + f_(1) rArr T + f_(2) = M omega^(2) r_(2)`
(where `f_(1)` and `f_(2)` are frictional forces for the two block and the surface)
`f_(1) = mu_(1) mg rArr f_(2) = mu_(2) mg`...(i)
The above two equations get reduced to
`T = m omega^(2) r_(1) = miu_(1) mg`
`T + mu_(2) Mg + M omega^(2) r_(2)` ...(ii)
Subtracting Eq (i) from Eq(ii),
`mu_(2) Mg = M omega^(2) r_(2) - m omega^(2) r_(1) - mu_(1) mg`
`omega^(2)= (g[mu_(1) m+mu_(2) M])/(Mr_(2) - mr_(1)) rArr omega = [(g[mu_(1) m+mu_(2) M])/(Mr_(2) - mr_(1))]^(1//2)`