constan angular accleration = (alfa)                       (given)

w  =  w_o + (alfa)t                    

w = (alfa)t                                     (w_o = 0)

normal reaction bw rod & ring is ma ...                                          (a is linear accleration of rod at any time)

a = (alfa)r

N = m(alfa)r                 .......1

friction force (f) = uN = um(alfa)r     ...............2 

when this friction force is equal to centripital force then ring will slip

F_c(centripital force) = mw²r         .........3

equatin 2 & 3

 w² = u(alfa)             ...........4

w = (alfa)t so above eq becomes

(alfa)²t² = u(alfa)

      t² = (u/alfa)

     t = (u/alfa)^1/2

Approved