A uniform disc of mass m and radius r is rotated about an axis passing through its centre and perpendicular to its plane with an angular velocity w. It is placed on a rough horizontal plane with the axis of the disc keeping vertical. Co-efficient of friction between the disc and the surface is u. Find:

(1)the time when the disc stops rotating

(2) the angle rotated by the disc before stopping

consider a elemental ring of radius r...

a small retarding torque is acting on this element say dT due to friction...

dT = df(r)                                              (df is friction force acting on ring)

    =u(dm)gr        ..................1                                  (dm is the mass of ring )

mass per unit area = M/piR²                                  (M is mass of disk)

mass of elemental disk (dm)= (2pir dr)(M/piR²)                              (dr is the thickness of ring)

                                       dm  =2Mrdr/R²               ................2

now putting dm in eq 1 we get

 dT = 2uMgr²dr/R²

integrating both sides

T = 2uMgR/3       .....................3

T = I(alfa)

alfa = 4ug/3R              ..........5                                            (alfa = angular accleration)

w = w_o - (alfa)t

finally disk stops so w=0

t = W_o/(alfa) = 3wR/4ug 

@ = W_ot - (alfa)t² /2    

put t in above eq u will get the angle rotated before coming to rest....

approve my ans if u like