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 = wo - (alfa)t
finally disk stops so w=0
t = Wo/(alfa) = 3wR/4ug 
@ = Wot - (alfa)t2 /2    
put t in above eq u will get the angle rotated before coming to rest....
approve my ans if u like