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SOHAN SARANGI Grade: 12
        

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

6 years ago

Answers : (1)

vikas askiitian expert
510 Points
										

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

6 years ago
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