# a disc of mass M and radius R is rolling purely with center's velocity v0 on a flat horizontal floor when it hits a step in the floor of height R/4 .the corner of the  step is sufficiently rough to prevent any slipping of the disc against itself.what is the velocity of the disc just after the impact?

148 Points
12 years ago

Dear akash

initially it is purelly rorating

so w=v/R

initial angular momentum about point A

Li = I (Vo/R)  + mV0 (3R/4)

= MRVo/2  + 3/4MRVo

= 5/4MRVo

just after the impact sphear start rorating about point A

Lf  = MV1R

so

5/4MRVo  = MV1R

V1 =5/4V0

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Karan Sharma
22 Points
3 years ago
Sir this is not giving the correct answer................................................................................................................................................................................................................................................................................................................................................................................................

Karan Sharma
22 Points
3 years ago
The initial angular momentum about the point A
L = (mr^2)w/2 + m(3r/4)v = (5mrv)/4

FInal angular momentum about point A , assuming the new angular velocity to be W
L=  (mr^2W)/2 + mr(Wr) = (3mWr^2)/2

Now equating both angular momentums;

5mrv/4 = (3mVr^2)/2

V=5v/6