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a disc of mass M and radius R is rolling purely with center's velocity v 0 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?

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?

Grade:Upto college level

3 Answers

Badiuddin askIITians.ismu Expert
147 Points
11 years ago

Dear akash

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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
2 years ago
Sir this is not giving the correct answer................................................................................................................................................................................................................................................................................................................................................................................................
 
Karan Sharma
22 Points
2 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

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