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If we assume disc and step are inelastic then we can use
conservation of linear and angular momentam
Just prior to impact, the disc is instantaneously rotating
about its point of contact with the floor.Let F be that point, S the tip of the step, and O the centre of the disc.The impulse is orthogonal to the line FS, and will bring the
disc to a halt in that direction.S will be the new instantaneous centre of rotation.Moment of inertia of disc is.Prior to impact, moment of disc about its centre is MR^2/2.Moment of disc about F, I = MR^2/2 + MR^2 = 3MR^2/2Vertical distance from S to O = R-R/4 = 3R/4Moment of disc about S, I’ = MR^2/2+M(3R/4)^2 = 17MR^2/16
Angular speed before striking is w = V/R and after striking is w’ = V1/(3R/4)If just after impact disc has rotational speed w (and is in rolling contact with S) then by conservation of angular momentum about S:\
Iw = I’w’
(3MR^2/2) *(V/R) = (17MR^2/16)*(V1/(3R/4))
V1 = (36/34)VThereafter it still has to climb the step (energy being conserved from here).Thanks & Regards,
Nirmal Singh
Askiitians faculty
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