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# a disc of mass M and radius R is rolling purely with center`s velocity v 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 centre of the disc just after impact in terms of v?

Nirmal Singh.
7 years ago

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^2/2
Vertical distance from S to O = R-R/4 = 3R/4
Moment 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)V
Thereafter it still has to climb the step (energy being conserved from here).
Thanks & Regards,

Nirmal Singh

Devang
33 Points
7 years ago