Q. A uniform disc of radius r lies on a smooth horizontal floor. A similar disc spinning with angular velocity w is carefully lowered into the first disc. How soon do both the discs spin with the same angular velocity ?  The friction coefficient between both the discs is mu.

Rates assured.

forces on the first disc= N=mg.....1

frictional force= nmg.......2

torque due to frictional force is found out by

nmgr(2pi rdr)/(pi R²)    a=0 to b=R    =2/3nmgR    (if no silly mistake)

for the first disc.......2/3nmgR=1/2mR2 A..........A=alpha...........1

A=4/3ng/R

for disc1.....................W= 4/3ng/R *t................1)

for disc2.....................W= w - 4/3ng/R*t............2)

we get t=3wR/8ng.......

might be a silly mistake but the concept is very much the same.........

rate if useful......

Friction is the force that retards the first disc and accelerates the second disc. Hence, to find the frictional torque on a disc, consider an annular ring of thickness dx. Its area = =

Hence, mass of the disc = (Assuming areal distribution of mass, and assuming m is the mass of the disc)

Hencr frictional force =

Torque of friction about this elemental ring about center =

Total torque of friction =

Now, |angular acceleration| for either disc is found by using:

For the discs to acquire a common angular velocity, we have:

Solving, we get t = .

(excuse me for calculation mistakes , if I have made any)