17.7  (Rigid Body Work Energy Equation)

Disk A is of constant thickness and is originally at rest.  It is then allowed to fall in the slot and come into contact with the moving belt BC.  (Belt BC's velocity is v, a constant.)   Once the disk comes in contact with the belt, there is a period of slipping while the disk's outer edge velocity comes up to the belt's speed.   During this period of slipping, there is a constant frictional force between the belt and the disk.  The coefficient of friction between the belt and disk A is m.  Write an expression (for q, in terms of v, r, and m) giving the number of revolutions of the disk necessary to bring the disk up to a constant angular velocity, w.

Comments/Hints:
1.  The only work on the disk A is the work of the friction force, F, and it acts over a distance equal to the arc length of the periphery of the disk A.  So, F = mmg, and arc length = rq.  
2.  Calculate I_A = 1/2mr².
3.  Use the work-energy equation:  T₁ + SU_1-2 = T₂
4.  Kinetic energy stored in a body in rotation is  T = 1/2·I_G·w².
5.  You should get q = v²/(4rmg) radians = v²/(8prmg) revolutions.