Disk A, with a mass of 2.0 kg and a radius of 60 cm , rotates clockwise about a frictionless vertical axle at 20 rev/s . Disk B, also 2.0 kg but with a radius of 40 cm , rotates counterclockwise about that same axle, but at a greater height than disk A, at 20 rev/s . Disk B slides down the axle until it lands on top of disk A, after which they rotate together. After the collusion, what is their common angular speed (in rev/s) and in which direction do they rotate?

ω _f = -18 rev/s       (clockwise)

Explanation:

Givens:  

m_1 = 2.0 kg  

R_1 = 60 cm = 0.6 m  

ω _i1=-20 rev/s=(-20 rev/1s)*(2π/1 rev)=-40 π rad/s

m2_1=2.0 kg

R_2 = 20 cm= 0.2 m

ω _i2= 20 rev/s =(-20 rev/1s)*(2π/1 rev)= 40 π rad/s

Because the axle of rotation is frictionless, so there is no external torque that might act on the two disks. That means we can consider the two disks as an isolated system. So that the angular momentum of the system will be conserved and constant. By conservation of angular momentum, the initial angular momentum of the system is equal to the final angular momentum of the system.  

L_i = L_f  

I'll choose counter-clockwise to be the positive direction of the rotation.  

I_1*ω _i1+I_2*ω _i2=I_1*ω _f1+I_2*ω _f2

the two disk, finally, will rotate together as an object

ω _f1=ω _f2=ω _f

I_1*ω _i1+I_2*ω _i2=║I_1+I_2║ω _f

ω _f=I_1*ω _i1+I_2*ω _i2/I_1+I_2                      (1)

now we are going to find I_1 and I_2

I_1 =m_1*R_1^2/2

     =0.16 kg m^2                                             (2)

I_2 =m_2*R_2^2/2

     =0.04 kg m^2                                             (3)

put (2) and (3) into (1)

ω _f=I_1*ω _i1+I_2*ω _i2/I_1+I_2

ω _f = -36π rad/s

ω _f = -18 rev/s       (clockwise)

answer; i believe the correct answer is (40,000 j) good luck