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 collision, what is their common angular speed (in rev/s) and in which direction do they rotate?

Explanation:

I1 = 0.5 x m1 x r1²

= 0.5 x 2 x 0.6²

= 0.36 kg.m²

m2 = 2 kg

r2 = 40 cm = 0.4 m

w2 = 20 rev/s (counter clockwise)

I2 = 0.5 x m2 x r2²

= 0.5 x 2 x 0.4²

= 0.16 kg.m²

let wf be the final angular velocity of the two disks.

Applying conservation of angular momentum ;

(I1 + I2) x wf = I1 x w1 + I2 x w2

wf = (I1 x w1 + I2 x w2) / (I1 + I2)

= (0.36 x (-20) + 0.16 x 20 )/(0.36 + 0.16)

= -7.69rev/s

The negative sign indicates the disks rotate clockwise direction.

we need some answer choices, or something like that

explanation:

0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

360seconds i 1 hr