I'm studying for a test and I can't get this concept right in my head
If I bring both disks closer to each other, what exactly makes it so they're not slipping?
The question is this
Two disks are spinning freely about axes that run through their respective centers (see figure below). The larger disk (R1 = 1.42 m) has a moment of inertia of 1140 kg · m2 and an angular speed of 5.4 rad/s. The smaller disk (R2 = 0.60 m) has a moment of inertia of 920 kg · m2 and an angular speed of 8.0 rad/s. The smaller disk is rotating in a direction that is opposite to the larger disk. The edges of the two disks are brought into contact with each other while keeping their axes parallel. They initially slip against each other until the friction between the two disks eventually stops the slipping. How much energy is lost to friction? (Assume that the disks continue to spin after the disks stop slipping.)
The Attempt at a Solution
I know momentum is conserved.
Initial momentum = I1 ω1 + I2 ω2
Kinetic energy is NOT conserved, so Energy Lost = Initial K - Final K
I would use K = 1/2 (I ω)
I know the angular speed also not the same at the end because they have difference radius.
How do I figure out when they're not slipping? What exactly does it mean when it says no slipping?
My best guess was that their tangential velocity is the same, because they're touching.
Initial momentum = Final momentum = 13516
13516 = I ω1f + I ω2f
R1 ω1f = R2 ω2f
ω1f = R2 ω2f / R1
13516 = I (R2 ω2f / R1) + I ω2f
ω2f = 9.66
ω1f = 4.05
Energy Lost = Initial K - Final K
Energy Lost = 0.5(1140 * 5.42 + 920 * 82) - 0.5(1140 * 4.052 + 920 * 9.662) = - 6213.4 J
I end up with a negative energy lost, which seems very wrong.