Two disks of identical mass but different radii (r and 2r) are spinning on frictionless bearings at the same angular speed ? 0, but in opposite directions. the two disks are brought slowly together. the resulting frictional force between the surfaces eventually brings them to a common angular velocity.
a) what is there final angular speed?
b) what is the change in the rotational kinetic energy?

a) 3/5*ω₀ b) -4/5*m*r₁²*ω₀²

Explanation:

Assuming no external torques acting during the time while both disks are brought together, total angular momentum must be conserved.

⇒ L₀ = Lf

The initial angular monetum can be expressed as follows:

L₀ = I₀₁ * ω₀₁ + I₀₂*ω₀₂ (1)

The moment of inertia of a solid disk, regarding an axis going through the center of the disk, is expressed as follows:

I = m*r²/2

If m₁ = m₂ = m, r₁ = 2*r₂, and ω₀₁ = - ω₀₂, replacing this values in (1), we get:

L₀ = (2*m*r₁²-1/2*m*r₁²)*ω₀ = 3/2*m*r₁²*ω₀ (2)

The final angular momentum can be expressed in this way:

Lf = (I₁+I₂)*ωf = 5/2*m₁*r₁²*ωf (3)

Solving for ωf, from (2) and (3):

ωf = 3/5*ω₀

b) The initial rotational kinetic energy can be written as follows:

Krot₀ = 1/2*I₁*ω₀² + 1/2*I₂*(-ω₀)² = 1/2*(2*m*r₁²+1/2m*r₁²)*ω₀²

⇒ Krot₀ = 5/4*m*r₁²*ω₀²

The final rotational kinetic energy, considering the value of ωf we calculated in a), can be expressed as follows:

Krotf = 1/2*(2*m*r₁²+1/2m*r₁²)*(3/5*ω₀)² = 9/20*m₁*r₁²*ω₀²

The change in the rotational kinetic energy is just the difference between Krotf and Krot₀, as follows:

ΔKrot = Krotf-Krot₀ =( (9/20)-(5/4))*m₁*r₁²*ω₀² = -4/5*m₁*r₁²*ω₀²

the results is a decrease in air temperature with an increase of altitude.

explanation:

0.19 m/s and 0.45 m/s