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