Answer:
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
