Final answer:
The angular velocity of the combination is 0.67 rad/s. The percentage of the original kinetic energy lost to friction is 97.32%.
Step-by-step explanation:
To find the angular velocity of the combination, we need to use the law of conservation of angular momentum. Since the cylinders couple and rotate about the same vertical axis, the total angular momentum before the coupling is equal to the total angular momentum after the coupling. The angular momentum of a cylinder is given by Iω, where I is the rotational inertia and ω is the angular velocity.
Let's calculate the angular velocity:
Angular momentum before: I1 * ω1 + I2 * ω2
Angular momentum after: (I1 + I2) * ω
Set the two equations equal to each other and solve for ω:
I1 * ω1 + I2 * ω2 = (I1 + I2) * ω
Plugging in the values, we have (2.0 kg m² * 5.0 rad/s) + (1.0 kg m² * -8.0 rad/s) = (2.0 kg m² + 1.0 kg m²) * ω
Simplifying, we get 2.0 kg m² * 5.0 rad/s - 1.0 kg m² * 8.0 rad/s = 3.0 kg m² * ω
Calculating the left side, we get 10.0 kg m² rad/s - 8.0 kg m² rad/s = 3.0 kg m² * ω
So, 2.0 kg m² rad/s = 3.0 kg m² * ω
Dividing both sides by 3.0 kg m², we get ω = 2.0 kg m² rad/s / 3.0 kg m² = 0.67 rad/s
Therefore, the angular velocity of the combination is 0.67 rad/s.
To find the percentage of the original kinetic energy lost to friction, we need to calculate the initial and final kinetic energies. The initial kinetic energy is given by KE = (1/2) * I₁ * ω₁², where I₁ is the rotational inertia and ω₁ is the initial angular velocity. The final kinetic energy is given by KE = (1/2) * I * ω², where I is the rotational inertia of the combination and ω is the angular velocity of the combination.
Using the given values, the initial kinetic energy is (1/2) * 2.0 kg m² * (5.0 rad/s)² = 25.0 J
The final kinetic energy is (1/2) * (2.0 kg m² + 1.0 kg m²) * (0.67 rad/s)² = 0.67 J
Therefore, the percentage of the original kinetic energy lost to friction is (25.0 J - 0.67 J) / 25.0 J * 100% = 97.32%