Question

URGENT: A cylinder with rotational inertia I1 = 3.4 kg · m2 rotates clockwise about a vertical axis through its center with angular speed 1 = 5.4 rad/s. A second cylinder with rotational inertia I2 = 1.1 kg · m2 rotates counterclockwise about the same axis with angular speed 2 = 6.4 rad/s.

(a) If the cylinders couple so they have the same rotational axis, what is the angular speed of the combination (in rad/s)?

(b) What percentage of the original kinetic energy is lost to friction?

Answer 1

The angular speed of the combination is 2.51 rad/s. 22.78% of the original kinetic energy is lost to friction.

Step-by-step explanation:

(a) To find the angular speed of the combination, we can use the principle of conservation of angular momentum. The initial total angular momentum is equal to the sum of the individual angular momenta:

I₁ω₁ + I₂ω₂ = (I₁ + I₂)ω, where ω is the final angular speed of the combination. Plugging in the given values, we have:

(3.4 kg·m²)(5.4 rad/s) + (1.1 kg·m²)(-6.4 rad/s) = (3.4 kg·m² + 1.1 kg·m²)ω

Simplifying the equation, we get:

18.36 kg·m²·rad/s - 7.04 kg·m²·rad/s = 4.5 kg·m²·ω

11.32 kg·m²·rad/s = 4.5 kg·m²·ω

ω = 11.32 kg·m²·rad/s / 4.5 kg·m² = 2.51 rad/s

(b) The initial kinetic energy is given by:

K₁ = 0.5 I₁ ω₁² + 0.5 I₂ ω₂²

Using the given values, we have:

K₁ = 0.5 (3.4 kg·m²)(5.4 rad/s)² + 0.5 (1.1 kg·m²)(6.4 rad/s)²

K₁ = 0.5 (3.4 kg·m²)(29.16 rad²/s²) + 0.5 (1.1 kg·m²)(40.96 rad²/s²)

K₁ = 0.5 (100.944 kg·m²·rad²/s²) + 0.5 (44.928 kg·m²·rad²/s²)

K₁ = 50.472 kg·m²·rad²/s² + 22.464 kg·m²·rad²/s²

K₁ = 72.936 kg·m²·rad²/s²

The final kinetic energy is given by:

K = 0.5 (I₁ + I₂) ω²

Plugging in the given values, we have:

K = 0.5 (4.5 kg·m²)(2.51 rad/s)²

K = 0.5 (4.5 kg·m²)(6.3001 rad²/s²)

K = 56.35044 kg·m²·rad²/s²

The percentage of the original kinetic energy lost to friction is:

(K₁ - K) / K₁ * 100

(72.936 kg·m²·rad²/s² - 56.35044 kg·m²·rad²/s²) / 72.936 kg·m²·rad²/s² * 100 = 22.78%