Tension provides the centripetal force: T = mv²/r = 0.5 kg * (4 m/s)² / 2 m = 8 N.

Using v = rω, we have ω = v/r = 2 m/s / 0.5 m = 4 rad/s.

Using ω = ω₀ + αt, we set ω = 0, thus 0 = 12 rad/s - 3 rad/s² * t, giving t = 4 s.

The solid disk has a lower moment of inertia than the hollow cylinder, allowing it to accelerate faster down the incline.

Using τ = Iα, we have α = τ/I = 10 N·m / 2 kg·m² = 5 rad/s².

Using KE = (1/2)Iω², we have 50 J = (1/2)(5 kg·m²)ω², solving gives ω = 10 rad/s.

Rotational kinetic energy is given by KE = 0.5 Iω² = 0.5 * 4 kg·m² * (3 rad/s)² = 18 J.

Using L = Iω, we find ω = L/I = 20 kg·m²/s / 4 kg·m² = 5 rad/s.

Using τ = Iα, we have α = τ/I = 10 N·m / 2 kg·m² = 5 rad/s².

Using the formula ω = ω₀ + αt, where ω₀ = 0, α = 2 rad/s², and t = 5 s, we get ω = 0 + 2 * 5 = 10 rad/s.

Using the formula θ = ω₀t + 0.5αt², where ω₀ = 0, α = 2 rad/s², and t = 5 s, we find θ = 0 + 0.5 * 2 * 5² = 25 rad.

90 degrees is equal to π/2 radians, which is approximately 1.57 radians.

Centripetal acceleration is given by a_c = v²/r = (20 m/s)² / 50 m = 8 m/s².

Centripetal acceleration a_c = rω² = 3 m * (4 rad/s)² = 3 * 16 = 48 m/s².

Work done W = τθ, where θ = 60 degrees = π/3 radians. Thus, W = 15 N·m * π/3 = 15 * 1.047 = 15.71 J.

Work done by torque is W = τθ, where θ = π/2 rad. Thus, W = 5 N·m * (π/2) = 5 * 1.57 = 7.85 J.

Torque is indeed the rotational equivalent of force, making this statement true.