Answer:
6.76 s
Explanation:
The angular impulse on the propeller is equal to the net torque (τ) times time (t), as well as the moment of inertia (I) times the change in angular velocity (Δω). Modeling the propeller as a thin rod, the moment of inertia about the center is equal to one twelfth of the mass (m) times the square of the length (L).
τ t = I Δω
τ t = (¹/₁₂ mL²) ω
t = ¹/₁₂ mL² ω / τ
Next, convert rpm to rad/s.
ω = 2000 rev/min × (2π rad/rev) × (1 min / 60 s)
ω = 209.4 rad/s
Now plug in the values.
t = ¹/₁₂ (40 kg) (2.2 m)² (209.4 rad/s) / (500 Nm)
t = 6.76 s
From Newton's second law of motion, the net torque (τ) is equal to the moment of inertia (I) times the angular acceleration (α). Modeling the propeller as a thin rod, the moment of inertia about the center is equal to one twelfth of the mass (m) times the square of the length (L).
τ = Iα
τ = (¹/₁₂ mL²) α
α = 12τ / (mL²)
α = 12 (500 Nm) / [(40 kg) (2.2 m)²]
α = 31.0 rad/s²
The angular acceleration (α) is equal to the change in angular velocity (Δω) over time (t).
α = Δω / t
t = Δω / α
t = (209.4 rad/s) / (31.0 rad/s²)
t = 6.76 s
