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PROBLEM 2.11 KNOWN: Data are provided for a disk-shaped flywheel. FIND: (a) Obtain appropriate expressions for the moment of inertia and the kinetic energy. (b) Using given data, determine the kinetic energy and mass for a steel flywheel. (c) Using results from part (b), determine the radius and mass of an aluminum flywheel. SCHEMATIC & GIVEN DATA: Steel flywheel: Ω = 3000 RPM R = 0.38 m w = 0.025 m Aluminum flywheel: Ω = 3000 RPM w = 0.025 m ENGR MODEL: 1. The flywheel is the closed system. 2. Motion is relative to the flywheel support structure. ANALYSIS: (a) Evaluating the moment of inertia. dI = ∫_V ρr²dV For the disk, dV = (2πr dr) w. Thus, since ρ is constant I = ρ(2πt)∫_0^R r³ dr = ρtWR⁴/2 ⟵ I The kinetic energy is KE = ∫_0^l (1/2 r²ω²) dV and V = rω, so KE = ∫_0^l (1/2 r²ω²ω²) (2πr dr) w = 1/2 ω²² (2πt)∫_0^R r³ dr = 1/2 (ρtWR⁴/2)(ω²²) = 1/2 Iω² ⟵ KE (b) From Table A-19, the density of steel is ρ_ST = 8060 kg/m³. Thus, the mass is m = ρV = ρ (wπR²) = (8060 kg/m³)(0.025 m)(π)(0.38 m)² = 91.41 kg ⟵ m Using the result of part (a), KE = 1/2 Iω²; where I = πρWR⁴/2 + π/2 (8060 kg/m³)(0.025 m)(0.38 m)⁴ = 6.6 kg·m² PROBLEM 2.11 (Contd.) Accordingly, KE = 1/2 I ω² = 1/2 (6.6 kg·m²) (3000 REV/min) [(2π rad/1 REV) (1 min/60 s)]² (1 N·m / 1 kg·m²/s²) = 32.57×10³ N·m (c) If ω, N and KE are the same for the aluminum flywheel as for the steel flywheel (KE)_AL = (KE)_ST (1/2 I ω²)_AL = (1/2 I ω²)_ST ⇒ I_AL = I_ST or (πρ_WR⁴/2)_AL = (πρ.WR⁴/2)_ST (ρ.R⁴)_AL = (ρ.R⁴)_ST R_AL = (R_ST / ρ_AL) ¹/⁴ ρ_ST With ρ_AL from Table A-19, ρ_AL = 2700 kg/m³ R_AL = (8060/2700) ¹/⁴ (0.38 m) = 0.5 m Then, the mass of the aluminum flywheel is m = ρV = ρ (wπR²) = (2700 kg/m³) (0.025 m)(π)(0.5 m)² = 53.01 kg ⟵ m PROBLEM 2.12 From Problem 2.11, KE = 1/2 I ω² ⇒ ω = [2 (KE/I)]¹/²; where I = 200 lb·ft² The change in potential energy for a 100 lb mass raised 30 ft is ΔPE = mg (z₂ - z₁) = (100 lb) (32.2 ft/s²)(30 ft) (1 lb·s²/32.2 ft·lb·g) = 3000 ft·lb·f Thus, with KE = 3000 ft·lb·f ω = [(2(3000 ft·lb·f) / 200 lb·ft) (32.2 ft·lb·f/s) / 1 lb·f]¹/² = 31.08 s⁻¹ In terms of RPM ω = (31.08 s⁻¹) (1 REV / 2π) (60 s / 1 min) = 297 rev/min ⟵