CH12 - Exercises 4. K = 1/2 I ω 2 , L = I ω , so K = L 2...

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Unformatted text preview: Exercises 4. K = 1/2 I ω 2 , L = I ω , so K = L 2 /2I 9. (a) τ = (m 2 – m 1 )gR = 1.57 N.m; (b) L = (m 1 + m 2 )vR + I(v/R) = 0.64v + 0.16v = 0.8v; (c) τ = dL /dt : 1.57 = 0.8a, thus a = 1.96 m/s 2 . 17. (a) L = mrR = 900 kg.m 2 /s; L f = (1/2 MR 2 + mR 2 ) ω = 990 ω , Thus ω = 0.999 rad/s. (b) 1/2 mv 2 = 750 J, K f = 1/2(1/2 MR 2 + mR 2 ) ω 2 = 409 J Thus ΔK = –341 J 34. 5g(0.3)cos30° + 2g(0.15)cos30° – (0.04)Tsin55° = 0, so T = 466 N 35. T(1.2 cm)cos30° + 2g(13.8 cm) cos30° – (50 N)(28.8 cm)cos30° = 0. Thus T =974 N. 37. (a) T = 200g = 1960 N; (b) ∑ F x : H – Tcos30° = 0, so H = 1700 N; ∑ F y : V–T – Tsin30° = 0, so V = 1.5 T =300g = 2940 N 40. (Left support) ∑ τ: 60g × 2.5 – V 1 (0.5) = 0, so V 1 = 2940 N down (Right support) ∑ F y : V 2 – (360 g) = 0, thus V 2 = 3530 N, up. 41. Fhcos θ = WHcos( α + θ ) = Whcos 51.8°, H = (D 2 + L 2 ) 1/2 /2 = 0.539 m; and tan α = 0.4 Thus F = 200 × 0.539 × cos 51.8°/(1.1 cos30°) = 70 N....
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This note was uploaded on 05/01/2011 for the course ELECTRIC MAT351 taught by Professor Ahmettantuni during the Spring '10 term at Abant İzzet Baysal University.

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CH12 - Exercises 4. K = 1/2 I ω 2 , L = I ω , so K = L 2...

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