# Sol-Ch12 - CHAPTER 12 E xercises 3 i 2 j 6 j = 18 k k g m...

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Exercises 1. l   1 = (3i – 2 j   ) × (6 j   ) = 18k kg m 2 /s : l   2 = (2i + 3 j ) × (1.5i – 2.6 j   ) = –9.70k kg m 2 /s l   3 = (–3i + j ) × (5.14i + 6.13 j ) = –23.5k kg m 2 /s; l   4 = (–4i – 2 j ) × (7.5i ) = 15k kg m 2 /s L = –0.2k kg m 2 /s 2. (2i – 3j) × (20i + 28 j ) = 116k kg m 2 /s 2 3. l = m v (a + d) – m v a = mvd 4. K = 1/2 I ω 2 , L = I ω , so K = L 2 /2I 5. l x = 0, l y = 2 l 6. a r = ω × v ; a t = α × r 7. v = ω × r , a r = ω × v 8. F : ke 2 /r 2 = mv 2 /r, so v = (ke 2 /mr) 1/2 . Use v in m v r = nh/2π, to find r = (nh/2π) 2 /k me 2 . 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 . 10. L = mvR + I ω = (m + M/2)vR, while τ = mgR From τ = dL/d we find a = mg/(m + M/2) = 4.9 m/s 2 11. v = 3At 2 i + (2Bt – C) j ; a = 6Ati + 2B j   . (a) L = r × p = MAT 3 (2C – Bt)k ; (b) F = ma = m(6Ati + 2B j ) 12. (a) To the left; (c) L = I ω , ΔL = LΔ θ = L(vΔt/r) = ω LΔtR/r Thus, ΔL/Δt = ω LR/r = I ω 2 R/r = 2KR/r 13 (a) L 1 = I 1 ω 1 = 0.024, L 2 = I 2 ω 2 = 0.0143 ω 2 From L 2 = L 1 , we find ω 2 = 1.68 rad/s. (b) K i = 0.024 J, K f = 0.020 J, thus ΔK = –4 × 10 –3 J CHAPTER 12

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14. (a) L = [3m (d 2 ) + m(2d) 2 + 2m(3d) 2 ] ω = 25md 2 ω ; (b) L = [Md/3 + M(d 2 /12 + (1.5d) 2 + M(d 2 /12 + (2.5d) 2 ] ω + 25md 2 ω = (9M + 25m) d 2 ω 15. (a) I i ω i = I f ω f leads to ω f = 6.4/1.8 = 3.555 rad/s. (b) K i = 0.144 J, K f = 0.128 J, thus ΔK = –0.016 J (c) Δ ω = 0.445 rad/s, α = 0.223 rad/s 2 , so τ = I α = 4.5 × 10 –3 N m 16. (a) L 1 =1/2 (1/2 MR 2 ) ω 1 2 ; L 2 = I 1 + m(0.1) 2 ω 2 Set L 2 = L 1 to find ω 2 = 1.64 rad/s (b) K 1 = 45 mJ, K 2 = 37 mJ, thus ΔK = –8 mJ 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 18. (a) Li = [ML 2 /12 + 2M(L/4) 2 ] (20 rad/s) = 50ML 2 /12 L f = [ML 2 /12 + 2M(L/2) 2 ] ω = 7ML 2 ω 2 /12 Thus ω 2 = 50/7 = 7.14 rad/s (b) 7.14 rad/s; (c) No centripetal force 19. (a) L 1 = (1/2 MR 2 ) ω 1 = 320 kg.m 2 /s, L 2 = (160 + 40x4) ω 2 Thus
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## This note was uploaded on 11/09/2011 for the course EE EE taught by Professor Ee during the Spring '11 term at National Chiao Tung University.

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Sol-Ch12 - CHAPTER 12 E xercises 3 i 2 j 6 j = 18 k k g m...

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