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

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

## 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.

### Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online