2031 03 2 t 4 1 2 0061 009t 4 1 2 a b c t

Info iconThis preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: / 2 (a) (b) (c) t t t 0: 2 s: 4 s: aB 0.06(1 0) aB aB aB 0.0600 m/s 2 0.0937 m/s 2 0.294 m/s 2 aB aB 0.06(1 0.09 24 )1/ 2 0.06(1 0.09 44 )1/ 2 Chapter 15, Solution 23 at at r (0.025 m)(120 rad/s 2 ) 3 m/s 2 t (120 rad/s 2 )(0.5 s) 2 (a) t 0.5 s: an r 60 rad/s (0.025 m)(60 rad/s) 2 an 90 m/s 2 90.05 m/s 2 2 aB at2 t 2 an 32 902 240 rad/s aB (b) t 2 s: an an 2 aB (120 rad/s 2 )(2 s) 2 r (0.025 m)(240 rad/s) 2 1440 m/s 2 at2 2 an 32 14402 aB 1440 m/s 2 Chapter 15, Solution 41 vB [ vB vA vB/A 30 ] [1.2 m/s [vB/A 25 ] Law of sines. vB sin 65 vB/A sin 60 1.2 m/s sin 55 vB v B/A 1.269 m/s 65 (b) 1.328 m/s 30 vB/A 1.269 m/s AB ( AB) AB (0.5 m) AB 2.538 rad/s AB (a) 2.54 rad/s Chapter 15, Solution 50 Arm AB : AB 90 rpm rAB AB 3 rad/s (15)(3 ) 4 rad/s 24 in./s 45 in./ s vB Gear A: A 120 rpm rA A vD (6)(4 ) Gear B : 24 vD vB 45 v D/B v D/B vD/B 21 in./s vD/B B rB 21 9 rad/s B 70 rpm Chapter 15, Solution 89 Locate the instantaneous center at intersection of lines draw perpendicular to vA and vB. Law of sines. AC sin[90 ( )] BC sin(90 l sin BC cos l sin cos( l sin cos l sin ( AC ) l ) AC cos( ) AC BC ) Angular veloci...
View Full Document

Ask a homework question - tutors are online