Ch16_Sec1-3_Ex-5-18-25-35-41-46

Ch16_Sec1-3_Ex-5-18-25-35-41-46 - .. A. .as “and... MN;...

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Unformatted text preview: .. A. .as “and... MN; .u . , 16-5. Due to an increase in power, the motor M rotates ’ the 5 ii A with an angular acceleration of a = '1 (0.060z ad/sz, where 0 is in radians if the shaft is initial] turning at (no = 50 rad/s, determine the angular velocit of gear B after the shaft undergoes an angular displac trient A9 = 10 rev. 1 l l l r r I mdw=iad0 2mm LI“ =10 0.06:?2 do l 2800) 1 w‘ imzLo =o'0203 o 0.5mz 1250 == 4961 ( l, a): 111.45 rad/s man 405’s (111.43%)(12) = mac) 'l mum’s Ans 16-18. The vertical-axis windmill consists of two blades that have a parabolic shape. If the blades are originally at rest and Eyegin to turn with a constant angular acceleratio o a. = 0.5 rad/52, determine the magnitude of the veloc ty and acceleration of points A and B on the blade after the blade has rotated through two revolutions. Angular Motipn : The angular velocity of the blade after the blade has rotated 2(2):) = 4n rad can beobtained by applying Eq.16— 7. 1 to2 = afi+2a,(9—90) a): = 0‘ +2(0.5) Hit—0) a) = 3.545 rad/s Motion of A and B r: The magnitude of the velocity of pointA and B on the blade can be determined using Eq. 16 — 8. ' v‘ = mrA = 3.545(20) = 70.9 [1/5 Ans v, = air, = 3.545( l0) = 35.4 ftls Ans The tangential and normal components of the acceleration of pointA and B can be determined using Eqs. l6— ll and 16— 12 respectively. WA = a" = 0.5(20) = 10.0 {:152 (am = min = (3.5452) (20) = 251.33 (Us: (0.). = ar. = 0.5(10) = 5.00 {us2 (an )5 = 01'» = (3.545’) (10) = 125.66 {Us2 The magnitude of the acceleration of points A and B are (an F ‘/(a,)i+(a,,)i =./to.o=+251.332 = 252%2 Arts (a), {fllmfiumg ‘=/5.oo=+125.662 = 126 ms2 Ans m .u -~- mn— -..... -._... n... .V-_ _.._ ‘ .— D..-. accele ation oz= (1001/3) rad/$2, where 0 is in radians Deter me the magnitudes of the normal and tangential compo ents of acceleration of a point P on the rim of the diskw ent==4s. a = 106; ‘ a; mda): ads 3 j”mdm= j‘ “109* :10 y 0 0 i I ‘ I i —af = 1069!} = 7.591 0 ; = I _ ._ a" da 10,1154: ash/Ts} o= 2.152? ,.. =137.71 _d9 i 2| _ w— d‘as‘455: "‘-l03.28 (up). = ah = (103.28)1(o.4) = 4257 ml;2 An: 1UP): = 0" (10(137.7i)§)(o.4) = 20.7 ml;1 Am 16-35. The 2-m-long bar is confined to move in the horizontal d vertical slots A and B. If the velocity of the slider bl k at A is 8 m/s, determine the bar’s angular velocity and the velocity of block B at the instant 9 = 60°. x=2c058 ‘ ‘ 1 1 VA =—2 $an l i it 8: -2 si‘hé‘bfiu a): -4. 1% = -452 rad/s = 4.62 rad/s ‘5 Ans y= 25in 3 i 6 V» = 2c sQaJ 3 Jr“ Vs ='2(c 360°)l—4.6l9) J 7 '° ‘ v. = —4.l62 I'n/s = 4. 62 mls 1 An: F"‘i' ' "‘ l 4 , t 1644 Arm AB has an angular velocity of w and an - 'g angul r acceleration of a. If no slipping occurs between '5 the d SR and the fixed curved surface. determine the angul r velocity and angular acceleration of the disk. l ds=(k-r) d9=-rd¢> ‘(R-r to a} =‘ ) Ans ‘ I' :R—fia 0" =‘% An: “pend mm.“ a . 16-46. 1116 rank AB is rotating with a constant angular velocity of 5 ad/s Determine the velocity of block C and the angulari v locity of link BC. Pas ition Cdordimue Equation : From the geometry. ; x = 0.6cos 9+0.3cos a 0.6sin0=0.15+0.3srn¢ Eliminate ¢ J‘rom Eqs.[l] and [2] yields 1: = 0.6cos 9+0.3‘/25in 9—45in19+0.75 Time Den' att'v es : Taking the time derivative of Eq. [3]. we have d: .- , 0.15(2cos 6—4sin 29) d9 7- == —0.6s1n 9+ 1. «25in o-4sm29+o.7s d! ‘ d9 However, ’ E vc and E = a)”. then from Eq.[4] ‘ . 0.15(2cos 9-4sin 29) DC = —0.6stn 8+ at ‘llsm 6-4sin30+0.75 1 At the instant 0 = 30". a)“ = 5 rad/s. Substitute into Eq.[5] yields 1 1 0.1m 30°—4‘ 50° vc = Harmon °°S 5‘" ) i ‘ ‘/2sin 30°—4sin:30° +0.75 (5) = -3.00 m/s Taking the derivative of Eq [2]. we have ‘ d9 dd) i 0. — = , __ ‘ (Secs 9 d! 0 Sims (0 dt dd d9 However. 7'- = m” and 2-!- = a)”. then from Eq.[6] i i i w _ 2cos 9)w 1 ac- co”) AB At the insram 9 = 30", [mm E412], o = 30.0°. From Eq.[7] 2cos 30" = 5 = .0 a)“ [cos 30.00} ) 10 rails F‘Wwwflwmgflwnpfiuyn vywrrqu V, .wfir t, +1.. ._ Note : Negative #ign indicates that vc- is directed in the opposite direction to that or‘ pdsruvc x. ...
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Ch16_Sec1-3_Ex-5-18-25-35-41-46 - .. A. .as “and... MN;...

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