hw17 - PROBLEM 1T.4E A uniform 168-11 cube is attached to a...

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Unformatted text preview: PROBLEM 1T.4E A uniform 168-11: cube is attached to a uniform lSfi-ib circular shaft as shown and a couple M of constant magnitude 13!] 1121'111. is applied to the shaft. Knowing that r = 4 in. and L = [2 in... determine the time required for the angular velocity.r of the systern to increase from moo rpm to 21101} rpm. SOLUTION 2 2 Moments of inertia. Cube: imp} + L2) = LEHE] + [E] ] = [1.86951 llzn-sE -fi 12 12 32.2 12 12 1 1155 4 L. l: ‘3' 12 2 J = {1.26915 lh-sz-ft Total: F 1.13s? [basin Angular speeds. all = Itlflt'} rpm = 113;}?! rada's (5:12 2 EDGE! rpm = 29:3 rada's Applied couple. M = 181} loin. = 15.0 1131-11 Syst Momental + first Ext lmpHg = first Moments:3 Moments about cylinder axis: fail + M! = in); — M {LIBSTHEUU * rung _ M _ 15.13 I PROBLEM 17.53 Each of the gears A and B weighs LS lb and has a radius of gyration of i=3; I ._H [.6 in.. while gear C 1weighs 8 lb and has a radius of gyration of 4 in. _ .. 5. fl” .: Assume that kinetic friction in the bearings of gears A, B and C." produces "v fll‘r'?f_______.. ;" couples of constant magnitude 0.1 lh-ft, 0.] lb-fi. and 0.2 lb-fl, .~ respectively. Knowing that the initial angular velocity of gear C is I: xffilu ' 2000 rpm, determine the time required for the system to come to rest. fiin ._-F ‘32. $2.. ”'13- ASL-Ii“. '.« .-,*-_!,"-"\": .i' .. ;: aoLUTIoN _ _ 3 1.5 4 3 Moments ofinertia. I2 = In. = ”210' = {—M = 322.15 x104" lb-sz-ft g 22.2 _ 22.22. {a} a}: . . if = C '1 =—'* =21605x10" lbs-fl g 32.2 Kinematics. {art )1 = 2000 rpm = 209.44 radfs {4.224 = £34223 = .F'L-fl-‘rt- [rs-AL = {2223}. = ${mcjl = $909.44} = 523.60 Iadx's ’.-i Georgi: I202 ‘i' = @ O 52.. a 536st |"i"IIi~mei:Ita1 + Syst Ext ImpH 1 = Syst Momentaz 1'tjrnolnents about A: L (“J-4L + if4 ff'gro'r — (Mr-L! = 0 _: ,, 2.4 [322.15 2.10 ~}[52s.ea} + E" [Em-ctr — an or 0.1.t — 0.2ij4fidt = 0.43362 {1} Gear 3: By a similar analysis 0.1: — 0.2 [FM-cf! = 0.43362 [2} Let x = java: + [are and add Equations [1} and {2}. a2: — 0.2.: = [1.36224 {3} Syst Mnmentu. Syst Ext Imp: fl = first P'l'IIlnwmmI Jmoments about C'.’ IE [arch -— [ML-Li — r5 [Fxfcdr + r(.ij~d.' = I] (21505x 10-3)[209.44}— 0.2: — fiflacm + [Eggs] [LEI + 415:: = 5.7'316 Solving Equations. {3] and {4} simultaneously, r = 11.368, 1: = 1:12 [bag PROBLEM 1 15? Show that, when a rigid slah rotates about a fixed axis through 0 IV. perpendicular to the slab, the system of the momenta of its particles is equivalent to a single vector of magnitude mFa-J, perpendicular to the line 0:}, and applied to a poi_nt P on this line, called the center of percussion, at a distance GP = .t 2 I? from the mass center oi" the slab. SOL UTIDN Kinematics. Point 0 is fixed. System momenta. Components parallel to Hi: Moments about G: ...
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