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hw11 - PROBLEM 1 5.4 The motion of an oscillating flywheel...

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Unformatted text preview: PROBLEM 1 5.4 The motion of an oscillating flywheel is defined by the relation 3 = Ens—m” sin 4m, where d' is expressed in radians and I in seconds. Knowing that Eu = +14 rad, determine the angular coordinate, the angular velocity, and the angular acceleration of the flywheel when {a} .r = H.125 s, {b} r= cc. SD LUTION s = age-”"5 sin 4st at = % = §fl[—%e'mm simian + 41:94am“. ere-54:1!) r so 4931 23273 4m 23s? _ .dr se a a = — = H“. c'lmlfi sin 4m — e cos4rr£ — é a ”"5 eos41rr — 1632e'm'rfi sin 4m] = —Hl,s'?""'fi [[16 — E] 23in4m‘ — €412 comm] {a} 3o = D.4rad, r= DJESs {MD-*5“ = 0.63245, 4m = — a = {o.4}{e.53245]{1} = D25293radiana e = dream-:14 m = [o.4}[o.sss45} 5—” [1} = 41.92am radfs a: = 43.92:» radfs 4 s a = —[o.4}[e.ess4s} ts — E a2 [1} = —ss.551raar53 or = oas- rad-’52 4 as {Teri =o 9:04 o:- = [H fl!=fl“ PROBLEM 15.1!) In Prob. ISA}. determine the veltneit},r and acceleration of comer C, assuming that the angular velocity is 10 radfs and decreases at the rate of 20 radfs‘. SOLUTION rM = (500 mm}1 — {225 mm}j + (500 111101)}: = [0.5 m]1—[0.225m}j+(0.5 11-11111 1,”; = 1.30.53 + 0.2252 + 0.52 = 0.525 m Angular vefueflj’ vector. .55 = firm = $110.51 — 0.2251 + 0.51:] 3.01 0.525 = (E radfsji — [3.6 radfs}j + (4.3 ratifsjlt Aflgufa? meter-aria” vecmr. 0'. = 3-00.: = 13(055— 0.225j + 0.5k} (.0: 0.525 = ~(lfi radfszh + ("122 radisflj — (9.6 radfsz)k 145(00ng a!" C. “C = {'3' 9‘ r1315 115-3 = —{5flfl 1'1'1111}i = —{fl.5 In}1 i j I: 0,; = s —5.0 4.0 = —2.4j 2 1.31: —D.5 [l fl *0 = 42.40 m-“sfl— {1.000 mafia! Aceeferm‘ion af ('3. aIf = 01. X r1713 + 0} x v5 i j k i j k : -IE 12 —9.|5 + 3 —3.6 4.3 —U'.5 [I D U —2.4 —l.3 = 4.8j + lfik +13i+l4.4j— [9.215 = IBi + 19.2j — 15.611. ac = (10.00 1:053)1+(10.20 «052]1—(15550 mafia 5! PROBLEM 15.15 The earth makes one complete revolution {in its axis in 23 h 56 min. Knowing that the mean radius of the earth is 637”.) kin. determine the linear velocity and aeeeleratien (if a paint on the surface of the earth {a} at the equator, (£2) at Philadelphia latitude 410° nerth, [c] at the North Pole. SOLUTION 23h56min = 23.933h = 36.16 x it)" e. lrev = 2;: rad 221' = m = 22.923 >< mrfi man‘s . X (it? a = e320km = 6.32 K 1:3" m {a = raj, r = Reeagai + Rainer-j v = mxr = wReesrpk = -[22.923 x lfl'fiufiji x iatlcusak = —[464.53ees+;a might a = mx VP = wj x {—chuse-H = —m3Reeacpl = —[33.3T6><10'3eese=na'52)i {a} Equator. [:21 = {1"} new = LUUU v = —{465 nuejlr. v = 465 me's‘l a = —(33.9 x iii—3 mlazli a = [1.13339 units2 «II {in} Pat-imagine. [:2- = 4a“) ease: = 0.?{3604 1: = —[4fi4.52}[fl.?efifl4]k = -{356 m‘sjk v = 356 me! a = —(33.ET6 x 1e'3][0.2aae4}i = —[e.223 x 10-3 na’szli a = 0.132591an 4 {c} New!!! Pain. [gi- = 90“) were = fl PRUBLEM1524 A gear reduction system consists of three gears A, E. and C. GearA starts h.t:~-"-'~"“'--'-.-. _.-‘ from rest at time t = ti and rotates clockwise with constant angular 5ft “1111 3 1.1“ 5‘ . " .H ' ~.-.l‘ m ~_£‘i' (I; 1:“. a '1'. EL 9 E 3-. -.~".:- w T 1 50mm ltflmtlu Winn” . acceleration. Knowmg that the angular velocityr of gear A is fiflfl rpm at '1- time I' = I 3, determine [a] the angular accelerations of gears B and C‘, [It] 3 the accelerations of the points on gears E and (3 which are in contact when r = llfi s. SOLUTION {000323} 2 213a WIFE " w J (a) Attimet=23, wA=fiflflrpm= W :I N, cori ZEL'AI', aA=TA=1flxradfrj Let D be the contact point bcIWeen gears A and B. {an}I = rungs = [50mm] = Sflflrr n'trnr‘s2 I, {an l, 2 500a —=5zradfszj an. =15J1radt‘szji rm” lflfl RH: Let E be the contact point between gears B and C. [ta-E]: = rmag = {30){33} = 3303100053 1 {as}, = 350:: = 1.00033: was!) a: = 3-24 mars-2} 4 rat arr: (h) fit I = this. For gear Rm); = czar = {SEEKS} = 2.51: rails: (11,.) = this); = {512]}{1521‘}: = 3.0343 x103 mmfsz = 3.0343 1112's: 4— 'H {as}, = 250a wants 1; = 335.40 tom's: l = 0.333110%; 3 = 33.03433 + 0.33343 = 3.13 ms2 _ 0.3354 _ , ,0: = 14.29“ a... = 3.133032 3.14.3031 3.0343 PROBLEM 15.24 CONTINUED 0.: = 3.; 4133003003} = 0.333333 1-333} [0,.) = M305. = [100]n[0.333333]|2 = 10330353103 001053 = 1.02303 ma‘51_.. H {.05}r = 0.3354 W32 f {—4 “E = ding}: + {“3}: = W = 1.204.113g _ 0.3354 —3?4 _ 1.03303 _ 43 = 1.234133- ; 30.4 q anfi‘ PROBLEM 15.32 A simple fi'iction drive consists of two disks A and 3. Initially. dial: A has a clockwise angular velocity of Silt] rpm and disk 3 is at rest. It is known that disk A will coast to rest in 6|] 3 with constant angular acceleration. However, rather than waiting until both disks are at rest to bring them together, disk 3 is given a constant angular acceleration of 2.5 rants;r counterclockwise. Detennine [a] at what time the disks can be brought together if the}.-r are not to slip, to} the angular velocity of each disk as contact is made. —flj = 52.3w rada’s t- nasiune uniform angular deceleration ot'disk A during coasting. +3 I} = [mfln + a” = —52.:tso watts} .—.:..' : -.:-. a, = natzss mas“; 1.4—. to". = (sign + air = —52.3sa + oatzssr Let C be the contact point between the gears. For gear A, are = “a; = {sa}[ss.ssa — 3.?2661') 1 ac = {3141s — 52.3sar}mmts L For gear 3. a3 = 2.5 rants: 1'} {HR = as? = 2.5! in: = was}; = {lflflfllfit} = lfiflrmmfs i {a} For condition at no slip, equate the two expressions for Vt" 25flr=3l4l.o—52.360r 1:10.393! ”C“ = 250.’ = 2591st {a} Corresponding angui’or veioctri'ss. is, = ”—r = 259” = 43.293 mas so as“. = 413mm; ‘1 -i — 25916 = 25.9?6 radt's - too (as =24Brp1n34 ...
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