HW11_solutions_ - Problem 117E“ The 2-l-tg sphere A is...

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Unformatted text preview: Problem 117E“ The 2-l-tg sphere A is moving toward the right at —1 mI's when it sttiltes the end of the fi-ltg slender bar 3. The coefficient of restitution is c = [1.4. The duration of the impact is [LUBE seconds. Determine the magnitude of the average horizontal force exerted on the bar by the pin support as a result of the impact. Solution: Stater- angular momentum is oonserved about point I9. The coefficient of restitution is used to relate the relative velocities before and after the impact. We also use the linear impulse momentum equation for the ball and for the bar. 1 mgt-‘g;L = mg1'_{-_L + Empif {LI}. c1311 =t't31L — t'gl. L Fe: + as: =me;:. a. mgt-‘g; — =m_{t'g;. Solving we find {1 + c]m_{mg l'_-11 1.4{2 ngIIE- lutgHi m-‘sII __ R = — = — = mu N 2(3m_i+mgjflt 2(11 kg]l[fl.[2fl2 s} R =1.2'-I 1L3". Problem 113.3 The length of the bar is l m and its mass is 3 kg. Just before the bar hits the smooth floor. it has angular 1s'elo-cit'f.r to and its center of mass is moving downward at 4 mfs. The coefficient of restitution of the impact is c = EL—‘l. 1|What 1value of to would cause the bar to have no angular velocity after the impact‘.‘ Solution: Given L = 1 an. m = 2 kg. to" = I] srnocth floor. Angular momentum about the oontact point L 1 . L nit-G: cos EDD + Earl‘s:- = mills“? cos sc“ co = 4 III-“s. c= t].-1_ CDETfit‘iEILI: of restitution L a' I. c us—chosE-D =|:|—t-'G Solving we find a.- = —24 radJs to = 2-1 rad-'5 cloclttvise 1m Qt. tt-I fill:l Prehlem 21.3 The mass elf the disk is 2 kg and the mess Elf the slender he: is [H- ltg. Determine the length L Elf the he: st: that the elf smell Elseilletitlns Elf the pendulum is 1 3. Strategy: Draw a graph Elf the value Elf the fur a range Elf lengths L to estimate the value Elf L entre— sptlntElng to e peritltl elf 1 s. Selutien: ‘We have md=21tg. ms=U.-1Ltg. r =[I.E|3 In. 1:15. The mmettt of inenie of The system about the piTDt paint is 1 .. 1 ,. . I = Emkl‘ + 3mg?” +md[L +r]'. The equatitm of metiectt fm‘ EII-EJJ. amplitudes is 1 a 1 . , . ' I. (Emu? + :mer‘ +1??de + r3")? + (m: +meIL + r1139 = D Thus. the gated is given by 1 -. 1 ,. . Emsl‘ + Emer‘ +de-L +J’]' 1:23 L . (ml:+:+md[L+r]}g This is e eumphested equetiett III solve. Yeu can drew the graph and get an unrest-ate fi-Dlutifllll. Dd? 3m eett use a rant salve: in yeur calculate: Elt Elclt your EtIIII'ILPUIIEI. T_‘s:i.ttg B. 1'Eltllt selves. we find L =l].ED3 111. St] tum Problem 21.31 A slender bar of mass m and length F is pinned to a fined support as shown. A torsional spring of constant it attached to the bar at the support is unstretcheci when the bar is amt-ital. Show that the equation governing small I.-'i1:ural:ioris of the bar from its 1vertical equilibrium position is Solution: The system is consers'afive. The pivot is a fixed point. The moment of inertia about the fixedpoirit is I = mL1f3. The kinetic energfs of the ration of the bar is The potential energy is the sum of the ene-rgjtr in the spring and the gravitatimel energ' associated with the change in height of the center of mass of the her. my}. l'II'I.= lJ||—|I Hi: — III—costi'}. .h For a consers'aflve 551m. 1 at?” t 1 . t T+V=cocnsL= :I{F} +:.i'ti" — “if [1—costi'} Take the time derivative and reduce: HE" mL: '51?" mgL — — —. kti' — ' ti' =[|. (den: 3 (_dr*_)+ 2 5m] (it? Ignore the yossihle solution E =l]. from winch L: Tilt?" L m {—.)+1—a — “‘9 mad. 3 dr- 2 For small amplitude vibrations sinti' —} ti'. and the canmical form (see Eq. (2191:” of the equation of motion is Problem 21.2.2. The initial conditions of the slender bat in Problem 21.21 are I9 = I] I = U — = If? . d: a La) If It 5 %mgf, show that I9 is given as a function of time by s k—% 1 I9 = —sinm.r. where col —[ 1"?ng to 1mi- thj If It +-: %mgf, show that I9 is given as a function of time by 1 1 — F— It I9 = —u{e'l” — f_'l”}.. where L" = [1,13 ‘ J 2!: -mi" Strategy: Io do part {1:}, seek a solution of the equation of motion of the form x = Ce“, where C and l are constants. Solution: Write the equation of motion in the feast :9 + pqti' — D he — _ e _ 1 1'...- 1-E _ (If: P Define Broil- F2: _ mgL _ _ mgL . _ ifIt-r. Iffmwhlt‘hp=a.l:lfk} _=a:l'Ld-p=£hl:lfk-='. mgL ‘- . l..— . . fl _'i':h.E-1'E£ = 9—1.Assuzlnesso1u11onofthefortnti' =Asinpt+ ‘ are Euros pt. The time derivative is E = FA cos p: — p3 sin pt. Apply ii: the initial confiifims st r=|ZI_ to obtain E =D. and A =—If:|:|31'|'l P Whinh the solution is 5' = — sin pr. P m L I {:3} Substitute: if It 2:- f I . - a.- m L . (b) HI: -:: ‘E I 5' = %Sifll:£hf]. From the definition of the hype:- /_ ? bolit: sine_ sinh F2: sini'hr 1 I: J: = _|[£.kr e—hr} h tit 21’: fiocnt which the solution is fil = ink“ _ E—M] . [Check An alternate solution fun.“ part based on the suggested strategy is: mgL Fork-=1“ Wiite the equation of motion in the from and. assume a general solution of the form 5' = Cc” + E's—1". Substitute into the equation of motion to obtain [.5 —h:}ti' = III. from whinh J.=:I:Ft. and the solution is 5' =C€M+Dc_kl where the positive sign is taken without loss of genetslitj'. The time derivative is cit? E = FIFE“ — .tzDe-‘M. Apply the initial eotuiitions at t = [I to obtain the two equations: III = C + D. and ti'n. = Fall" —hD.Sol1.'e: ti: c =_. 2:: :3 hi. 2:: from Which the solution is 5' = its“ 2h _ (“1' check] Plubleln 2.1.37 The disk rotates in the horizonffpiane with constant angular 1..relll:|-I:it5P fl = 12 radars. The mass m=2kg slides in a smooth slut in file disksnd is attached In a spring with. constant I: = Bfil] Him. The radial position Elf the mass whm the spring is unstretehed isr20.2111.Atr:l],fl1emassisinfl1ep:lsilim1r= [IA-m and Jr'de = I]. Determine the pusitinn r as a function Elf time. Solution: Using pols: coordinates. Newman's seem law in the r diree'liml is EF, : —Hr — r5.) = mEF — r93). F+(£—fl1)r=£rn in RI f +£15.51 mass)“: = 36 NE. I'luzsulufim is d'r r = Ashlin? +Beesmt+fl.301m. E = Amnesmt — Be: sinand'. Putting in the emdiflms, we have rfiI' =uj=B+n.3u1n-.=u.4m=:- E =n.us93 .11. Thus the equaflnn is r = (DDSSE In.) nesflfifi taxi“: t] + [0.301 m). ...
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This note was uploaded on 03/29/2009 for the course MECHENG 240 taught by Professor Perkins during the Spring '09 term at University of Michigan.

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HW11_solutions_ - Problem 117E“ The 2-l-tg sphere A is...

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