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Unformatted text preview: Problem 117E“ The 2ltg sphere A is moving toward
the right at —1 mI's when it sttiltes the end of the ﬁltg
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'elocit'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. 1What 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
nitG: cos EDD + Earl‘s: = mills“? cos sc“ co = 4 III“s. c= t].1_ CDETﬁt‘iEILI: of restitution L a' I.
c us—chosED =:—t'G Solving we ﬁnd a. = —24 radJs to = 21 rad'5 cloclttvise 1m Qt. ttI 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.E3 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‘ EIIEJJ. 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‘ +deL +J’]' 1:23 L .
(ml:+:+md[L+r]}g This is e eumphested equetiett III solve. Yeu can drew the graph and
get an unrestate ﬁDlutiﬂlll. 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 amtital. 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 energjtr 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'... 1E _ (If: P Deﬁne
Broil F2:
_ mgL _ _ mgL . _
ifItr. Iffmwhlt‘hp=a.l:lfk} _=a:l'Ldp=£hl:lfk='.
mgL ‘ . l..— . .
ﬂ _'i':h.E1'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=
[IAm 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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 Spring '09
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