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Exam 3 2004

# Exam 3 2004 - ES 223 —— Rigid Body Dynamics Exam ILI...

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Unformatted text preview: ES 223 —— Rigid Body Dynamics Exam ILI March3l,2004 NAME solg‘ijgﬂé Closed book and closed notes. Show all work and enclose your ﬁnal answer with a box. Problems will be graded for partial credit only if the work is neat and all intermediate steps are clearly shown and pmented in an easy to followr format. Equations: Uniformly accelerated rectilinear motion: Accelgmﬁon due to gravity: _ 2 v=vo+m g—9.81m/s 2 1 _ g = 32.2 ﬁ/sec s -- so +vot+§m2 v2 = v02 + 2a(s — so) Normal/tangential math! coordinates: 17 = vé, ﬂ v 2 . . A (for circular motion) a = —e,' + vet ' P v = r9 2 v - - 2 . . an=——:Pﬂz=vl3 an:V—=r62=v9 P r 0:213:3’ arzfizré‘ a = on 2 + 61,2 Polar coordinates: guadratic mtion: 17=fér+rééa ax2+bx+c=0 i=(F—ré2)é,+(r§+2f9)éa xZ—bixlbz—4ac 20 (for circular motion) v, = 0 Power: v9 = r0 P: F47 _ . 2 a, — ”r6 Panama: £19er "’ P Work-energy: Coefﬁcient of restitution: 2; +Um = 2; e z (14),. —(v;),. 3T1 + Vg, +Vﬂ +U{_,2 = T2 +ng + Vﬁ2 (v1)" “(v2)" U : IF-df T =imv2 2 1 . U|—>2 = -§k(x§ _ x12 ) (Spnng) Selected intggEals: 1 V =mgh K=~kr2 g 2 j d" :wl—ln(a+bx) a+bx b j sin(a + bx)abc = —-ll;cos(a + bx) Impulse-momentum: :2 jocs(a+bx)dx= %Sm(a+bx) E, +I2th: G; ‘s §=m17 ‘2 Ho. +IEA70dt= H02 Rotation about a ﬁxed point: Relative motion —- translating axes: v = rm L; = Es + 2% V2 a - a + a aﬂ=rc222=—=va) ”II—4i —% r , = rat 2 = Q X E Relative motion — rotating axes: gn:_x (Bxﬂ 211233 +QXI+XM Que—ml: 9,; =23+Q><£+Q>< wle+2ag><2mr+2rex E; = Q X 1: Law of Sines: C a a _ b c sin A sin B sin C b 1. (25) Rod BDE is partially guided by a roller at D which moves in a vertical track. Knowing that at the instant shown the anguiar velocity of crank AB is 5 rad/s clockwise and that B = 25 °, determine (a) the magnitude of the anguiar velocity of the rod and (b) the magnitude of the velocity of point E- Use the method of instantaneous center of zero velocity. "If:— no guy»! r' -“63 wBDE' "“ I‘C—B ET: =‘IOOLOS'AS: (934.4mm 1:56 2 100 Sm '25: 84-5 MM 2. (30) In the engine system shown below, the crank AB has a constant clockwise angular velocity of 210 rad/sec. For the crank position shown, determine (a) the angular velocity of the connecting rod BD (rad/sec) and (b) the velocity of point D (in/sec). Point D is attached to the piston which is constrained to move in the horizontal direction. Express your answers in vector form. Relax-HR M 0*10‘11 '? Translaﬂ-mra Am 54in hm AngleB=14° RH“ ______ 0—“) 1;, o 15B j) ~Dzﬁ+yple 1313:1153 A 3333 2 £989 x [BIA—- -l\ohx(3cmoz+3smvo;) -— 4mg- +4051 EDIE: [533,0 KID/B? @Bn’x‘txpwsmt-a 5»: I43) A 2 730901503: +'Iq+ (.039 ’L’ 7M“ “mm (D A MD}: -.-. #833 +WO§2 +7.74. wmgnnqwmp Equategi 0: ~U£3 +7. 76 can => W 1,233!) = £2.31: (Ml/sec Equale'ﬁf ”Us ‘- so; +l:°l'-l(£'bl) '2: 31b y)?- 5am m] I Sec. 4 3. (3 5) Block B of the mechanism is conﬁned to move within the slot of member CD, IfAB is rotating at a constant angular velocity of 3 rad/s, determine the anguiar acceleration of member CD at the instant shown. Express your answer in vector form. R QVA‘VQ rho-H 0“ At the instant considered men = 0.75 rad/s CW To'h’l +‘% 0046’!) s a! u! '7‘: 6“- - P-Lommcies tax-HA E but} IS diadneﬁ I ‘H: CD ‘ Must soWe Udocpiu‘ eauaihm 4'0 X . fixer} 13rd . 200mm yﬁg11r¢+k9¢uxfp +1J'ml/i: G) .913 =- nggx ”In = ink): (noncosbot- IOU 5m 60%) :-—[50§'-—3LBOL own/5 ' 1.31: O A A ﬁne? “CF/6: ~0a75’4aXQ002 = 450% mv‘IS Then from CD Agog-wool = o ~150§+V1~e\2‘: : mammal: ~1to0 Now aue\em:mw. 95=ﬁ+gmch-wfo EPILJ‘" 139.1%?er 'i 53 rel ' \$5: 2%; ﬁfe/A“ L0; [5,“ : O 5310» (054351 - 1005:0603) = -‘t801+77‘i’a\ 111% {ram ® ‘ 4502mm =0+0fc32x zoo’[email protected]§'a 0078 to) 60053044602) + am 17 A. — Hsol‘niqé‘ -. 100‘?ch— msi} 851034 a”; A. fat/«ﬁe? 77%: 1100‘?“ +3qo =bo‘co-7- HMS lax“): HHS]; tact/51' 4. (5) ' State in words Newton’s Second Law. TM ac¢e{era+/6Y\ VP 0. mQSJ In" proﬁﬂr‘lLIMQ/ +0 +he. resuif-n‘nﬁ #60ch amt-m on ,1— “A Is m We dlwdrm o-P +th ﬁerce 5. (5) As a rigid body changes its position, the instantaneous center also changes its position in space. The locus of the instantaneous centers in space is known as the Sng Q ﬁg 3 iﬂd g (two words). ...
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Exam 3 2004 - ES 223 —— Rigid Body Dynamics Exam ILI...

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