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Exam2-2007-solutions

# Exam2-2007-solutions - ME 3212 Mechanisms Exam#2 Fall 2007...

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Unformatted text preview: ME 3212: Mechanisms Exam #2 Fall 2007 1. For the inverted slider crank shown below, (a) Diﬁ‘erentiate the velocity equations to get the acceleration equations and put into matrix form. (b) Solve for F and 93. (0) Setup and differentiate the position equations to get expressions for the velocity of point D. ((1) Setup the acceleration equations of point D (you don’t have to solve the equations). D ' . where 02A.=_28mm, r: 100 mm, CB 2 70 mm, CD = 40 mm, e =12 mm, d = 79 mm, 62 2 120°, 93 = 22°, 02 = a); = 20 rad/sec (constant), 93 = —4.4 rad/s (CCW), and f = 345 mm/s. The loop closure equations are as follows: 02A cos 6; + rcos 93 261 02A sin 62 — rsin 03 + e =0 __ and the velocity equations are ‘ ~02A023in02+fc0s93—r03sin93 = 0 02A02c0s62—i'sin93—r03c0563 = 0 “53 ‘ QAQngazwrsmeg—Ms U136: CO Q Dfﬁméméf “Ni 25”“ >49 gﬁﬂ eﬂffftgg/ymg {W aﬁee‘mé/m [*9 O 3/4 92 COEQL_ 05/} 9: 3&9 V6)” 93— (‘93 C0593: ~ CO gtogQB —‘ r9 5C6§93 4 (~932'1‘5ah293 g 03/492926 + 02/} éz 240592 ifl/L [/Vb/C/‘FL‘K ﬁrm; +2 (‘91? 521093 +(‘Qs 30593 ‘ ‘ PM 93 C059; j“; Z- ‘r (as 63 -\$»n€93 Y‘ .J 19} 9/4095;me hwmbar's war ndwrtml QMEMA, a 6 - no 0 4. 255(10) cos: no 2 #100 Sm Z2 cos 31 93 , fZ(B"15‘)(‘H :95 “5133 g * + Ioo( Whowsczw ‘OeraCOSeZ +Ozﬁéif 51016: +Zr‘93 COSQ *(‘éSZ 5més I 1 1 , j ' O +3420) gm(tzo°> ; , + 252490 who“ 22") i i 1 O . g a i i K I I | i i k i i s i I . ! 1 i . 1 i I i i . I x . l i i i a I i , i } 1 1 — Ioo< 4/ 45‘ SMCZU) [47.ch 0.6mm é; AWL; 639fé‘5ﬁ g , 3 => j 24%)? 5L-a\1;;|45 0.37%! {4 él57-3 F 3 ,9; 2 394,6, ”rd/52 1‘:- éié’???) ”AM/52 E Fat” €093 j 5W” Z 5W“ rip/i 0 Ah ’5‘ A}, v ‘v [/5 C) [0061/‘Im 0? {Sbmvﬁ D, — 113x” 9k ~CBcOst—vcbsmeg E - *€_+Cf3\$1n93“¢“3¢0593 D3 - \/)( 3 ”“39 914.93 “ CD65COS§3 i V? : (43\$ Coséj +CJ>93S SW93 1 + CDégz SUN 393 i I “ 11 c. 3+ (439‘: C0393 29f E bx: 9460392 + (““033 C05 93 ~ Cb SMQS i W 2 3 Ac, , D3 : 03 A 37716; "' CPuCB) SINGS "(_ CO C08 @3 E \lx:'0z’4:: “‘9: + (“00393 ’—<(>CB)935\AQE ~w9§ 9959‘ iA— _, ,OZAéaSIhGZ ~02149:[email protected] +r(0383 ~5é3\$n65 , X "" 39:9th “(r 336% Sn93 “(V (margegcogeg ~CD93 Cosés + CD \$32 5M9: W3 03,49? C056: .~ OLAQ: 9‘19: ‘ CSquG: “F QSCdSQg- (“93603533 : “(r4335 903503393 ‘E’ ("”133 63 5h 63 + CD 935.9% +CDQ? ZCostE . (20 pts) Consider the following cam follower motion: Rise in 60 degrees in cycloidal motion to a plateau at 20 mm Dwell for 90 degrees Rise in 45 degrees in cycloidal motion to 60 mm Dwell for 135 degrees Return to zero in the remaining rotation degrees using cycloidal motion. (a) Sketch f (6), f (6), and f” (9) curves for the cam follower on the next page. (b) Identify the equations describing the proﬁle of the follower for each segment. (0) Calculate the overall maximum velocity and acceleration (absolute values) for the follower with a cam rotation speed of 1725 rpm. ((1) True or False: The acceleration for the same rise and rotation is greater for harmonic motion than for cycloidal motion. .6) kéamwc W2C C5043 UA: nQéw) / t3 : 35(ng (,0: ”ZS, I3"— y Zxraw X JILL.“ ': [\$0, éq rad/{yL W“ rev 90 sec ; l i mow mm me m We Fame 6?th I w - z y )3 T1“ V “C 6>MGVX : ZZZ—J} _, é0)/ MM ));M<5.7[’ﬂ ( jg? ‘— i‘TLTT/gﬁ ‘375—\ Amt 1:731:31: 5‘1- f (9) "‘1', 60 4O 20 0 3O 60 90 120 150 180 210 240 270 300 330 360 f”(9) 3. (20pts) Consider the constrained SolidWorks sketch shown below. (a) Label all members of the sketch that have equal lengths using a letter. (i.e. place a letter on each member that should have the same length. (b) Identify the points for the newly synthesized 4—bar linkage that would have no relative trans- lational velocity. OZ .L. 04’ (0) Draw one position of the 4—bar linkage on the sketc ._ (d) What kind of linkage is this? (Crank-Rocker) or Double-Rocker r (Double-Crank) ...
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Exam2-2007-solutions - ME 3212 Mechanisms Exam#2 Fall 2007...

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