HW_6 Solutions

HW_6 Solutions - ME 212 — Dynamics Due 5/15/09 Self SO9...

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Unformatted text preview: ME 212 — Dynamics Due 5/15/09 Self SO9 Homework #6 — Due May 11, 2009 1. The crank-slider mechanism is one of the most common in engineering. Determine the linear acceleration of point A for 0= 90° if é = lrad/ s 65 = 3rad / s2 . ' 2. (worth 10 points) The net force exerted on the piston by the exploding fuel-air mixture and friction is 5 kN to the left. A clockwise couple M = 200 N—m acts on the crank AB. The moment of inertia of the crank about A is 0.0003 kg-mz. The mass of the connecting rod BC is 0.36 kg, and its center of mass is 40 mm from B on the line from B to C. The connecting rod’s moment of inertia about its center of mass is 0.0004 kg—mz. The mass of the piston is 4.6 kg. The crank AB has a counterclockwise angular velocity of 2000 rpm at the instant shown. What is the piston’s acceleration? (Neglect the gravitational forces on the crank and connecting rod.) Note: Don’t spend too much time with the algebra — okay if you just set up the equations and unknowns (might lose 1/2 point if don’t solve out). You should be able to find the angular velocity of the connecting rod: Partial Ans: 0313c = ~66.4 rad/s. ‘96"? 3. Prob 6/90 in your book. Partial answer: does not slip, oc= 0.1121 rad/s2 -- may want to wait until after Wed class to attack this one. 4. Problem B/l in your book. (a: Gwam 9:909, “é?! @953 hawk; F;wa lak _. _, _ , . TYPE" 9‘3 V Kwémmw—S flag-Tm: _ 77> we, Eséx—F‘F ' 4mm " , . ’ ' L " ’ J r A “3‘ w a + ‘2 xr » (,0 53% w 6% - ac.‘ Alé war/4w“; —= _ 0,43 A 0 .3 .->- 2 2,. 3A '3 ?% 9'59/9w; ‘5‘ fiwxfigg “’ fiié W; a 2 =~— 3£xv(a,4’a‘)_ bH’C ( 9‘»- ? oéeafi xii—03C +90%?) sfg'og, 31 4» mafia); ' A (D " 2. fig? 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(9324 43‘; mzoa wag 0.03%5mfi a» @2032) wag: + c. was a“ + 0,932,! 75w um fi’va: ArDDéD Pm VNWefiww~ 9.296;, as; My: 20w % x1; = WW “3/5 % 35500 . a»: 35500 ' (pjz “QM W 59% mag? 3&3, Vmfl’ 5%»? m.» v3 «2: x521. » Mg g “Fm 2-: gagww(a.osgaz+mufl ! “1767:: mmaxm + 39.3;an m/fi “dc; T? 4" _ v ’A - <7; = 'lmz'lat + 80mg 4— wgéribXCO'mogfi'O'ng‘fi V \lc} Vaxfi+§gfi *3» “(admhay-Béazzfi‘ «I» @298 wfij + 0.07224ka ,,, “WA H . V _ 66W; J D: 8.9,az+a.xaa& Wm I mafs Mm??? EELME; azééfigéj «mm, )waaflés I; CAN be .WQfi-TTF B We, <2: -- 1:" £99056 «2.; [C345 "30) Q25, 3ao '9' am v (“é/c Ozéx I‘: “H @2653“ 1 Zia 4* dmfixéafigflwsmfii%@.oesgm,qfi> ’ ~ +‘ 0005553fl!")5qfi\> "A L ' 0% 1 ‘3!“ gaucazyaagaff -— 0.0.29 (Kw/D + owallébwv ’8 7/ “ aozfléwa) j‘ We?” £35) 343% " ' ' 0% : ac“ —-Q,§>2fi «1,30 +0,@62;.r(fw,w)z’ i" ‘ y L V (1&3 C 9% ~ 0082/0950— 0,024‘I(-«%H) . 25% Maw" MM ‘7};‘15 mpg. fl Lgfié 72%». A.» 42?: f _€;E.N&£$§‘§% w Elam 39M .iéwéwmza Ma. 'Nfiwiwé, ' @an WM 0mg. WE aw [25% 777%: \ I ' 'szcas “BM 1313} AWQ @755 “1% b53¢fl aura, QM; AW” )3 'm 4,7 7‘7“%§ mmm 445w gnaw? pus DETéKJMI/vfé may: m “mg uflw/i gm ' 35500 ' (a) ’2 m0 ’3 X O; 200; aCx Cy N 4.6 0 0 0 —1 1 0 0 0 0 1 0 0 0 0821 0 0 O 0 .0 0 0 0 0 0 1 0 0 0 0 0 BX aZGx By aGZy alt 0 0 0 0 0 0 0 0 —1 -0.36 0 O 0 0 —1.0 —0.36 0.01029 0 0.03866 0 —0. —0.0321 0 0.0383 0 0 0 0 0 0. 0 0 0 0 0. 0 —1 0 O —0. 0 0 0 1 O. O.O383*WAB(3)A2+O.1208*WBC(3)“2; O.O321*WAB(3)A2—0.O321*WBC(3)A2; -O.0821*WBC(3)A2; —O.0219*wBC(3)02; X=M\A -4043.8 N -207.86 m/s.2 914.67 N 914.67 N -3918 N -349.48 m/s2 1629.1 N —1984.5 m/s2 22995 rad/52 -39459 rad/s2 “BC alf AB 0 0; 0 0; 0 0; 0 0; 0004 0; 0 —0.0003; 0321 —0.0321; 1208 0.0383; 0219 O; 0821 O; ] éz/v) Find lxx for the shape to the right Need to set up a differential mass element. What is the definition for Ixx? L I’M-:S r aim Have r and dm — really want to rewrite dm in terms of r so we can integrate. How can we do this? Define p = mass/unit volume Use an “annular ring” for am a f L (2/er Ar. differential element T \ Circumference Thickness Now just do the integration. What are your limits? 1:,“ :2- S (2 fl.(zrrrr\clr’ R 9 I“ = 2mm {0 r cit” ll : Z’le’L. 1T 0 2: zirij EW?LR We need to somehow get mass out of this — which means density times volume. m = ALp = 7:12sz Sub this into the equation above 1 1 1 I =-z LR4=— flRzL R2=— R2 xx 2 p 2( p) 2 Reflection: For most design problems, you would simply find your mass properties using your CAD program. ...
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This note was uploaded on 12/10/2009 for the course ME 212 taught by Professor Staff during the Spring '05 term at Cal Poly.

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HW_6 Solutions - ME 212 — Dynamics Due 5/15/09 Self SO9...

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