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Unformatted text preview: 164%. A! a given instant the top 3 of the ladder has
an acceleration a3 = ZfI/sz. and avelociiy of '03 = 4 Ms, both acting downward} Determine the acceleration of the 1“ ‘
bottom A of the ladder, and thc.1addcr‘s angular 4 0238
  .  ﬂ.) = : _
acceleratlon at this Instant. 16mm" 675mm W 4%;
“A = ‘3 + a x rm ' warm "Uh —aAi = —2j + (ak)x{—16cos30°i  lésin30°j) — (6.288675)2(16c053£)°i — Iﬁsin30°j]
a‘ = 8a+ 1.1547
0: 2—13.856u + 0.666? a = —o.o952mm= = omsmwﬂ An: «A = 4.335%: = (Lassa/52a Ans *16112. A! a given instant the whee! is rotating with the anguiar velocity and angular acceleration shown.
Determine the acceleration of block 3 at this instant. V3 = VA + VBM Us = 0.6
J, “’7 + w(0.5)
\f 45“ —> 0 = —D.chs30° + 03(0.5)cos45° co = 1.470 rad/s “A! = 3.4 + 33.54 a,1 = 1.8 + 1.2 + (1.470)1(0.5) + 1110.5) ~L “7 3°.“ 45° 7 ms" (in o = f1.Scos3D° — 1.2511130” — 1.08sin45° + ar(ﬂ.S)cos4S° {+ $1 .25 = 1.8sin30" — 1.200530“ + 1.0351114? + (1(0.S)Sin45° a a 8.266 radis’ ﬂew ' ‘Jt m = “a = 3.55m2 Ans f5  d 0“ Also: “HOW/3 VB : VA +.:z)xrBM 18"")?
mvﬂi = (—0.600830°i — 0.6sin30°j) + (—aJk)x(0.Soos4S°i + 05311145?)
0 = —0160083D° + 0(05311145") w = 1.470 radfs 2
25=aA£UI'BM+IIXI‘BM 703] = (71.25in30°i + 1.200330°j) + (—LSoosﬂtPi — 1.Ssin30°j)
+(ﬁﬂk) X (0.500345°i + D.Ssin45°j) u = —1.Zsin30° — 1.8:;0530" 7 (1.470)2((}.Scos45"] + MU.SSin45°)
may = 1.2mm" — 1.8311130" — (L470)2(0.551n45°) — mummy) a = 8.266 radis as = 3.5511115: Ans *16116. The hoop is cast on the rough surface such M w=4mdjs
that it has an angular velocity to = 4 rad/s and an angular A “a
acceleration a = S rad/32. Also. its center has a velocity
of v0 = 5 .m/s and a deceleration a0 = 2 m/sg. Deter
mine the acceleration of point B at this instant. xx 0!: 5 rad/s2
‘\ as = 30 + 33/0 as = [g] + [Sign] +[<4):53.3)] 23 = [4.333] + [4:55] to = 4 rad/s 0 ® 0.3 m )fa=5,radls'2 B as = 62] m/s2 Ans 7 4.45
9 = tan"1 = 45.83 A Ans
Also: as = 30 + a X rum —wzr3/0 33 = —2i + 5k x (0.3 cos 45°i ~ 0.3 sin45°j) ~— (4)2(0.3cos 45°i — 0.3'sin'45°j) at, = [—4.333i + 4.455j} m/s2 an = 6t21nu'32 Ans 4.455 a
9 = turf" = 45.8 A Ans *16120. Rod AB has the angular motion shown.
Determine the acceleration of block C at this instant. W: = W; + ch II [as] g; + “gm 30‘ 49‘ [3+] 0 = —1.5 sm30° + 0.6%.51114? am=L768radls
8c=an +8015 I
War
(I; [5‘75
[115]: 5(o.5) + (3)1(05) + (1.768)=(o.6) + ac3(0.6) q
l
“’7 30'); 4’4! . ‘13? ($1) 0 = —2.5 c0560” — 4.5 c0530" +1375 00545“ + ac,(o.5)ms45° (+ T) —ac = 2.5 311160" + 4.5 511130“ +1.875 sin45°  auto6151M?”
ac = 2.41 rnis2 iv Ans ac, = 9.01 radix: ) Also.
vc = v; 1 GM 1'05
—vc.i = (—1.5 sin30°i~ 1.5 m§30°j)+ (an k)>< (415 sin45°i‘°5 “545"”
[1+] 0 = 4.5511130“ + an (Deﬁnes45°)
at, = 1.768 rad/s 3 ac = In +a><rcm —02'c.'ah
—acj = (—4.5 cosSO°i+ 4.5 sin30°j)+ (—2.5 c0560°i—2.5 sin60°j) Hagan); (—0.6 sin45°i— 0.5 cos45°j)—(1.763)1(—o.6 513491—06 cos45"j)
[10 o = «4.5 m530° —2.5 c0560“ 4 a“ (0.6 cos45°)+ (1.7621)2 (0.6 531114?)
(+ T)  ac : 4. 5 mac“ — 2. s 511160“  am ('0. a sin45°) + (1.71513)2 (1). 6 cos45°) ac = 2.41 1111152 J, Ans ac. =9.o11a11.1s2 5' 16126. The disk rolls withmit slipping such that it has
an angular acceleration of a = 4 {fad/82 and angular
velocity of a) = 2 rad/s at the instant shown. Determine
the accplerations of points A and B on the link and the
link’s angular acceleration at this: instant. Assume point
A lies on the periphery of the disk, 150 mm ffom C. ThEICiS area, so w=0_ i1. =8c+axrmc‘0)2'l'uc an = 0.6i+ (—410 x (o. 151) — (2)1“). 15j) m,l ={L20i—0.6j} m/s“  mm, M a
m= {1.20)2+(—O.6)2=1.34mlsz Ans no *8: 4 {7' I < A» E
I _. 0:. == “(945): abwt—fs OBmlsa
as = M +0Xl'sm ‘012l‘nm aai=120i0.6j+aﬂkx(0.4iO_3j)—0
(—3) an =1.20+0.3a”. (+ T) o = —0.6+0.4a” 05.43 = 15 lad/52 j Ans a3 = 1.65 111/52 > An; 16133. The collar E is attached to. and pivots about,
rod AB while it slides on rod CD. If rod AB has an
angular Velocity of 6 rad/s and an angular acceleration
of 1 rad/32, both acting clockwise, determine the angular
Velocity and the. angular acceleration of rod CD at the
instant shown. «4+ 44+
+S° 4);. Fix axes ID ED. Q = Clka 0 = arcDI: rap = 4j VEm = Vsmj 35m = “Elsi VE = 6(4)j = —24j as : —{5)1(4)i—1(4)j= —144i— 41
W; = VD + 9X l'EID + (Va/D)”:
7243 s 0 + (1ka x 4j + ngj
"241' = “401Cﬂ1' VEIDj Thus COCD =0 Ans as = 31) +er£fp +0x€ﬂ>< rg3)+2£1>< (VEDLH +(3BDLH
‘14454j = 0+acnkx4j+0+0+agpj " 144i “ 4i = —C¢co(4)i + amu' use = — = 35 radfsz A.“ 16139. Rod AB rotates counterclockwise with a constant angular velocity at = Brad/s. Determine the
velocity and acceleration of point C located on the double collar when 0 = 45°. The collar consists of two pin
connected slider blocks which are constrained to move
along the circular path and the rod AB. rm = {0.40m + 0.4005} vC = vvci ll vc VA + Q x rm + (Vol.4)xyz _vci = 0 + (3k)x(0.400i + 0.400.» + (ch “545% + V“ $111450»
_vci = o — 1.2m + 1201’ + 0707chi + 0.707vm1' vc = —l.20 + 0.707VCIA 0 = 1.20 + 0.707ch vc = 2.40 m/s “‘5 vm = —1.697 m/s ac = “A + Q X '04 + ‘2 X(Q X rem) 'l' 29 X(VCM)xyz + (“C/A)”: (2.40)2. . .
04 j = 0 + 0 + 3k x [3k x (0.4: + 0.41)] + 2(3k) x[0.707(—1.69 + 0.707(—1.697)j] + 0.707%,“ + 0.707%,” ‘(ac)ti _ —(ac),i — 14.40] = 0 + 0 —3.60l — 3.60j + 7.20i — 7.20j + 0.707ami + 0.707%
—(ac), = —3.60 + 7.20 + 0.707 am —14.40 = —3.60 — 7.20 + 0.70mCM am = —5.09 m/s2 (“(3): = 0
Thus, (2.40? 2
ac = (ac),, = 0—4 = 14.4 m/s ac = {14.4j} m/s2 Am 16142. The “quickretum mechanism” consists of a crank
AB, slider block B, and slotted link CD. If the crank has
the angular motions shown, determine the angular
motions of‘the'slotted link at this instant. v3 = 3(0.1) =0.3 m/s (11.), = 9(0. 1) :09 m/s2 (a, ),. = (3)1(0. 1) = 0.9 m/s2
V» = Vc + 9X l'lI/c + (VB/C)”: 0. 3cos60°i + 0. 255ij = 0 + (web k) x (0. 3i) + valci va/c = mls cocp=0.866 rad/s ‘) Ans as = ac +QXl'a/c +9X (QXl'a/c) +29>< (Vs/c)”; Han/c)”: 0.9cos60°i — 0. 9cos30°i + 0.9sin60°j + 0.9sin30°j = 0 + (acpk) x (0. 3i)
+(0.866k) x (0. 866k x 0. 3i) + 2(0. 866k x 0.15i)+a5,cl —0. 3294i + 1.2294) = 0.31am j —0. 225i+0. 2598j +a3,ci
ag/c = —0. 104 III/82 aw = 3.23 rad/52 '} Ans *16148. A ride in an amusement park consists of a rotating arm AB that has an angular acceleration of a“ = l rad/s
when a)“ = 2rad/s at the instant shown. Also at this instant the car mounted at the end of the arm has a relative I . . .
angular acceleration of a’ = {—0.6k} rad/s2 when u’ = Va = (0.43 x ram = 2" X (8561 + 5]) = {—1001 + 17321} NS
{—0.5k} rad/s. Determine the velocity and acceleration of ' the passenger C at this instant. a _
B _ r,“ = (1000530°i + lOsin30°j) = {8.66i + Sj} ft 0‘43 X ram — wiB'B/A = (1k) >< (8.66i + 5j) —(2)2(8.66i + Sj) = {—39.64i  11.34j} {Us2
9 = (20.5)k = 1.5k
5'2 = (l—O.6)k = 0.4k vc = Va + Q X rem + (Va/5):” = 10.0i + 17.32j + 1.5k x(—2j) + 0 {—7.00i + 17.3j} ft/s Ans ac = 33 + Q x rem + Q X“) X'l'cm) + m X("cm)xyz + (aC/B)xyz l —39.64i — 11.34j + (0.41:) x (—2j) + (1.5k) >< (1.5k) x(—2j) + o + o {—38.8i — 6.84j} ft/s2 Ans 16149. The cars on the amusementpark ride rotate around the axle at A with a canstant angular velocity Vc =VA +ercm +(Vmim “1
MA}! = 2 rad/s, measured relative to the frame AB. At
the same time the frame rotates around the main axle ac =aA +erm +Qx(n><rc,,)+znx(vm)m Ham)”: [2]
support at B with a constant angular velocity
m, = I rad/s. Determine the velocin and acceleration of Magm of Mon0,, gfcwim mm
the passenger at C at the instant shown. moving "game m moving Mama
rcm = {‘SE} ﬂ 9. = {3k} rad/s (vm )3” = 0 0‘10 (armhﬂzu Motion ofA : VA = ‘0 x9413
=(1k)x(1500530°i+355il130°j) = {7.51» 129%}st 1
3A = “X [Am "'3 H13 = u — (1)2(—15ms30°i+ 15 sin30°j)
= {12.99i —7.5j} ms1
Subsn'tuu: the dam into Eqs.{1] and [21 yields :
vc : (w7.51712.99j)+(3k) x(—8i) +0
= {—7.5i —37.0j )st Ans aC = (12.99i—75j) +0+(3k) xI(3k) x(—8i)1+0+0 = {85.0i —7.5j} fL’sz Ans ...
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This note was uploaded on 02/13/2010 for the course ME 324 taught by Professor Neptune during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Neptune

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