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Unformatted text preview: *164. Just after the fan is turned on. the motor gives the
blade an angular acceleration a = (20e‘0‘6') rad/52, where
t is in seconds. Determine the speed of the tip P of one of
the blades when t2 3 5. How many revolutions has the
blade turned in 3 5? When I = 0 the blade is at rest. Jo): a dr Hdw = f ZOe'D 6' d! 0
1' I _ _ 0.6r
0.33.3(1 e ) 20
____F—0 m
0,6 (L): m: 27.82 rad/s
v,p = ar:27.82(l.75)=48.7 ftls Ans d9: wdr 3 _ ' _ ~05.
jade—£33.30 e )d: B = 33‘3(l+(0—’l—6)e'°'6')’: = 33‘3[3+(0__16)('—06m_ 1)] 9 = 53.63 rad = 8.54 rev Ans 165. Due to an increase in power, the motor M rotates
the shaft A with an angular acceleration of a =
(0.0692) rad/52, where 6 is in radians. If the shaft is
initially turning at (00 = 50 rad/s, determine the angular
velocity of gear B after the shaft undergoes an angular
displacement A9 = 10 rev. wdw: 01:19 [” wdw=[”"°’o_069’ d8 50 0 51021:) = 0.0293l2m) 0
0.5a} —1250 = 4961
a): lll.45 rad/s
0140 = (”are
(iii.45)(12)=wa(60l we =22.3radls Ans ”R7 1617. For the outboard motor in Prob. 1616, determine
the magnitudes of the velocity and acceleration of a point
P located on the tip of the propeller at the instant I =
0.75 s. (all), = 285.8 in./s2 (am = wlr ={100(o.75)~‘/211(2.2) (apt, = 9231 in./sZ ' w ,
/ du) = / lSOJi m
up = ,/(285.8)1 + (9281)2 " °
, = 1 ,1/1
= 9285.6 = 9.29 x no1 in./s' Ans w 00' Up = (or = [100(0.75)3/11(2.2)
01AM = “NH Up = 143 in./s (300Ji)(0.7) = (”(1.4) (up), = or = [150J0.75(2.2) a, = ism/i 1618. Starting from rest when .r = O, pulley A is given
an angular acceleration a = (60) rad/s , where 0 is in
radians. Determine the speed of block B when it has risen
s = 6 m. The pulley has an inner hub D which is ﬁxed
to C and turns with it. CIA = 66,1 9~=——=oa
‘ 0.075 8” 9.4 (0.05) 2 80(0. [5) 8,; = 240 rad adl) =wdw 1“" But 9,.(50) = 150(ec) a“
69.4 (19,1 =/ (DA 1111),;
t) (I 10.1 = [6(240)3l”2 = 587.88 rad/s (587.88)(0.05) = wants)
”(I "’(3
w(=l95.96 f, 66‘1”“: A “’“d‘” 2__ 2
69C_wc v3 = l95.96(0.075) = 14.7 m/s Ans 6
too ._ 0*.075 — 80 rad Also. 01A = 66A
log = Jam) = 195.96 BUI (“(50) = 1500c
U}; = (195.96)(0.075) = 14.7 m/s ADS an = 30C 3(15 = 69A (1C = 26A *1624. The disk starts from rest and is given an angular
acceleration a = (106w) rad/52, where 9 is in radians.
Determine the angular velocity of the disk and its angular
displacement when t= 4 s. a = 106'" wdw: mil? in 9 
dew=l 10:?1 :16 u D 1a} = 1069;): 7‘56;
2 4 and—6 =/ﬁ9i dt [”9‘ide=j'./Ed: U 0 3ei = /T5:
9: 2. 1521’,=.. =138 rad Ans d
(02—? = 6.45511 = 103 rad/s Am [=4 1625. The disk starts from rest and is given an angular
acceleration a = (1091/3) rad/52, where 0 is in radians.
Determine the magnitudes of the normal and tangential Components of acceleration of a point P on the rim ofthe
7 disk when t = 4 s. 11:106)‘l wdm: ado {lemma}: j: 109i de a? = 1069?) = 7.59; 7! = 645511 = 103.28
{=4 “2' = (103.2a)1(0.4) = 4267 m/s‘ Ans “"(10(137.71)i)(0.4) = 20.7 mm Am 7/11 *1628. Rotation of the robotic arm occurs due to liner
movement of the hydraulic cylinders A and B. If cylinder
A is extending at the constant rate 0.5 ft/s while B is held
fixed, determine the magnitude of velocity and
acceleration of the part C held in the grips of the arm.
The gear at D has a radius of 0.10 ft. Angular MnIi'vn : The angular velocxry of gear D must be determined firSL
Applyutg Eq. 16 — 8. we have u = wDrD; 0.5 = a)“ (0.10) (on = 5.00 rad/s Motion of Part C : Since the shaft that arms the robot‘s arm is attached to gear
D. then the angular velocity of the robot's arm (0,, = (DD = 5.00 rad/s. The
distance of pan C from the rotating shaft is 'c = 4cos 4S°+ Zstn 45“ = 4.243 {L
The magnimde of the velocity of pan C can be determined usmg Bq. 16 — 8. DC = (0er = 5.0004243) = 212 ft/s Ans The tangential and normal components of the acceleration of part C can be
determined using 5415,16 — ll and )6 — l2 respectively. a, = arc = 0
11,, = wﬁrc =(5i001)(4‘243) = 106.07 {1/51 The magnitude of the acceleration of point P is up = Jag”; = 401+106.07= = 106 stz 1 _ .
62:? At the instant shown. gear A is rotating with a
‘ 5 ant angular velocity of w,‘ = 6 rad/s. Determine the largest angu v ' f
 ‘ lill' GIOCHV H 63 ‘
Of ' l. ‘.  g l' B and the maxrmum speed Valw = m )M. = soﬁ mm ”ﬁlm”. = (r4 )..,,. = 50 mm When r m '
,t u ax . r5 5 min. I.—
l00 mm "—4 M 1633. The bar DC rotates uniformly about the shaft at
D with a constant angular velocity w. Determine the
velocity and acceleration of the bar AB, which is confined by the guides to move vertically. y=lsin9
y‘: u, = {00309 y' = a, = 1(aosaé—sinoé’) Ham v, =vA.. a, =a“.and é=m, §=¢=o_ v“ =lcosolm)=wlcoso Ans a“ = {casem— momf] = —alllxin8 Ans 1634. The scaffold S is raised hydraulically by moving
the roller at A toward the pin at B. If A is approaching
B with a Speed of 1.5 ft/s, determine the speed at which
the platform is rising as a function of 0.The 4—ft links are pin—connected at their midpoint. Position coordinate equation : x = 46050 \ = 4 ‘
Time derivatives: ‘ ““9
x = 4sin69 However, .i = —vA = ~15st
— l,5=—4sin0€ 05:03”
sine 0.375 5m V=Vy =46059€=46059( )= LSCOIB Ans 1635. The mechanism is used to convert the constant
circular motion to of rod AB into translating motion of
rod CD. Determine the velocity and acceleration of CD for any angle 0 of AB.
x=lcoa9 x' v, = lsin9é i=a, = —1(sin96+cosaé’)
Hem v, =ch, a, =acn, and é=m. 5=a=0. ch = —lsin 9(a)) = —m lsina ADS
ac» = {sin0(0)+cost9(a))1]= ~0210059 Ans Neglive signs indicate um bothvcp and a“ u, dim“: opposite wpasiﬁvé 1639. At the instant 0 = 50°, the slotted guide is
moving upward with an acceleration of 3 m/sz_ and a
velocity of2 m/s. Determine the angular acceleration and
angular velocity of link AB at this instant. Note: The
upward motion of the guide is in the negative y direction. y=0.30059
”in
y‘ = v, = —0.3$in oé . "“1 7
y"=a, =—o.3(sin9é'+cos9é‘)
Hue v,=—2m/s. a,=3m/sz,and é=w.é=a,9=50°. ” u
‘2 = '035iﬂ50°(01) a) = 8.70 rad/s Am ’Ymbf/x
ﬂ urn/1"
—3 = o.3[sin50°(a) +cosso°(s.70)’] a: —5o.5 rﬂlaz An: A W—A »
 *1640. Disk A rolls without slipping over the surface
of the ﬁxed cylinder 8. Determine the angular velocity
of A if its center C has a speed v( = 5 m/s. How many
revolutions will A have made about its center just after
link DC completes one revolution?
As shown by the construction, as A (0115 ihrough the arc .r = 9‘ r,
ihe CCnlEr of ihc disk moves through [he same distance 5' = I. 
Hence.
I = HA I
I = 9 r _
A 5  2r 6“,
‘ f
5=mA(0.l5) 9‘ 5:9“.
(5‘
m4 = 333 rad/s Ans ‘D
Link ;
3' = IIBED :5 = 9‘“
20CD = 9A
Thus, A makes 2 revolutions for each revolution of CD. Ans 1641. Arm AB has an angular velocity of a» and an
angular acceleration of a. If no slipping occurs between
the disk and the ﬁxed curved surface. determine the c
angular velocity and angular acceleration of the disk. dI=IRr)de=—rd¢ “wet4:4)  _(R~,)w
w _ $ Ans
. (R ;
a =_ Ha
\ Ans , 349 *1644. The pins at A and B are confined to move in
the vertical and horizontal tracks. If the slotted arm is
causing A to move dOanard at VA. determine the
velocity of I} at the instant shown. Pusmtm coordinate equauon . NI: Timc denvatives : 1645. Bar AB rotates uniformly about the ﬁxed pin A
with a constant angular velocity w. Determine the
velocity and acceleration of block C, at the instant
: 0 = 60°. An: Lou9+Lcnu¢ =L cosa+oos¢=l unob+sin¢¢=o (l) maul!)2 + sins '0 + sin¢ ll + cow (W = 0 (2)
Whﬂlﬂ = 60°, 45 = 60", thus. 9 = —d> = m (from Eqm) ‘é  o 3 = 4.155121 (from Eq.(2)) Also, :c = Lsin¢ — Lsinﬂ yc=Lcos¢3¢—Lm599 II: = —Lsin¢(¢)’ + Loos¢(6) — Loosen?» + Lsinaw)‘ A19=60°. ¢=6°° .IC = 0
"c = ucos60°)(m)  LoosGO°(m) = —Lm = LuT
ac = —L “cm—m): + LooséO°(—1.155m‘) + o + L newt»): ac = —o.s77 Lm’ = 0.577 La;z T 351 Ans Ans 1654. The shaper mechanism is designed to give a slow
cutting stroke and a quick return to a blade attached to
the slider at C. Determine the velocity of the slider block
C at the instant 6 = 60", if link AB is rotating at 4 rad/s. V(' =V3 +wxrtflﬂ ‘ .
v(~i : —4(0.3)sin 30°i + 4(0.3)cos 30°j + wk X (—0.125cos 45°i + 0.1255m 45°J) —v(~ = —l .0392 — 0.008839cu
0 = 0.6 — 0.08839w Solving, a) = 6.79 rad /s vr = L64 m/s Ans. (DAB : 4 rad/s 1655. Determine the veloc it of th '
at the instant 6 = 45° y e 5M“ block at C , if link AB is rotating at 4 rad/s. vc=v.+mxrc,. —4(0.3) oos45°l + 4(0.3)sin45°J + a) It. x(—0.125ws45°l + 0.1258in45°]) ycI =
vc = —0.8485 — 0.0883941)
”L
0 = 0.8485 — 0.08839 a) Q, ;o.l154~‘
45° 5
Solving.
a) = 9.60 rad/s
V: = 1.70 mls Ann *1656. The velocity of the slider block C is 4 ft/s up the
inclined groove. Determine the angular velocity of links
AB and BC and the velocity of point B at the instant shown.
For linkBC
vc = (—4oos4s°l+4sin45°]} ﬂ/s v. =—v,t a): (Duk
I r“. =(1l) ﬂ ”,5
1 ft *1 "C = 4 ft/s
vc=v,+mxrm B
—4ms45° +45in45°' = v. i +(mu k) x(1)
—4cos45°l +4sin4s°j = v,l+ mu]
1 ﬁ Equattng Ihel and J components yields:
—4cos45° = —v, v, = 2.83 Ms Ans
4sin45° = a)” In"; = 2.83 rad/s Ans For linkAB : Link AB routs about the ﬁxed poinlA. Hence Va =wu’u 2.83=m‘,(1) a)” =2.83 mas 356 *1660. The rotation of link AB creates an oscillating
movement of gear E If AB has an angular velocity of
(1),”, = 6 rad/s, determine the angular velocity of gear F
at the instant shown. Gear E is rigidly attached to arm
CD and pinned at D to a fixed point. kinematic Diagram : Since link AB and arm CD are relaxing about the ﬁxed
points A and D respectively, then vaand vcare always directed perpendicular
their their respective arms with the magnitude of v, = tour” = 6(0.075t '= 0.450 m/s and DC = mcgrw = inseam. A; the instant shown. v, and vC
are directed toward negan've .r axis. Velocity Equation : Here. ram = {0.lcos 30°i +0.lsin 30°] } m = (0.08660i+0.05j} m. Applying Eq. 16 16. we have Vc = V5 +mBC chza
—0.450i = —O.15wwi +(msck) x (otosséoi +0.05j)
—0.450i = —(0.05w,c +0.15wCDJi + command Equaiing i andj components gives 0 = 0,0866%)” a)” = 0
—0.450 = —[0.05(0) + 015mm,] mm, = 3.00 rad/s Angular Motion About a Fixed Point : The angular velocity of gear Eis
the same with arm CDsince they are attached together. Then. (05 = mm = 3.00 rad/s. Here, (uErE = coFr,r where (or is the angular velocity of gear F. r5 100
w, = mE = (_
r, 25 )(100) = 120 rad/s Ans 1661. At the instant shown, the truck is traveling to the
right at 3 m/s, while the pipe is rolling counterclockwise at = 8 rad/s without slipping at B. Determine the velocity
of the pipe’s center G. \ 8 rad/s um: 1.5(8) = 12 m/s Also: v(, = v” +w x raw
vui = 3i + (8k) x (1.5j)
v1; = 3  12 v,; = ——9 m/s :9 m/s ‘— Ans 1662. At the instant shown, the truck is traveling to the
right at 8 m/s. If the spool does not slip at B, determine
its angular velocity so that its mass center G appears to
an observer on the ground to remain stationary. mete4:3? .5; 15 = 5.33 rad/s Ans U) 2 Also:
VG : VB +60 X [0/3
0i = 8i + (wk) x (.5j) 0=8—l.5w 1 15 = 5.33 rad/s Ans a): 1677. Mechanical toy animals often use a walking
mechanism as shown idealized in the figure. if the driving
crank AB is propelled by a spring motor such that
in“, = 5 rad/s, determine the velocity of the rear foot E
at the instant shown. Although not part of this problem,
the upper end of the foreleg has a slotted guide which is
constrained by the fixed pin at G. 3°. ”:5 if 3i“soP» .
(:b) vccos30° = 25mm" + o v. mayy,
vc = 2.21 in]:
mm = $ = 2.2111111:
v: = (2.21)(2) = 4.42 in./l Am
Also:
'1 = ”Al x rm Vc = mac x 'cm
vc =v, +mxra,
(m,ck)x(coa60°1 + sin60°j) s (5k)x(05co550°i + 053mm) + (mu31);
—0.866¢.u,c = 1.915
0.507.: = — 1.607 — 3a)
a“ = —221rId/a a! = —0.167 rid/a v,  2(2.21)  4.42 111.1: Ans AHA 1689. The wheel rolls on its hub without slipping on
the horizontal surface. If the velocity of the center of the
wheel is vc = 2 ft/s to the right, determine the velocities
of points A and B at the instant shown. in. VC=ﬂTruc
2:41)
12
cu=8rad/s
_ 11
va—arg,,c=8(ﬁ)=7.33ﬂ/s —» Ans
B
I
_  3/5 J 7'5
VA (time—8(W]=2.83ﬁ/s Ans '2 “3
B un"(3)—4° K14 £4
A E — 5 \: Ans IL “I;
3 a
12"“ 535' H
‘7. 1690. lf link CD has an angular velocity of a)“, =
6 rad/s, determine the velocity of point E on link BC and
the angular velocity of link AB at the instant shown. vc : (inﬁrm) = (6)(O.6) =_3.60 m/s w 3.60 a)“ : —— : ———— = K139 rad/s Fry/c 0.6 m30°
, v ‘ 0.6
,. a — mural/c = (10439)( )= 720 "1/5 R cos30° 605‘
_ V5 _ 7.20 _
m _(W‘)_6mdls 3 Am «r 39' 0'°£"“3°.""
sin30° B 7’; "E = marry” = [0,39 (0,6 lan30°)Z + (0.3)2 = 4.76 m/s Ans i 5
D 3 0:” 9=Ian" 0.3 _ o
(0.6m30°)_ 40'9 A Ans ’3’7’7 1697. Due to slipping, points A and B on the rim olthe
disk have the velocities shown. Determine the velocities
of the center point C and point E at this instant. 1'”: l0st B 5: = l6 ~10:
x = 1.066671! 10
1 .(m7 = 9.375 rad/s m: "c = ”(Vicc)
= 9.375(1.06667  0.8) 2.50 ft]; An: "5 = ”(Vic—1:)
= 9.375 (0.8): + (0.26667): = 7.91 Na All WM” 1698. The mechanism used in a marine engine consists
of a single crank AB and two connecting rods BC and
BD. Determine the velocity ofthe piston at C the instant
the crank is in the position shown and has an angular
velocity of 5 rad/s. vs = 02(5): l m/s —) Member BC : rC/IC = 0.4
sin60° sin45° ’CIIC = 0.4899 in ”NC __ 04
sin75° ‘ sin45° r3”: = 0.5464 m = 1.830 md/s “’6 = 0.5464 vc = 048990.830) = 0.897 m/s / Ann ’29“ 16102. The epicyclic gear train is driven by the rotating
link DE, which has an angular velocity mm; = 5 rad/s. If
the ring gear F is ﬁxed, determine the angular velocities
of gears A, B, and C. mcs_’ 0.03 1.6 W‘E 0.05 16103. The mechanism produces intermittent motion
’ of link AB. lf the sprocket S is turning with an angular
velocity of 015 = 6 rad/s, determine the angular velocity
of link AB at this instant. The sprocket S is mounted on
a shaft which is separate from a collinear shaft attached
to AB at A. The pin at C is attached to one of the chain
links.
’KlIlemalit Diagram : Since link AB is rotating about the ﬁxed pointA. then v,
is always directed perpendicular to linkAB and its magnitude is u, = a)” ’45
= 0.20)”. At the instantshown. vgis direcred at an angle 60° with the
horizontal. Since point C is attached to the chain. at the instant shown. it moves
vertically with aspeed of '05 = [1)er = 6(0.l75) = 1.05 m/s. instantaneous Center : 'nie instantaneous center of zero velocity of link BC
at the instantshown is located at the intersection point of extended lines drawn
{perpendicular from v5 and vE. Using law ofsines. we have ram. _ 0.15 sin 105° sin 30" rm = 0.239% in rent _ 0.15 = 0.2121
Sm 45° sin 30° rm m e angular velocity of bar BC is given by 1.05
at" = if— = — = 4.950 rad/s ram 02121
US. the angular velocity of link AB is given by
”it = wBCrBIlC 0.20.)“g = 4.950(0.2898)
a)” = 7.17 rad/s Ans a) =
' 0.05565 v, = 016(5)  0.8 ml! = 26.7 ml: v, = (0.06)(26.7) = 1.6 m/s [.6 0.45 T 10.04 x = 0.05565 m =_ 28.75 rad/s = [4.0 ml: Ans Ans v,” = 28.75(0.080.05565) = 0.700 m/s 6— Ana 0.? "Vs ’ 0.05m 16109. The wheel is moving to the right such that it has
an angular velocity (u : 2 rad/s and angular acceleration
a = 4 rad/s2 at the instant shown. If it does not slip at A,
determine the acceleration of point B. Sineenoslipping
 (E 0
ac = (1r: 4(1.45) = 5.80111? (0.70% " '
l‘Hl‘f WW?
'5 = ‘c + "we 39' land/5’ a, = 5.80 + (2)1(1.45) + 4(1.45)
) g Y” V30
l \ /
(1») (4;), = 5.80 + 5.02 + 2.9 = 13.72
(«(1) (12,), = 0 — 2.9 + 5.02 a 2.12 a, = ((13.72)! + (2.12): = 13.9 n/s’ Ans 2.123
e  " __ = _
m (13.72) 880‘ a A“ 1
"a =ac+axr,,C—mr,,c — 5.80i + (—411) x(—l.45ws30°i + 1.45sin30°j) —(2)’(—1.45ws30°i + 1.45sin30°j) J"
l a, = (13721 + 2.123j) 1115’ a, = 1/(13.72)z + (7.123)2 = 13.911151 Ann _ _, 2.123 _
9 — tan (—13.72) — 8.80“ 41' Ans
16110. At a given instant the wheel is rotating with the
angular motions shown. Determine the acceleration of
the collar at A at this instant. all
See Prob. 1659. °‘
05‘» I+.1(,r'/3 .,  4.16 mil5 ‘0'
(m:u§:5)7\(onﬂ =(g)*(o.15‘)=‘l‘ ‘ — I. + IA]. “A 14 +9.6 +(4.16)’(o.5) +1105)
4— L” T” 60'7 “our (in —a,.  “moo" + 9.6mm" — 8.6500860“  11(05):me w=8radls a = [6 rad/V (+ T) 0 = 2.4mm? — 9.6:!!30" — 8.65dn60° + ((0.5) 00860" a  40.3 mm") a.  12.5 ml.‘ 4— An
Also: .‘  n, + a x r“, — aft”, .1 can301
a‘i .. (3)1(o.15)(cu30°)1  (8)‘(0.15)dn30°1 + (16)(o.15) IMO“ + (10(0 5) ,
+ o_5m60°1) +(ak) x(0.500660° + 0.5m60°j) — (4.16)’(0.500560°l
n‘ a 8.314 + 1.1K!) — 0.433(1 — 4.326 O a —4.800 + 2.0735 + 0.25a — 7.4935 (1  40.8 mil/s1 ‘) a‘ = 12.5 mls‘ « Am 388 16115. The hoop is cast on the rough surface such that w=4mdls it has an angular velocity u) = 4 rad/s and an angular 2»
acceleration a = 5 rad/52. Also, its center has a velocity 3 a: 5 rad/s2
v0 = 5 m/s and a deceleration (I0 = 2 m/sz. Determine ‘* the acceleration of point A at this instant. 3,1 = 30 + Elva ad = [g] + [many] +[ may] . .__
= [as] +[418l a): 4 rad/s
a = 5 rad/s2 Also: 3.4 = no — (Uer/l) +a X l'A/a a,‘ = —2i — (4)3(03j) +5k x (0.3j)
a,‘ = [—3.5i — 4.8j) m/sz m = 5.94 m/s2 4.
6 = tan" (75.2) : 53.90 7 *16116. The hoop is cast on the rough surface such
that it has an angular velocity (u = 4 rad/s and an angular
acceleration a = 5 rad/52. Also, its center has a velocity
of 110 = 5 m/s and a deceleration (10 = 2 m/sz. Deter
mine the acceleration of point B at this instant. ﬂu = 30 +33“) as = [g] + [5gb] + [(49103)] an = [4.233] + [4155] \w = 4rad/s , (13 = 6.2] m/s2 0.3 m g = tan” (4455) ~ ,0: = 5 rad/$2 —— :4.“
4.333 583‘ 3 Also: “8 =30 +01 x ram — (U1I'g/0 .as = —2i + 5k x (0.3 cos 45°i — 0.3 sin 45°j) — (4)2(03 cos 45°i — 0.3 sin 45°j)
‘ {—4.333i + 4.455j) m/s2 ~6.2l rn/s2 Ans 4.455
tan“l E = ~
( 4.333) 45.8 , 16126. At a given instant, the gear has the angular
motion shown. Determine the accelerations of points A
and B on the link and the link’s angular acceleration at this instant. Forthegear
”A = (tn/1c = 6(1) =6in./s'
.0 = —12(3)i= (—36i)in./sz rm, = (21) in. a: (12k) rid/:1
n‘ = no +aerO—w’ruo
= ~36i + (12k) x (21) — (6)2 (21) = (12l+72.l) in./s2
ai = v/(12)‘+72z = 73.0 mm A... e = tan" (11%): 305° 3 All! For link AB TheICisuw. so n)" =0. ie.. I. =a.i a“ =—a”k rm =(8c0360°l+83in60°1)in. ‘l = “A H1,” )1er ' wall/.4 ani= (—12i+72])+(—auk)x(8co.160°l+ smw‘n—o
) a. =—12+ssineo°(la)= 113 ian’ —» An T) 0=72—8c0360°au a“ = land/s1) Ans 403 16133. The man stands on the platform at 0 and runs
out toward the edge such that when he is at A. y = 5 ft.
his mass center has a velocity of 2 ft/s and an acceleration
of 3 ft/sz, both measured with respect to the platform and
directed along the y axis. If the platform has the angular
motions shown, determine the velocity and acceleration
of his mass center at this instant. v4 = v0 + 12x rm} mm)"; v4 = 0+(O.5k)x(5j)+2j VA = l2.50i+2.00j) {1/5 An: IA = no +11er +Q><(ﬂ><mo) +2ﬂx (VA/0):): + (IA/a)”;
.A =0+(O.Zk)x(5j)+ (O.5k)x (0.5k x5])+2(0.5k)x(2])+3j
IA =ll—l.25_2l+3j aA =(3.00i+ 1.751} hisz Ans 16134. Block B moves along the slot in the platform
with a constant speed of 2 ft/s, measured relative to the
platform in the direction shown. If the platform is rotating
at a constant rate of w = 5 rad/s, determine the velocity
and acceleration of the block at the instant 0 = 60°. i + zj = {1551+ 21) a Two = IAnGO’
' = Vo + n X '510 + (7,,0)". ‘0 x
V, = 0+ 5k X (1.155i + Zj) — 2i v. = {—12.0i + 5.77j} ﬁ/s A... h = “a + n X 1"", + n X(n X l.510) + 2‘“ X('nlo):,; + (“l/0)”: I. = 0 + 0 + 5: XI(5k) x (1.155i + 2m + 2(5k) x(—2i) + o a, =0+0~28.87i50j—20j ., = {—28.9i —70.0j} ms: Ans ; 408 16137. A girl stands at A on a platform which is rotating
with a constant angular velocity w = 0.5 rad/s. If she
walks at a constant speed of v = 0.75 m/s measured
relative to the platform, determine her acceleration (a)
when she reaches point D in going along the path ADC,
d = 1 m; and (b) when she reaches point B if she follows the path ABC, r = 3 m. (I) [1] .D = .0 m x rm, +n x(0 um) +7.ox(vp,a),,l Hanna," Motion 0f Marion of D with respect
moving rzfemlce to moving refzmlcz
~ 30:0 rua={li}m
’ n  {0.51:} ndls (5,0),” = (0.75j)m/s
{1 =0 (swam =0
Subsliullc the dam Into Eq.[l]: I. = 0 +(0)x(li)+(0.’>k) x[(0_"k) X(1i)]+ 2(0.5k) X (0.75j) +0 = Hum/s1 Am G!) I. = no +ﬁxrm, +nx(0 xr,,a) +7.0): Una)": +0.10)", [2l Maxim: of I Motion of}? with raped
moving reference to moving reference rm, = {3i)m , to = 0
Q = (05k) my; Una)", = (0.75j) m/s
(1:0 (.510)," =‘(nua).l+(ﬂsla),j =—(£~;—")i
= (—0.1875l) m/s Substitute the data into Eq.[2] : I. = 0 + (0) x (3i) + (05k) x [(05K) x (31)] +2(0.5k) x (0.75j) + (—0.1875i) = (—1.1s9iimls2 Am }
a;
_ 16139. Rod AB rotates counterclockwise with a constant angular velocity in = 3rad/s. Determine the
velocity and acceleration of point C located on the double
collar when 6 = 45°. The collar consists of two pin
connected slider blocks which are constrained to move along the circular path and the rod AB. Vc Vc vcl = 0 + (3k) x(0.400i + 0.400j) + (vm cos45°i + ch sin45°j)
—vcl ‘VC = —vci rm = (0.400: + 0.4001} VA + a x rc,‘ + 070‘)"z 0 — 1.201 + 1.201 + 0707ch + 0.707%“) = — 1.20 + 0.707vcM o = 1.20 + 0.707%m vc = 2.40 m/s V cm .c = I4 + a x "cu + n x(ﬂ x ram) + 20 x000," + (“e/.4)":
—(ac). — —(ac),l — 14.40] = 0 + 0 3.60l — 3.60] + 7.2m — 7.203 + 0.707ami + 0.707%, 3
A = — 1.697 ml: (2.40)2
0.4 j = o + o + 3]: x (3k x (0.41 + 0.41)] + 2(3k) xl0.7o7(1.597)i
+ 0.707(—l.697) 11 + 0.707ami + 0.707%“) ~(ac), = —3.60 + 7.20 + 0.707 am — 14.40 = 3.60 — 7.20 + 0.707%“ “CIA = —s.o9 m/s1 (ac); = 0 Thus, ac _ (2.40)1
(ac). — 0.4 (—14.41) m/s1 14.4 m/s2 Ans ...
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 Fall '08
 Hudyma
 Dynamics

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