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Unformatted text preview: *17~4. A semiellipsoid is formed by rotating the shaded
area about the x axis. Determine the moment of inertia of this solid with respect to the x axis and express the
result in terms of the mass m of the solid. The material has a constant density p. revolving the shaded area the radius of gyratiogn k,.
al is 7 = 3801b/ft . 175. The solid is formedby
around the y axis. Determine .
The speciﬁc weight of the materi The moment of inertia of the solid : The mass of the disk element . «in = pmfdy = ﬁpw‘dy. The mass of the solid :  mwalummmm < 1715. ”Die wheel consists of a thin ring having a mass
of 10 kg and four spokes made from slender rods and
each having a mass of 2 kg. Determine the wheel‘s
moment of inertia about an axis perpendicular to the page and passing through point A. [A = 10 + mdZ
{zHiMHUZ} 10(0t5)2]+ 173((15)z = 7,67 kg m1 Ans *1716. The pendulum consists of the 3kg slender rod
and the 5kg thin plate. Determine the location y of the
center of mass G of the pendulum; then calculate the
moment of inertia of the pendulum about an axis
perpendicular to the page and passing through G. Hm l(3)+2r25(5l §=—'——= =l.781m=l.78rn Ans
‘ in: 3+5
:5: 2i¢+md2
= {Emmi + 3(l.7Bl — l)1+ l_12(5)(0,52 + 11)+ 5(2.25— 1.781)1 = 4.45 kg m2 Ans 1717. Each of the three rods has a mass m. Determine
the moment of inertia of the assembly about an axis which
is perpendicular to the page and passes through the center point 0. 1 3 asin60° 2 1
07:3[67710 +m( 3 )]=ima2 Ans 1718. The slender rods have a weight of 3 lb/ft.
Determine the moment of inertia of the assembly about
an axis perpendicular to the page and passing through the pin at A.
_lﬂ 1 13(3) 2 3(3)
4  [3922”) l + [12(EX3) + (mum
= 2.17 slugt:2 Ans w— i.5&—+— l.5ft—~l 1719. The pendulum consists of a plate having a weight
0f l2 lb and a slender rod having a weight of 4 lb.
Determine the radius of gyration of the pendulum about an axis perpendicular to the page and passing through
point 0. [U =215+md2 l—lt—JHH =4.917siug «1
4 12
= _ + ._ =0.496951
"' (32.2) (32.2) “8 (10 {4.917
= — = — =3.l5 fl All!
b m 0.4969 ' *1720. The pendulum consists of two slender rods AB
nd 0C which have a mass of] kg/m.The thin plate has
a maSS of 12 kg/mz. Determine the location 7 of the
center of mass G of the pendulum, then calculate the
ﬁtment of inertia of the pendulum about an axis
perpendicular to the page and passing through G. ,i = 0 + 0.75(3)(1.5) + “0.3)1(12)(1_g)  no.1)1(12)(1.8
3(0.8) + 30.5) + maxim)  ) 140.0%”)
= 0.8878 = 0.888 In An:
+0.4 m+o.4 m4 l 1 l
l = ~ _ l
I la 4, 12[3(0.8)](0.8) + 3l3(15)1tl.5)1 + 5[12(;r)(0.3)2](0.3)1 2 1 l
+ (1200(03) 10.8) — El12tz)(0.l)1)(0.1)1  (12(x)(o.i)11(1,s)1 = 13.43 kgva 1.5 m
j g = 16 + (9.9is)(o.ss7s)‘
I 1,; = 5.51 kg~m’ Ans Q A”: l
= __ 1 I
% ”ﬁloWW3) + 3l°v8>t03878>1 + ”(30.9105)” + 3(1.5)(o.ss7s—o.75)’ I
+ EHZMKOJVNOJV + [12(R)(0.3)1](1.3  1
0.8878):  5(12(x)(o.i)‘](c_x)’
— tlztrr)(o.1)’](i.3 — 0.837s)’  2 '
— 5.6K kgm A.“ 490 1726. The Z—Ib bottle rests on the checkout conveyor
at a grocery store. If the coefficient of static friction it;
,th = 0.2, determine the largest acceleration the conveyor
can have without causing the bottle to slip or tip. "Hie center of gravity is at G.
mu. 2
3U; = 01(06):? F5 = mac +TLI§=m(aG),; N. 2 0 2
(+2510 = 130400: 1’1 = 327%“) Assume botﬂe is about to slip. ’N, = 21b. F, = 0.2(2) = 0.4 lb 0.3 = 6.44 Ms‘, x = 1.6 in. > 1.5 in.
Boule will tip before slipping. Set): = 1.5 in. N, = 21b. a6 = 6.0411/52 Ans F; = 0.375 lb < 0.4 lb (OJCI) 1727. The assembly has a mass of 8 Mg and is hoisted
using the boom and pulley system. lfthe winch at B draws
in the cable with an acceleration of 2 m/sz, determine the
compressive force in the hydraulic cylinder needed to
support the boom.The boom has a mass of2 Mg and mass center at G.
s, + sz = 1 3A a, ' 'zaL 15'—
2= —2aL a; = —1rn/s2 Assembly: +Tm; =ma,; 21—3(io’)(9.31) = 3003)“) T = 43.24 kN
Room
@2214, = o; Fcua)  2(10’)(9.31)(6cosso°)2(43_z4)(1o’)(12coseo°) = ,0
F“, = 239»: Ans 432 1738. "llre sports car has a mass of 1.5 Mg and a center
of mass at G. Determine the shortest time it takes for it
to reach a speed of 80 km/h, starting from rest, if the
engine only drives the rear wheels, whereas the front
wheels are free rolling. The coefficient of static friction
between the wheels and the road is m = 0.2. Neglect the
mass of the wheels for the calculation. If driving power
could be supplied to all four wheels, what would be the
shortest time for the car to reach 21 Speed of 80 km/h? «LZF.=m(ao),: 0.2NA+0.2N5=1500aG (1) minimum»; NA+N51500(9.81)=0 (2) (+WC =0: —NA(125)+N5(0t75)(0.ZNA +0.2Na)(0.35)=0 (3) (Ifoo)(q.81) N FU‘ rear ._ wheel dnve: .
1500 QC Stl the friction force 0.2M =0 in Eqs. (1) and (3) 035‘m Solving yields 1 N4=.<,18kN>0 (OK); N329.53kN; aG=l.27lm/s" Since v = 8' "Mn/h = 2222 m/s. then . «mbamtwnmmm.m s~=n~ .. 1—») vv=~o+acl 2.1 32=0+ 1.27lt 1: I75 5 Ans Since v2 = 80 kmlh = 22.22 m/s, then F<ir4— wheel drive: v1 = v, +1161: 2222 = 0+ 1.962: NA=SOOKN>0 (OK); N,=9.71kN; aG=L962mlsz 1:11.3s Ans 1739. The sports car has a weight of4500 lb and center
of gravity at G. If it starts from rest it causes the rear
wheels to slip as it accelerates. Determine how long it
takes for it to reach a speed of 10 ft/s. Also. what are the
normal reactions at each of the four wheels on the road?
The coefficients of static and kinetic friction at the road
are m = 0.5 and #k = 0.3, respectively. Neglect the mass
of the wheels. 525 = Mac»: 0.3%) = $20 +T£€=Mtaa),: 7N. +21»; —4500= 0 Solving. N, = [393 lb An,
N,  857 lb A“
0a = 3.68 m1 (:0 v = vn‘let
10 = O + 3.68: t = 2.72 s Ans 112R 1745. The van has a weight of 4500 lb and center of
gravrty at Gr. It carries a fixed BOOlb load which has a
center of gravity at G,. If the van is traveling at 40 ft/s
determine the distance it skids before stopping. The
brakes cause all the wheels to lock or skid.The coefficient
of kinetic friction between the wheels and the pavement
15 {‘k = 0.3. Compare this distance with that of the van
being empty. Neglect the mass of the wheels. 1746. 'lhe crate has a mass of 50 kg and rests on the
cart having an inclined surface. Determine if the crate will
tip over or slide relative to the cart when the cart is
subjected to the smallest acceleration necessary to cause
one of these relative motions. What is the magnitude of
this acceleration? The coefﬁcient of static friction between the crate and the cart is p; = 0.5. Equation of Motion : Assume that the crate slips, then P; = MN = (+ W. = EMMA: 50(9.8l)cos15“(x)50(9.8l')sin 15mm
= 50(1205 15°(0.5)+SOasin l5°(.\')
+ ZF‘.=m(aG)vt2 N—50(9.81)c05l5°=50asin l5”
+25. =mtaolu 50(9,8l)sin 15°—0.5N=—50a;0$ 15° Solving Eqs.[l], [2] and [3] yields N=447t81 N x=0,250m
a= 2.01 m/s2 Smcex < 03 m. then crate will not Lip. Thus. the crate slips. . 5“)
4—2};  max; 0.3N +0.3N  ﬂ 4 ‘ ‘ 32.2“Ea (I)
+Tg=m; N.+N‘‘VL—45m=0 (2) SaWL = 8001b in Eqs.(1)uxi(2) NA + N, = 53m y; wsoolb
:44»
N
a = 9.66 1115’ '_' ' m. “
03Ne‘_ «can. 0—0 i?
‘ .
(—>) V’ = v: + ZachIo) N” N“ o = (40)‘ + 2(—9.66)(.r—0)
.r  82.811 Ans
For emptyvui WL = Oin F4541) ukHZ)
NA +N, = 4500
a = 9.66 ﬂ/s’
Thu, 382.8ﬂubefue Ans 0.5N, [ll
[ll [3] Ans Ans 1749. The arched pipe has a mass of 80 kg and rests on
the surface of the platform.As it is hoisted from one level
to the next. a = 0.25 rad/s2 and co = 0.5 rad/s at the
instant 0 = 30". If it does not slip. determine the normal
reactions of the arch on the platform at this instant. +1‘2p; =m(ﬂc),i NA 4 N, — 80(93!) = 20:13:60" — 2000:60“
NA +N, = 792.12 g+2MA = £(M, )A; N,(l) — 80(9.81)(0.5) = 2000560102) + 20$in60°(0.5)
—20cos60°(0.5) + 20ﬂn60"(0.2) rotarnu (0(°.15)(l):2¢, as
/ 7° 4.5...”5 ”(0.9%): 2, N, = 402N Ans NA 391 N Ans ll 1750. The arched pipe has a mass of 80 kg and rests on
the surface of the platform for which the coefficient of
static friction is u, = 0.3. Determine the greatest angular
acceleration a of the platform, starting from rest when
8 = 45°, without causing the pipe to slip on the platform. as = (co), = (ma)
‘4.“ = mph”; N,(l) ~80(9.8l)(0.5) = 80(la)(sin45”)(0.2) + 80(1a)(COs45°)(05) (in; = "1016).: 0.3m + 0.3N, = 80(1a)sin45° +T2p; =m(aa)’; N4 + N, —80(9.81) = 80(1a)oos45" a). 0:
LM m C W, a = 5.95 rad/:1 Ans
NA = 494 N mfiffllN
N, = 628N 1751. The crate C has a weight of 150 lb and rests on
the truck elevator for which the coefficient of static
friction is #3. = 0.4. Determine the largest initial angular
acceleration a, starting from rest, which the parallel links
AB and DE can have without causing the crate to slip.
No tipping occurs. 150
$25 = my. MNc = —(a)cos30°
32.2
150 +T2F = ; N — 1 = _ ~ o I may c 50 32.2(a)sm30
NC = 195,011:
a = 19.34 ﬁ/s1 15045
19.34 = 20: F ~o a = 9.67 rad/s1 Ans ‘ 'fN; T \————————_ *1756. The drum has a weight of 80 lb and a radius of
gyration k0 = 0.4 ft. If the cable, which is wrapped
around the drum, is subjected to a vertical force
1) = 151b, determine the time needed to increase the
drum’s angular velocity from a). = 5 rad/s to
wz = 25 rad/s. Neglect the mass of the cable. 80
(+21% = log; 15(05) = immune: (I) a  18.87 mus’ so
((+)n) = a), + a: u,
j o as
7.5 = 5 + 18.87: x 1‘ H
I = 1.06 3 Ann 93 P 1757. The spool is supported on small rollers at A and
B. Determine the constant force F that must be applied
to the cable in order to unwind 8 m of cable in 4 5 starting
from rest. Also calculate the normal forces at A and B
during this time. The spool has a mass of 60 kg and a
radius of gyration k0 = 0.65 m. For the calculation
neglect the mass of the cable and the mass of the rollers
at A and B. (i+) s = So + v01 + Elatrl 00
ll 0 + 0 + —la€(4)z
2 at = l m/sz l
a = —_ = 1 ‘
08 L25 rad/s (+21% = 10a; P(O.8) = 60(O.65)1(L25) P = 39.6N Ans “EH = max; N‘sinIS" — Ngsin15° = 0 N, coslS" + N5c0515° — 39.6  588.6 = O 3
l
3
II 325 N Ans [176—1. The 20kg roll of‘ paper has a radius of gyration
.A — 90 mm about an axns passing through point A It is
pinsupported at both ends by two brackets AB if the
roll rests against a wall for which the coefficient of'kinetic friction is ,uk = 0.2 and a vertical force F = 30N is applied to the end of the paper d ’ ‘
. , etermine the an ul 4 ' o.2Jv o.
acceleration of the r0ll as the paper unrolls g ar A 125) + 30(0.125) = 20(0.09)2a 7;,sin6738" — 0.2Nc — 20(9 81) 30
. — = o B 1762. Cable is unwound from a spool supported on
small rollers at A and B by exerting a force of T = 300 N
on the cable in the direction shown. Compute the time
needed to unravel 5 m of cable from the spool if the spool
and cable have a total mass of 600 kg and a centroidal
radius of gyration of k0 = 1.2 m. For the calculation,
neglect the mass ofthe cable being unwound and the mass
of rollers at A and B. The rollers turn with no friction. Equation of Motion : The mass momentof inertia of the spool about point Ois given by ’0 = mk; = 600( ll?) = 864 kg  m2. Applying Eq. l7 Hi we have (+ [Mo = ID a; — 300mg) = —364a a = 0.2773 rad/s: , l
_ 2
equation 9 — 00 + wnr+ Em , we have I 1
+) 6.25=0+0+§(0.2778)r‘ I=6.7ls 1
I" = ﬁm12+md2 = 3—,(70)(L2)’ +70(0.6)2 = 33.6kgm’ ”(4, = 41a? 2(509) = —33.6( a) a = —2.97629
Ma) = ads
l:aadm= —l¥2.97629d9 a): 2.71mi]: '1773. The disk has a mass of 20 kg and is originally
spinning at the end of the strut with an angular velocity
of a) = 60 rad/s. If it is then placed against the wall, for
which the coefﬁcient of kinetic friction is p." = 0.3,
determine the time required for the motion to stop. What
is the force in strut BC during this time? L» m; = m(aa)xi Fusin30°—NA = o + TzF, =m(a,;),; chcos30°—20(9.8l)+0.3NA =0 9 EM, = 1,, a; 0.3M (0.15) =[§l(20)(0.15)2]a NA =96.6 N
ch = 193 N
a: 19.3 rad/s2
(+0): at, +art
0:60+(l9.3)l i=3.lls Ans 1774. The disk has a mass M and a radius R. lfa block
of mass m is attached to the cord, determine the angular
acceleration of the disk when the block is released from rest. Also, what is the velocity of the block after it falls a
distance 2R starting from rest? (+310 =”Mao; ng =§MR‘(a)+m(aR)R
2mg 0‘ = R(M+2Jn) A” a=aR v2 = v5 + 20(5—30) v1 = o+2( )(ZR—O) 8
R(M+2m) 8mg]?
(M + 2m) 7'75 ' The two blocks A and B have a mass mA and m3,
espectively, where mg > mA. If the pulley can be treated
a disk of mass M. determine the acceleration of block A. Veglect the mass of the cord and any slipping on the pulley. a=ar 1
(+2510 : HMO)? ”‘30) — MAg(r) = (£M,1)a + Mn’a + M‘r a 8(Ml—MA)
g __________
a 191+”: +M‘)
“Ml—MA)
 ___________
a (%M+ M. + MA) "F 1789. The semicircular disk having a mass of 10 kg is
rotating at w = 4 rad/s at the instant 0 = 60". If the
coefficient of static friction at A is 11,: 0.5, determine if the disk slips at this instant. Equation of Motion : The mass moment of inertia of the semicircular disk about 1 . .
iLs center ofmass is given by 15 = 5(101(042) — 10(0.1698') = 0.5118 kg ~ m'. From the geometry, rGM = #016982 +0.42 — 2(0.1698) (0.4)cos 60" = 0.3477 In.
A]. ' 1 r ' Sine ““600 9 2501° 1 Eq 17 16 we
so. 1n aw , = , = . . in .  . "S g 0 5m“ 0.1698 0.3477 AP” 3 have (+£MA =ziM1hl 10(9.81)(0.l698si1160°)=0.5118a
+ 10(00), cos 25.01°(0.3477) + 10(ac)_,sin 25.01“(0.3477) ll]
(—21: ”1026),; 1;:10(ae). [2]
+Tﬁ. =m(aG)y; N—10(9.Sl)=—10(ag)y l3) Kinematics : Assume that the semicircular disk does notslip atA. then (a‘ )’ = 0.
Here. r5” = {—0.3477sin 25.01°i +0_3477cos 25.0l°j) m = {—0.1470i +0.315lj)
rn Applying Eq.16 18. we have as =aA +aXl'C/A —a)zr5,‘
my, i —(ac)yj = 6.40j+akx(—0.14701+0315”)—4z(—0.14701+0.3151j)
(aG), i—(advj = (2.3523—O.3151a)i+(l.3581—O.l470a)j Equaung i andj componean. we have (ac)x =0.3151a—2.3523 [41
(as), =0.1470a— 1.3531 [5] Solving Eqs.[l]. [2], [3], [4] and [5] yields a: 13.85 rad/52 (as)! = 2.012 m2 (as), =O.6779 m/sz
17:20.12N N=9L32N Since I; < (5)me = ,u,N = 0.5(9l.32) = 45.66 N, then the semicircular disk
does not slip. Ans 464 1797. The spool has a mass of 100 kg and a radius of
gyration kc; = 0.3 m. lfthe coefﬁcients of static and kinetic
friction at A are [1, = 0.2 and #1: = 0.15, respectively, determine the angular acceleration of the spool if
P=600 N. $25 moan: 600+FA =100¢16 NA —100(9.81)=0 + TZF, =m(aa),; (ﬁzMG = 16a; 600(0.25) a (0.4) =[100(0.3)1]a Msume no slipping: a5 = 0.4a a: [5,6 rad/51 Ans ag=6.24m/sz NA=981N FA=24.0N Since (mm = 0.2(981) = l96.2 N > 24.0 N 0K 1798. The upper body of the crash dummy has a mass
of 75 lb, 3 center of gravity at G, and a radius of gyration
about G of kg = 0.7 ft. By means of the seat belt this
body segment is assumed to be pinconnected to the seat
of the car at A. If a crash causes the car to decelerate at 50 ft/sz, determine the angular velocity of the body when
it has rotated to 0 = 30°. 75 75
Hug = 2mm: 7503:1119) = [(323)(o.7)’]a + [553(06):]0‘9) +_/ ‘0 = IA + (lam) + (Ia/A): (ac), = ~50e059 + 0+ (a)(1.9) 7‘"
142531119 = 1.1413a — 2212730059 + 8.4mm
142.5sm9 + 221.273oose = 9.5497a :"
mda) = at“? A It n I‘mdm = I’°'(14.922;ine + 23.17cos9)ds n U
10,1 = —14.922(cos30° — c050“) + 23.17(sin30° — sin0°)
2 a) = 5.21 rad/s All: 468 17107. The 16lb bowling ball is cast horizontally onto
a lane such that initially w = 0 and its mass center has a
velocity v = 8 ft/s. If the coefficient of kinetic friction
between the lane and the ball is pk = 0.12, determine the
distance the ball travels before it rolls without slipping.
For the calculation, neglect the ﬁnger holes in the ball and assume the ball has a uniform density. , lb
4 21'} = “(00);: 0.12M = m“
+T2F,=m(ac),: NAl6=0 216 Q» {Mo = (Ga; 0.12NA(0.375)=[g(m)(0.375)1]a Solving. v = v0 +a,r
NA = 16 lb; as = 3.864 h/s‘; a: 25.76 mil/:1 9.660r= 81864: When the ball rolls without slipping v = a)(0.375). r=0.592 s
( (+) w = as, + at: 1 z y S=Io.+Vof+§a,f O, 375 =0+25.76l’ l
v: 9.660: : =0+8(0.592)—5(3.864)(0.592)2 s=4.06ft Ans ...
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 Fall '08
 Hudyma
 Dynamics

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