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Unformatted text preview: 185. At the instant shown, link AB has an angular
velocity (1),“; = 2 rad/s. if each link is considered as a
uniform slender bar with a weight of0.5 lb/in., determine
the total kinetic energy of the system. 6
“’5: = z = 1.5 rad/s v: = 1.5(4/5) = 8.48531n./s
550 s ((2)2 + (4)2 = 4.472 “a = 1503.472) = 6.7082 in,/s 8.4853 “Dc ‘ T = 1.697 nd/s l I 3(0.5) 3 1 4(05) 6.7082 1 l
T=__‘_z 3 _ 4(0.5) 4
2% 32.2 ’92) "2’ + 2'\l‘~” * E’E‘W"E
335(05) s , ,
1‘ 213(Tl2 )(E) “1.697) = 0.0188 11!!! An )’1(1.5)’ 186. Solve Prob. 1758 using the principle of work and
energy. Tr +zUl1 =75 0 + 30(8) = %[(%)(L30)’]af a): 4.78 rad/s An: 18'7 Solve Prob. 17—59 using the principle of work and
energy. 7i +zvl2 = 75 dz 1 l 10
a = _ _ __ 2 2
0+In 5 do 2i12(32.2)(2)]“’ 3(5)’ = 0.0517“;1
2 2 m = 10.9 rad/s 1814. A motor supplies a constant torque or twist of . M = 120 lb  ft to the drum. If the drum has a weight of M: 120 lb . ft
30 lb and a radius of gyration of k0 = 0.8 ft, determine the speed of the 15lb crate A after it rises s = 4 ft starting from rest. Neglect the mass of the cord. Fru Body Diagram : The weight of the crate does negm’ve work since it acts
in the opposite direction to that of its displacementsw. Also. the couple moment
M does positive work as it acts in the same direction ofits angular displacement
0. The reactions 0,, 0) and the weight of the drum do no work since point 0 does not displace. Kinematic . Since the drum rotates about pomt 0, the angular velocity of the ’U ’U
drum and the speed of the crate can be related by (DD = —‘ = 1—; = 0.66671)". ’17
When the crate rises s = 4 ft, the angular displacement of the drum is given by Principl! of Work and Entrgy : The mass moment of inertia of the drum )(of) = 0.5963 slug  {8. Applying Eq. 30 about point 0 is IO = ink}J = (32 2 18 13, we have 7.1+ZUi—2=T2 l l ,
0+M0— Wcsc = 51002 +§mcvg l , i l5
0+120(2.667)—15(4) = 5(0.5963)(0.6667UA)'+5(— 2.2 11‘ = 26.7 {1/5 1815.
Determ
revoluti The hand winch is used to lift the 50kg load. ine the work required to rotate the handle ﬁve
ons. The gear at A has a radius of 20 mm. 75f I30 mrn 20m) = 9.030) thnad = Srevi =10}: 9, = 4.8332 rad Thus load moves up .1‘ = 4.8332(0J m) = 0.48332 in U = 50(9.8l)(0.48332) = 37 J 1823. 1116 20—kg disk is originally at rest, and the spring
holds it in equilibrium. A couple moment of M : 30 Nm
is then applied to the disk as shown. Determine how far
the center of mass of the disk travels down along the
incline, meaSUred from the equilibrium positron, before it
stopsThe disk rolls without slipping. +ZMA=01 —E(0.2)+20(9.81)sin30°(0.2)=0
F, =9s.1 N n =384E=0i654m 150 r
When G moves 5, the disk rotates 6 = 6—2 T: +ZUi—z =Tz 1
0+ 20(9,Bl)(.r)sin30“ + 30(5—2 ) {% (150m +0, ssaf — i (150)(0r654)2] = 0 243‘ I: = 75 (:2 + L308: + 04277) — 3208 248.1 = 75: + 98.1 5:2.00m Ans *1824. The linkage consists of two 8lb rods AB and
CD and a 10lb bar AD. When 9 = 0°, rod AB is rotating
with an angular velocity wAB = Zrad/s. lf rod CD is
subjected to a couple moment M = 15 lb  ft and bar AD
is subjected to a horizontal force P = 20 lb as shown,
:etermine wAB at the instant 6 = 90°. 7i +EU._2 = 72‘ 1 l
ztH—s— 2 2 1 10
2 3 322)”) "2) 1 + $323)“): + (20(2) + “(3') —2<s)<i) 40(2)] = 8 1 1
2[§{(— 3 32. m = 5.74 rad/s A,“ 1825. The linkage consists of two 8lb rods AB and CD
nd a 10—lb bar AD.When 0 = 0°, rod AB is rotating with
n angular velocity a)” = 2 rad/s. lf rod CD is subjected « f1 Couple moment M = 15 lb  ft and bar AD is
bjected to a horizontal force P = 20 lb as shown,
Etermine wAB at the instant 6 = 32.2 I 10
2mm“?! + gtmxzm)’ :IA) .11.”. 2 1 1 ‘0
2t2(3( )(2) }(2) 1 + 5975)“? + i20(2§in45°) + 15(f) _2(8)(1A 1 1 8
= 2  — __
[2{3(32.2 a) = 5.92 ndls )(2)2}¢D2] + l(_1_0_
2 32.2 Ans cos45°) —10(2 .. 2W4?” )(Zm)2 «J
“5:290 m :51 .. a 1829. The two 2kg gears A and B are attached to the
ends of a 3kg slender bar. The gears roll within the ﬁxed
ring gear C, which lies in the horizontal plane. If a 10
N  in torque is applied to the center of the bar as shown,
determine the number of revolutions the bar must rotate
starting from rest in order for it to have an angular velocity
of a)“, = 20 rad/s. For the calculation, assume the gears
can be approximated by thin disks. What is the result if
the gears lie in the vertical plane? Energy equation (where G refers to the center clone of the two gears): M6=Tz r t , 1
109 = 2 (Elwin) + 2 (5mm) (0.200%3)~ + 5143:1135  l 1 _ 2
Using mgm 2 2 kg. I0 = 5(2)(0.150) _ 0.0225 kg . m . When em = 20 rad/s. i 200 a
1A,, = EOXOAOOV = 0.0400 kg . ml, and calm" = ﬁlm, 9  560 Tad = 0.891 rev. regardless of orientation Ans mg” + 2(0.200)1wj,, + 0.0200“); 1830. The assembly consists of two lSlb slender rods
and a 20—lb disk. If the spring is unstretched when (9 = 45°
and the assembly is released from rest at this position,
determine the angular velocity of rod AB at the instant
6 = 0°. The disk rolls without slipping. 7' +Eur—2 = 7: 0+0) +2(15)(1,5) sin 45° — %(4)[6 * 2(3) cos4s°13 l l 15
2 _ _ W 2 :
[2(3(32.2)w>) As] LAB =4.28 rad/s ’5 L5 sin 45° {1;} aThe uniform door has a mass of 20 kg and can be T, + Z UL—l = T:
sectezatthin plate havrng the dimensions shown. If it is
» 8 0 a torsional spring at A, which has a stiffness m+§ 1 1
e SP8“? .m/rad, determine the required initial twist 0+1 80" d9 = 5 beam8’2] ('2):
y of Igzin radians so that the door has an angular
ed at rild/OS when it closes at 9 = 0° after being ,7 2
' 9 = 90 and released from rest. Him: For a “[09” + ’1‘) "901] z 3072 31 spring M = k . . , 2
angle of [wisp 9, when k is the stiffness and 9 is n 90 = l.66 rad 1845. The two bars are released from rest at the
position 6‘ Determine their angular velocities at the
instant they become horizontal. Neglect the mass of the
roller at C. Each bar has a mass m and length L. Potential Energy : Datum is set at pOintA . When links AB and BC is at their L
initial posmon. their center of gravity is located :sin 6 above the datum. Their
gtavuauonal potenual energy at this posmon 15 mg Sin 0 Thus. the initial and final potential energies are
, mﬂ. .
ltI =2(—2—su\ 6)=mgLsin9 V2 =0 Kinetic Energy : When Links AB and BC are in the horizontal position. then
0,, = mABL which is directed vertically downward Since link AB is rotating about ﬁxed pointA. Link BC is subjected to general plane motion and its
insmnumcous center of zero velomry [5 located at point C. Thus. 1),, = maﬁa/,5 or w‘HL z mch, hence a)” = a)” = m The mass moment inertia for link
_2 i,L
ABmdmhmmwmAmmCEMNA=War=ﬁmt+mEJ l 1
: imL‘. Since links AB and CD are at rest initially. the initial kinetic energy is T1 = ()i The final kinetic energy is given by i , ,
7} =;(IAB)A wis+ (lunatic
_1C ij+lp E z
—2 3In  —m )(u
i ,2
=—mL'm
3 Conrervatian of Energy : Applying Eq.18— l8. we have n+n=n+n i ,
0+mysm9=§anﬁ+o k.
wu=mac=m= 1—st Ans 1846. An automobile tire has a mass of 7 kg and radius
of gyration k(, = 0.3 m. If it is released from rest at A on
the incline, determine its angular velocity when it reaches
the horizontal plane. The tire rolls without slipping. ya = 0.410
Datum til lowest pomL n+w=n+n l
0+7(9.81l(5) = émmW‘ * EU“) 3’21“; *0 m: 19.8 rad/s AH! 500 *1848. At the instant the spring becomes undeformed,
the center of the 40kg disk has a speed of 4 m/s. From
this point determine the distance d the disk moves down the plane before momentarily stopping, The disk rolls
without slipping. Datum at lowest point n+w=n+w I l 7 4 2 l
_  41 ~ _ _ 1
2[2( ))(O,3) ](OI3) + 2(40)(4) +40(9 8])d51n30° l
= 0+ 5(200)d‘ lOOdZ — 196211430 = 0 Solvtng for the positive root d=3.38m Ans 1849. The pendulum consists of a 2lb rod BA and a
6lb disk. The spring is stretched 0.3 ft when the rod is
horizontal as shown. If the pendulum is released from rest
and rotates about point D, determine its angular velocity
at the instant the rod becomes vertical. The roller at C
allows the spring to remain vertical as the rod falls Potential Energy : Datum is set at point 9. When rod AB is at vertical posm'on.
its center of gravity is located 1.25 ft below the datum. [l5 gravitauonal potenn'al
energy at this position is  2( 1.25) ft lb. The initial and final stretch of the spring
are0.3 ft and (1,25+0.3) ft: 1.55 it. respectively. Hence. the initial and final 1 1 ,
elastic potential energy are E(2) ( 0,31) = 0.091hftand 5(2)( 155‘)
= 2.40251brft. Thus, Vl =0.091brft V2 =2i4025+[—2(l.25)]=—0.0975Ibft Kinetic Energy : The mass moment inertia for rod AB and the disk about point D l 2 3 2 2 7
are (1”)0 = ) 1.25 ) = Oil 178 slug ft‘ and (ID)0
l 6 1 .
= 0.25‘) = 0005823 slugv ft't Since rod AB and the disk are initially at rest. the initial kinetic energy i571 = O. The final kinetic energy is given by l 7; = (IAE)0(U:+Z(]D)oluz [cl—N].— l
(0.1 1723):»2 + 5(000582301
.0617902 O Conservation of Energy : Applying quS— 18, we have ﬁ+K=E+K
0+0.09 = 0.06”ng + (—0.0975) to = 1.74 rad/s Ans 502 1857. At the instant shown, the 50Ib bar is rotating
downwards at 2 rad/s. The spring attached to its end
always remains vertical due to the roller guide at C. If the
spring has an unstretched length of 2 ft and a stiffness of
k = 6 lb/ft, determine the angular velocity of the bar the
instant it has rotated downward 30" below the horizontal. Datum through A. T1 +V1 =T1 +l€ l l 50 2 2 l 1 l[l 50 2] 1 1   — — 4— =_ _ _ _ _ _
2i3i32.2)(6) in) ‘2‘” 2) 2 4322)“) af+2(6)(7 2) 50(1.5)
w=2.30rad/s Ans 1858. At the instant shown, the 50‘lb bar is rotating
downwards at 2 rad/s. The spring attached to its end
always remains vertical due to the roller guide at C. If
the spring has an unstretched length of 2 ft and a stiffness
of k = 12 lb/ft, determine the angle 6, measured below
the horizontal, to which the bar rotates before it momentarily stops. T1+li=T2+lé 1 1 50 1 1
§[5(3—23)(611](211 +5112)(42)z = 07112114 +6s1ne2)1 50(35in6) 61.2671: 24(1+3s'1n19)2 —1505in6
37.2671: —6$in9+ 216511119 Set x = sine . and solve the quadratic equation for the positive root: sine = 014295 6: 25.4” Ans 506 ...
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 Fall '08
 Hudyma
 Dynamics

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