Dynamics_CH18_HW

Dynamics_CH18_HW - 18-5. At the instant shown, link AB has...

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Unformatted text preview: 18-5. 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 55-0 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)’ 18-6. Solve Prob. 17-58 using the principle of work and energy. Tr +zUl-1 =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 +zvl-2 = 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 18-14. 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 15-lb 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 18-15. Determ revoluti The hand winch is used to lift the 50-kg 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 18-23. 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 *18-24. The linkage consists of two 8-lb rods AB and CD and a 10-lb 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 zt-H—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,“ 18-25. The linkage consists of two 8-lb 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 18-29. The two 2-kg gears A and B are attached to the ends of a 3-kg 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 - 5-60 Tad = 0.891 rev. regardless of orientation Ans mg” + 2(0.200)1wj,, + 0.0200“); 18-30. The assembly consists of two lS-lb 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 beam-8’2] ('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 18-45. 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 18-46. 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) = émm-W‘ * EU“) 3’21“; *0 m: 19.8 rad/s AH! 500 *18-48. At the instant the spring becomes undeformed, the center of the 40-kg 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 — 196211-430 = 0 Solvtng for the positive root d=3.38m Ans 18-49. The pendulum consists of a 2-lb rod BA and a 6-lb 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) = 0.091h-ftand 5(2)( 155‘) = 2.40251brft. Thus, Vl =0.091brft V2 =2i4025+[—2(l.25)]=—0.0975Ib-ft 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 18-57. At the instant shown, the 50-Ib 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 18-58. 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)(4-2)z = 07112114 +6s1ne-2)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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Dynamics_CH18_HW - 18-5. At the instant shown, link AB has...

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