PROBLEM 17.1 ‘The motor of an electric motor has an angular velcity of 3600 spm when the load and power sre cut off. The '50-kg rotor then coasts to rest afler S000 revolutions. Knowing tha the Kinetic ction ofthe rotor produces a ‘couple of magnitude 4 N -m, determine ine centoidal rads o gyration oF the ror. ‘SOLUTION Angular velocities: _Angolar displacement ‘5000 rev = 10000. rad Principle of work and energy: T;+U_.2= 7 71.061x10°T Uy gq =-MO=-14 N-m(10000% rad) = 4000007 Nem 71.061x107T —40000r Centroidal radius of gyration, SOkg F=18.1mm «PROBLEM 17.14 ‘The dhuble gulley shown has a mass of 15 kg and 2 ‘cearcidal radius of gyration of 150 mm. Cylinder A ‘and block B are attached 10 cords that are wrapped ‘on the puleys as shown, The coefficieat of kinetic fiction bewweee block B and te surface 4s 02. Knowing thet the system is at rest in the postion “shows when a consiant force P~ 200 N is applied cylinder 4, determine (a) the velocity of cylinder A as it strikes the ground, (b) the total distance that Dock # moves before coming to rst. Kinematics. Let rbe the radus oft outer pally and ry that of te ines pulley no =ing Use the prigcipe of work and energy with position 1 being the inital est position agd postion 2 being when cylinder a shes we ground, qe le Long 250 130m)" ,0384%g-m"],2 (0.250m)" (0.250 m) =R.272kg Principle of Work and energy applied tote system consisting of Blocks ane Hand tne double pulley C. Work. Uyan = Poy +e sy — Fp Mpg 5p SiN 30" where jaziPROBLEM 17.14 (Continued) and p= By OO a 0250m “To fi Fuse the free body diagram of eek B. Qa ALO? EF =O; Ny ~mmgg cos =0 a0 Usa = 200 Nm) +S REABSL OM} CLM) ~ (25487 N05 m)~(15 kgh9.81 m/s*\(05 mind” £189,613) Wort-energy: 0+189.6131 =(8.272 kgyr} (a) VelocityofA. vy = 4.7877 mis ‘when the cylinder strikes the ground, (150m, 02501 _ATSI1 vs _ 0250 After the cylinder strikes the ground use the principle of work asd energy applied to & system ‘consisting of block B and double pulley C. Let T be itskinetc energy when A strikes the ground 4.7877 mis) = 2.8725 mis 19.1508rals| er 5 ithe aditonal travel of lok 152.305) -(98.062 9) % (0) Totaldstance: s+ =18i0m 4 Bas 356mPROBLEM 17.16 [A slender rod of length ! and weight Wis pivoted at oncend as shown. tis Felease fom retin horizontal position and swings freely (a) Determine the angular velocty of the rod as it passes tough a vertical postion ‘and determine the coresponding reaction at the piv, (b) Solve pat for W=USiband/=3 1. SOLUTION poston: poston? wo ona! mepreorwensocneay: 1+. =% Ld nyt O+me $= Zmioh (@ Expressions for angular velocity and reactions. a leo Application of data: PROBLEN 17.16 (Continued) Weism, I-30 Tras ) siia PROBLEM 17.18 A slender 9b od can rotate in a vertical plane shout pivot at 8. A spring ‘of corstaat = 30 Ib and of unstetched leegth 6 in. is allached tothe rod ala as shown. Knowing thatthe rod is released from rest inthe position shown, * ‘dpiermine ls angular Yell alr & as rotated through SA my =a Iho h Ras AREA Kits 1 Peston Spring:Gravity: Kinsic energy: PROBLEM 17.18 (Continued) Y Vs=V,4¥, =09r75Ih-f 1 os 219 ane 2 n= (2 Joy? coopstessng (Bae e 322 oly, mehr 83 + Tod (3 (20 jon) women 7, =0.0941380} TAM ah+Vs 0413438 =0.094138 of +0.9375 0, =1132rais )PROBLEM 17.25 A rope is wrapped around «eylinder of radius rand mast m as shown, Knowing thet ‘hecylinder is feleased from rest determine the Velocity ofthe center ofthe cylinder after ithas moved downward distances. ‘SOLUTION Pint Cis the istantancous contr. sla 5 Peston 1. At est. T=0 Peston 2. Cylnaer has rallea trough aisianee 5PROBLEM 17.28, Fo [A small sphore of mass mand rave released fre test at A and af rolls without sliding on the cuved surface to Point B where it leaves t the surface with a horizontal yelecity. Knowing that a = 15 m and L = 2m, determine (a) the speed ofthe sphere as it sik the ground AC, (0) te correxpending distance ‘SOLUTION Work: Una =mea Kinctic energy: Rolling mtioe st position 2. Principle of work and energy. TU For path to C the motion is projectile mation, Let ‘Vertical motion Oden at amet Y=W+OPo! Log At Point C, 0=b+0-Let tc -- [eee _oapssos asa Oye 981 ms" 0.494628) =~4.8522 misPROBLEM 17.28 (Continued) Horizontal motion: Let hex coordinate pot to the lft with origin below B. S849 ms @ Spedae. vex {mere t= (CRT ASDF Distance c=We €= (4.5849 ms(0.49462 5)PROBLEM 17.36 ‘The motion of the uniform sod AB is guided by small wheels of ‘negligible mass that roll onthe surface shown Ifthe rod is eleased, from rest when @=0, determine the velocities of A and B when ©. Postion 1. Posiion2. Kinematics. Locate the instantaneous center C. Triangle ABCis equilateral, yy=ty =Len Les s0830° ‘Moment of inertia, Kinetic energy. vy =0778 Ye vy =071SYRL omsyge > or <PROBLEM 17.39 “The ends of a 9:1b rod AB are constrained 19 move along sb eut ina ‘vertical plate as shown. A spring of constant &=3 bin. is attached 10 ‘end 4 in such a way tht its tension is zero whes. @=C. IF the rod is ‘eleased from rest when @= 50, determine the angular velocity ofthe ‘od andthe Velocity of end B when @=0. ©, ae le arian son visswEansoretie =+91m( 2 nso avin ys.03in? z =-85.18+11963 3.45 in 1DPROBLEM 17.39 (Continued) Conservation of enorgy Tay -T+¥, 042.7871 =0,202208 Trails”) 4 2B Jonsnay vastus} 4PROBLEM 17.47 ‘The 8O-mm-radius gear shown has a muss of 5 kg and a centroidal radius of gyration of 60 mm. The 4-Kg fod AB is altached to the center (Of the geat and 10 a pin at & Gat slides fiely in a verical slot. Knowing that the systomis released from rest when @= 60°, determine the velocity of the center ofthe gear when 6 = 20 ‘SOLUTION “Angular velecity of gear Poteniial energy: Use the evel ofthe center of gear Aas the datum, Wal Loose 7 Kinetic energy: Misses and moments of inertia:PROBLEM 17.47 (Continued) Conservation of energy Pasion vad 720 Postion fom | 2 aD 254140625 = 056524+01887993 = 4661243 (4838110320)6082° =-00985 Conservation of emp 1 0-a.1992 = 460124") -5.8958 vy =059225 ms? 1% =0770 ms 140.770. mkPROBLEM 17.49 ‘Three shafts and four gears ae used to form a gear train ‘which will ansmit 75 KW from the moter a Ato & machine tool at F. (Bearings for the shats are omitted from the stetch.) Krovong that the frequency of the movor is 30 Hz, ddtenmiae the magnitude of the couple which is applied to shaft (a) AB () CD. (©) EF. Kinematics Gears B and C. Gears D and F. = fetey: (7S my Oe rad/s) = (180 mm)(0ep) bey = ely: (7S mmy25z rails) = 180 mm} o%g,) E E @ © © E E ge = WAIGT ral Power=M yyy: 7500 W=M yy (60% rads) My=B8Nm 4 Power=Mepaiy: 7500 W=Mpy2S ras) Mey=9SSNim Power=MrOjy? 7500 W = Mpy (10167 rails) Myy=29N-mPROBLEM 17.52 “The rotor ofan electric motor bas a mass of 25 kg, and iis observed tha 4.2 minis required forthe rotor to coast to ret from an angular velocity of 3600 fpm, Knowing that kinetic friction produces + couple of ‘magnitude 12 N-m, determine the cenfoklal radius oF gyration forte row. ‘SOLUTION Coasting time: 1242 min=252s Iniuat angular veloc: (0) = 3900 pm = 1208 reais Principe of impulse and memeatum, @+"®-@ Syst Momenta, + ESystlmpulses..,. = —=—‘Syst. Momenta (Moments about axle A: Tax — Mr =0 Radius of gyratio 1791 mPROBLEM 17.62 ise sas an inal angular velocity o9, when iti brought into cootact with disk A whic i at rest. Show that the final angular velocity of disk B depends nly oa a, end therabo of the masses mand m, of the two disks. ‘Lat Points 4 and B be the centers ofthe two disks and Poiat Cbe the contact point between the two disks. Lat ey andy be the final angular velocities of disks A and B, respectively, and lev be the final velocity at ommen i both dists. Kinematics: No slipping Ve=Ha= he ‘Moments of inertia, Assume tht both disks are uniform cylinders. Lind Lad tyadmg —ty=4my} 2 am Principle of impulse and momentum. O + Qe O- Fe FE (2) 7 @ [Disk A) Moments about A: 04+ HFT = 140 ey Le mae Disk 8) Moments shout B: Amgen, ~ to) + pittPROBLEM 17.66 a ‘Show that, when a rigid slab stats about» firad axis through O perpendicelar ZL to the slab, the system of the momenta o its panicles is equivalent ta single vector of “0, perpendicalar 10 the line OG, and applied 19 a Point P on this line, called the centr of percussion, ata distance CF ‘rom te mass center ofthe sh. ‘SOLUTION Kinematics. Point O is Fixed ‘System mementa, ‘Components parallelto mF: ‘Moments sbovt G:PROBLEM 17.74 “Two uniform cylinders. each of mass m=6 ke and ratius ¢=125 mm. are ccnnected by a bel as stown. Ifthe system is released from rea ‘when 1=0, determine (a) the velocity of the center of cylinder B fa 1=35, () te tension in the portion of Delt connectig the two cylinders Moment of inertia, (@) Nolcity of hecenter of A Opler B: Syst. Moment +) Moments about B: Coline A ‘Syst Momenta, © Q Pt + Syst Extimp..: = Syst Momenta, 0+Pu=Tay a Pt et Tay . GO . >: vat wii, + SyshEstimp.y.9 = Syst Momenta,PROBLEM 17.74 (Continued) )otomests about: 0-2Ptr+ mgtr= nity +Toy, w tanned) So Log? T+ Joy = mgr & gi Jom (See Ln joy mgr ap tye fone Ly =at 2a=t ta V=84lmis| () Tension inthe bet FromEgs (1) and @), \6817N P=1682N 4PROBLEM 17.77 A sphere of radius and mass m is projected slong a rough horizontal surface withthe intial velocities shown I the final velocity of the sphere is tobe ze, express (a) the required magaitude of in terms of vp andr, () the time equlred for he sphere to come to rest in lems of Yp and CoerTckeat of Kinete fiction 4 ‘SOLUTION Momeat of inertia, Solid sphere. fio. 2 WAN on tm) =( es Ft wel ‘SjstMomenta, + Syst Ext.Imp., = Syst. Momenta, {3 components: M-Wr=0 | N=W=ng 0 1 components: ay =0 P= my ° 2) Moments sot 6: Tay - Fir =0 o (a) Sohing for a, () Time tocome wo rest. From Equation (2),2m PROBLEM 17.96 “The circular platform A is fitted with a rim of 200mm inner ratius and can ‘tate freely about the vertical shaf. Its known tha the pltform-cim unit has ‘amass of § kg and a radius of gyration of 175 mm with respect to the shaft. Atatime when the platform is rotating with an angular velocity of 50 spm. a [kp disk B of radius 60 mm Is placed on the patiorm wilh no veloc. ‘Knowing that disk 2 then sides until it comes to rest relalive tothe platform against he rim, determine the final angulr velocity of the platform. ‘SOLUTION Moments of inertia. gX0.175 my 153125 kg-m® Sule Disk BB atres. ‘Slile2 Disk # moves with platform A. Kinematics. InState 2, (0.12 may Principleof conservation of angular momentum. Q)MomeawabouD: Tre Tye, Tyas + (0.153125 kg-m)ay =(0.153125 kg-m")a + 6.6107 kgm? a +(OB}0.12 mon 0:1531250y =0.2059%0 0 = 0.743609, =0.7436(50 1pm) Final angular velocityPROBLEM 17.91 A small 419 collar C can slide freely om a thin ring of weight 6 Ib and eadive 10 in. The ring is welded toa short vertical shaft, which can roate freely ina fixed bearing. Inally the ring has an angular velocity of 35 rad/s and the collar is at the top of the riag (@=0) when it is given a slight nudge [Neglecting the effect of irctiog, determine (a) the angular veloity of the ring ‘the collar passes ough the positon @ =90°, (0) the eatesponding vebily ‘ofthe collar Featve tothe ring. SOLUTION ‘Moment of inertia of rng. Postion I Position 2 mph" = (my +2me) Fan 2, oPROBLEM 17.91 (Continued) Potential energy. Datums te cemer ofthe ring Vi=mesk Kinetic energy: 4 inp of consrvaion of nee: Mahe 1 nptted +mcen=(Lne +t | Rial + mer? c dintok mesh hina mer} 2) ats wena ors in=os33330 (@ Angulacvelocty. From Ba, (0 oy =15 00 als (©) Nelocity of cobar ceative wo ing From Ee. 2), [SE aE sme 30,50.43.3333- 16.984 +0061 oa ve saisPROBLEM 17.96 At what height h above its center G should ili ball of rads be ‘truck horizontally by 2 cue ifthe ball i to star reling without sliding? ‘SOLUTION ‘Momeat of inet. Principle of impulse and momentum, ()- Syst Moment; + ‘Syst Ext. Imp.s.2 = Syst. Moments: Kinematics. Rolling witout sliding. Poist C sth instantaneous center of rotation, + Linear components: 0+ Pat =m, mre, ( Moments about G: 0+ APAt=Ta o-sncnrany (ZonPROBLEM 17.102 T Ge] 4 pho te with» wlan of dn ae 029 non 088 HOS Q| | Sure pant ot ate p20 Kaowing ate panel stn ca FRx domino be regured dance ft the fon omponet LN | cre npaave ncn at Ast te a, () te comeing [yf Seti ft tof al immeslaf e f cme teat soLuTiON ‘my =0.085 tg mp =9kg Mg day? = 4902007 0.06 kgm? Kinomase. AMterimpact the pate is rotting aout the aed Point A with angola veocty © =") eed = Principe of impulse and momentum, To simplify the analysis, neglect the mass of the bullet after impact. ‘Aco Ay). eo t 5° dew - + = | Pat ues a Syst Momenta, + Syst Ext mpg = Syst. Momentay 4etmereomanm ——aayantssdemagem (te) (0.0851s00.05)=(2"0100)0 «=19:239 ral = (0100199238) =1 9239 als © ©) Moments (yy 00859 + (gy sin 5°) = 1p0r + my ))Moment about A: mgyjconS+ (npr sin5}2= Ica ¥o mftsn tons teatneo (0.085,400)heos5°+0,100sin5 [205+ 40,0200? Jassm 179315h+ 0.1509 =2.9885 © nzois92m, azissmm < (©) FromEg, (0), Yo =1992ms—> €® PROBLEM 17.113 “The slender rod AB of leagth & strikes the frictionless surface shown with a vertical velocity im forms an angle f= 30° withthe vertical as 2s and fe fo angular vebeity. Knowing thatthe coefficient of restituion between the fod ' ‘and he grcund is¢=OS, determine the angular Yelecity ofthe rod irmeditely i ster the impact. ‘SOLUTION Moment of inert. 7 a ‘Apply coefficient of restitution. [yl =e) k= 1 YAH OValt yb Oat TD Kinetis i, « nily ma, / pe 4 A Sad Syst. Momenta, + Syst.Ext.Imp.,, = Syst.Momenta, + nerizontat components: o+0=nm, — mt,=0 inmates reaver tan DyLI=1 HT lem), 1 sin Velocity somposent _) Moments sbout 4: anys p +0, Esinp + To £sinp=m{ Lersinf -eF |Esin, sn fanpae(op-ban (eepara ota siltersinPROBLEM 17.119 (Continued) 1380" 30° 7 rads)PROBLEM 17.115 “The uniform rectangular bicek shown is moving along a fictionless surface with a velosity ¥, when i trikes @ ‘small obsincton at Assuming that he impact between comer A and obstruction B is perfectly plastic. determine ‘he magaitule ofthe velocity ¥, for which the maximum ‘angle 6 through which the block will ate i 30 ‘SOLUTION Let mbe the mass ofthe block. Dimensions 0.200 m b=0.100.m ‘Moment of inet about the mass center. Let dbe one half the diagonsl Kinematics. Befow ingot ===, @=0 After impact the Hock sroating sot commer at =m) y=004/ Prinspl of impai nd moneston, a ((-Moments touPROBLEM 17.115 (Continued) = 3H a+)” “The motion after impacts a rotation about corner Postion 2(\mmestaety afer mpact). —_-¥, = day Angular velocity afer impact e: o Postion 3 (@=30"). p= ta! =a 95=20565° fiz dsin(f + HP) =0.1180sin56 565° =0.093301 Potential energy: Kinetic energy: ‘Principe of conservation of energy: T+Yy=T4%, 1 a? + ?ya2 altel Ama? +003 +e. ‘ ay +mgn 3g(2h—¥) _XOBDVO.18660-0.100) aj =28 ery ‘Magnitude of inital velocity. Solving Eo. (1) for % (210.200)? + (0.100)510. Gx0.100) y= 238musPROBLEM 17.119 [A Lox bllat is fred with 4 horizontal velocity of 750 mil into the 18.1 ‘wooden beam AB. The team is suspended from a collar of negligible mass that can slide slong a horizontal rod. Neglecting friction between the collar and the rod, determine the maximum angle of rotation of the beam during is ‘Subsequent mouon. ( ‘SOLUTION Mass of ballet Mass of beam AB. Since 8 iso small, the mass of the ballet wil be neglected in comparison with that ofthe bearn in determining the motion after the impect. =m? 2 “Memeat of inertia. Impact kinetics Saat imo, || To, Brot | eae Syst. Momenta, + Syst. Ext.Imp.;_2 — Syst:Momenta; + tinoar componcsts: Bony sO=nm, Hp =P 2) Momeni about B: 0402 Tamir,PROBLEM 17.119 (Continued) ‘Motion during sig. Postion 2, ust the impact, me (Gavmatiots Position 3. = Syst Momenta, 22/2100)? aa =09e021 y= 45.0"PROBLEM 17.127 Member ABC nas a mass of 24 Kg and Is attached 19 a pin support at 2. An 800: sphere D strikes the end of member ABC with a yertcal velocity ¥, of 3 mvs. Knowing that Z— 750mm ‘and that the coefficient of restitution betweea the sphere and member ABC is 05, determine immediately after the impact | @ the angular velocity of member ABC, (8) the velocity ofthe 2 osn7507 0.1125 kg Kinematics after impact. a re (0 8009(3,(0.1875)= (0.800)0.1875)%, + [0.1125 + (2.4,01875)" Jef O45= 015% + 019687500 oPROBLEM 17.127 (Continued) Cocificint of restitution, Vp -O187Sa/ =-(05)8-0) ¢ Solving Eqs. (1) and 2) simaltaneousy. (@ Angularvelocity, (®) Nelocity of D. rads”) € yp = 0938s) €