CHAPTER 16 VECTOR MECHANICS FOR ENGINEERS: DYNAMICS Ferdinnd P. Beer E. Ruell Johnton, Jr. Lecture Note: J. Wlt Oler Tex Tech Univerity Plne Motion of Rigid Bodie: Force nd Accelertion
Content Introduction Eqution of Motion of Rigid Body Angulr Momentum of Rigid Body in Plne Motion Plne Motion of Rigid Body: d Alembert Principle Axiom of the Mechnic of Rigid Bodie Problem Involving the Motion of Rigid Body Smple Problem 16.1 Smple Problem 16. Smple Problem 16.3 Smple Problem 16.4 Smple Problem 16.5 Contrined Plne Motion Contrined Plne Motion: Noncentroidl Rottion Contrined Plne Motion: Rolling Motion Smple Problem 16.6 Smple Problem 16.8 Smple Problem 16.9 Smple Problem 16.10 16 -
Introduction In thi chpter nd in Chpter 17 nd 18, we will be concerned with the kinetic of rigid bodie, i.e., reltion between the force cting on rigid body, the hpe nd m of the body, nd the motion produced. Reult of thi chpter will be retricted to: - plne motion of rigid bodie, nd - rigid bodie coniting of plne lb or bodie which re ymmetricl with repect to the reference plne. Our pproch will be to conider rigid bodie mde of lrge number of prticle nd to ue the reult of Chpter 14 for the motion of ytem of prticle. Specificlly, F m M G H nd G D Alembert principle i pplied to prove tht the externl force cting on rigid body re equivlent vector m ttched to the m center nd couple of moment Iα. 16-3
Eqution of Motion for Rigid Body Conider rigid body cted upon by everl externl force. Aume tht the body i mde of lrge number of prticle. For the motion of the m center G of the body with repect to the Newtonin frme Oxyz, F m For the motion of the body with repect to the centroidl frme Gx y z, M G H G Sytem of externl force i equipollent to the ytem coniting of m nd H. G 16-4
Angulr Momentum of Rigid Body in Plne Motion Angulr momentum of the lb my be computed by n H r v m G i 1 n [ r ( ω r ) m ] i 1 ω Iω After differentition, H G Iω Iα i i ( r m ) i i i i i i Conider rigid lb in plne motion. Reult re lo vlid for plne motion of bodie which re ymmetricl with repect to the reference plne. Reult re not vlid for ymmetricl bodie or three-dimenionl motion. 16-5
Plne Motion of Rigid Body: D Alembert Principle Motion of rigid body in plne motion i completely defined by the reultnt nd moment reultnt bout G of the externl force. Fx m x Fy m y M G Iα The externl force nd the collective ective force of the lb prticle re equipollent (reduce to the me reultnt nd moment reultnt) nd equivlent (hve the me ect on the body). d Alembert Principle: The externl force cting on rigid body re equivlent to the ective force of the vriou prticle forming the body. The mot generl motion of rigid body tht i ymmetricl with repect to the reference plne cn be replced by the um of trnltion nd centroidl rottion. 16-6
Axiom of the Mechnic of Rigid Bodie The force F nd Fct t different point on rigid body but but hve the me mgnitude, direction, nd line of ction. The force produce the me moment bout ny point nd re therefore, equipollent externl force. Thi prove the principle of trnmiibility where it w previouly tted n xiom. 16-7
Problem Involving the Motion of Rigid Body The fundmentl reltion between the force cting on rigid body in plne motion nd the ccelertion of it m center nd the ngulr ccelertion of the body i illutrted in freebody-digrm eqution. The technique for olving problem of ttic equilibrium my be pplied to olve problem of plne motion by utilizing - d Alembert principle, or - principle of dynmic equilibrium Thee technique my lo be pplied to problem involving plne motion of connected rigid bodie by drwing free-body-digrm eqution for ech body nd olving the correponding eqution of motion imultneouly. 16-8
Smple Problem 16.1 SOLUTION: Clculte the ccelertion during the kidding top by uming uniform ccelertion. At forwrd peed of 10 m/, the truck brke were pplied, cuing the wheel to top rotting. It w oberved tht the truck to kidded to top in 6 m. Determine the mgnitude of the norml rection nd the friction force t ech wheel the truck kidded to top. Drw the free-body-digrm eqution expreing the equivlence of the externl nd ective force. Apply the three correponding clr eqution to olve for the unknown norml wheel force t the front nd rer nd the coicient of friction between the wheel nd rod urfce. 16-9
Smple Problem 16.1 v0 10 m/ x 6 m SOLUTION: Clculte the ccelertion during the kidding top by uming uniform ccelertion. v 0 v0 + ( x x0 ) ( 10 m/) + ( 6m) 8.33 m/ Drw free-body-digrm eqution expreing the equivlence of the externl nd ective force. Apply the correponding clr eqution. F y F y N A + N W B 0 F x F x µ k F F ( N + N ) A A µ W k B µ B k m g ( W g) 8.33 9.81 0.849 16-10
Smple Problem 16.1 Apply the correponding clr eqution. M A M A N N N ( 1.5m) W + ( 3.6m) N ( 1.m) A B B 1 3.6 5W 17.16 kn W N B W + 1. g B 7.37kN m W 1.5 + 1. 3.6 g N rer F µ rer N 1 A k Nrer 7.37 kn ( 0.849)( 3.69 kn) N rer F rer 3.69 kn 3.13 kn N front F µ front 1 N V 1 k N front ( 17.16 kn) ( 0.849)( 8.58 kn) N front F front 8.58 kn 7.9 kn 16-11
Smple Problem 16. The thin plte of m 8 kg i held in plce hown. Neglecting the m of the link, determine immeditely fter the wire h been cut () the ccelertion of the plte, nd (b) the force in ech link. SOLUTION: Note tht fter the wire i cut, ll prticle of the plte move long prllel circulr pth of rdiu 150 mm. The plte i in curviliner trnltion. Drw the free-body-digrm eqution expreing the equivlence of the externl nd ective force. Reolve into clr component eqution prllel nd perpendiculr to the pth of the m center. Solve the component eqution nd the moment eqution for the unknown ccelertion nd link force. 16-1
Smple Problem 16. SOLUTION: Note tht fter the wire i cut, ll prticle of the plte move long prllel circulr pth of rdiu 150 mm. The plte i in curviliner trnltion. Drw the free-body-digrm eqution expreing the equivlence of the externl nd ective force. Reolve the digrm eqution into component prllel nd perpendiculr to the pth of the m center. F t F t W co 30 mg co 30 m ( 9.81m/ ) co 30 8.50 m 60 o 16-13
Smple Problem 16. Solve the component eqution nd the moment eqution for the unknown ccelertion nd link force. M ( ) G M G ( FAE in 30 )( 50 mm ) ( FAE co 30 )( 100 mm ) ( F in 30 )( 50 mm ) + ( F co 30 )( 100 mm ) 0 DF DF 38.4 F F DF AE + 11.6 F 0.1815 F AE DF 0 8.50 m 60 o F F F AE AE AE F n F n + F DF 0.619 0.1815 F W in 30 AE W 0 in 30 ( )( 8kg 9.81m ) 0 F AE 47.9 N T FDF 0.1815 ( 47.9 N) F DF 8.70 N C 16-14
Smple Problem 16.3 A pulley weighing 6 kg nd hving rdiu of gyrtion of 00 mm i connected to two block hown. Auming no xle friction, determine the ngulr ccelertion of the pulley nd the ccelertion of ech block. SOLUTION: Determine the direction of rottion by evluting the net moment on the pulley due to the two block. Relte the ccelertion of the block to the ngulr ccelertion of the pulley. Drw the free-body-digrm eqution expreing the equivlence of the externl nd ective force on the complete pulley plu block ytem. Solve the correponding moment eqution for the pulley ngulr ccelertion. 16-15
Smple Problem 16.3 SOLUTION: Determine the direction of rottion by evluting the net moment on the pulley due to the two block. M G ( 5kg)( 150mm) (.5kg)( 50mm) 10in lb rottion i counterclockwie. note: I mk k 6kg 3.ft W g 0.4 kg m 8 1 ft Relte the ccelertion of the block to the ngulr ccelertion of the pulley. A r α rbα A ( 0.5 m) α B (0.15 m) 16-16
Smple Problem 16.3 I A B 0.4 kg m ( 0.5 m) α ( 0.15 m)α Drw the free-body-digrm eqution expreing the equivlence of the externl nd ective force on the complete pulley nd block ytem. Solve the correponding moment eqution for the pulley ngulr ccelertion. M G ( M G ) Then, α A r A ( 49.1N)( 0.15 m) ( 4.5 N)( 0.5 m) Iα + mbb ( 0.15 m) ma A ( 0.5 m) ( 49.1)( 0.15) ( 4.5)( 0.5) ( 0.4) α + ( 5)( 0.15α)( 0.15) (.5)( 0.5)( 0.5) ( 0.5 m )(.374rd ) α A.374 rd 0.61m α B r B ( 0.15 m )(.374rd ) B 0.37 m 16-17
Smple Problem 16.4 SOLUTION: Drw the free-body-digrm eqution expreing the equivlence of the externl nd ective force on the dik. A cord i wrpped round homogeneou dik of m 15 kg. The cord i pulled upwrd with force T 180 N. Determine: () the ccelertion of the center of the dik, (b) the ngulr ccelertion of the dik, nd (c) the ccelertion of the cord. Solve the three correponding clr equilibrium eqution for the horizontl, verticl, nd ngulr ccelertion of the dik. Determine the ccelertion of the cord by evluting the tngentil ccelertion of the point A on the dik. 16-18
Smple Problem 16.4 SOLUTION: Drw the free-body-digrm eqution expreing the equivlence of the externl nd ective force on the dik. Solve the three clr equilibrium eqution. F x F x 0 m x x 0 F y ( F y ) T y W T m W m y M G M G Tr T α mr Iα 180 N - 1 ( mr ) ( 15 kg) α ( 180 N) ( 15 kg)( 0.5m) (9.81m 15 kg y α ).19 m 48.0 rd 16-19
Smple Problem 16.4 Determine the ccelertion of the cord by evluting the tngentil ccelertion of the point A on the dik. cord ( A) t + ( A G) t.19m + (0.5m)(48 rd ) cord 6. m x 0 y.19 m α 48.0 rd 16-0
Smple Problem 16.5 SOLUTION: Drw the free-body-digrm eqution expreing the equivlence of the externl nd ective force on the phere. A uniform phere of m m nd rdiu r i projected long rough horizontl urfce with liner velocity v 0. The coicient of kinetic friction between the phere nd the urfce i µ k. Determine: () the time t 1 t which the phere will trt rolling without liding, nd (b) the liner nd ngulr velocitie of the phere t time t 1. Solve the three correponding clr equilibrium eqution for the norml rection from the urfce nd the liner nd ngulr ccelertion of the phere. Apply the kinemtic reltion for uniformly ccelerted motion to determine the time t which the tngentil velocity of the phere t the urfce i zero, i.e., when the phere top liding. 16-1
Smple Problem 16.5 SOLUTION: Drw the free-body-digrm eqution expreing the equivlence of externl nd ective force on the phere. Solve the three clr equilibrium eqution. F y F y N W F x F x 0 N W F m µ mg µ g M G M G k k mg Fr Iα 5 µ g ( µ k mg ) r ( mr )α α k 3 r NOTE: A long the phere both rotte nd lide, it liner nd ngulr motion re uniformly ccelerted. 16 -
16-3 Smple Problem 16.5 g k µ r g µ k α 5 Apply the kinemtic reltion for uniformly ccelerted motion to determine the time t which the tngentil velocity of the phere t the urfce i zero, i.e., when the phere top liding. ( )t g v t v v k µ + + 0 0 t r g t k + + µ α ω ω 5 0 0 1 1 0 5 t r g r gt v k k µ µ g v t µ k 0 1 7 g v r g t r g k k k µ µ µ ω 0 1 1 7 5 5 r v 0 1 7 5 ω r v r r v 0 1 1 7 5 ω 0 7 5 1 v v At the intnt t 1 when the phere top liding, 1 ω 1 r v
Contrined Plne Motion Mot engineering ppliction involve rigid bodie which re moving under given contrint, e.g., crnk, connecting rod, nd non-lipping wheel. Contrined plne motion: motion with definite reltion between the component of ccelertion of the m center nd the ngulr ccelertion of the body. Solution of problem involving contrined plne motion begin with kinemtic nlyi. e.g., given θ, ω, nd α, find P, N A, nd N B. - kinemtic nlyi yield x nd y. - ppliction of d Alembert principle yield P, N A, nd N B. 16-4
Contrined Motion: Noncentroidl Rottion Noncentroidl rottion: motion of body i contrined to rotte bout fixed xi tht doe not p through it m center. Kinemtic reltion between the motion of the m center G nd the motion of the body bout G, t n rα rω The kinemtic reltion re ued to eliminte t nd n from eqution derived from d Alembert principle or from the method of dynmic equilibrium. 16-5
Contrined Plne Motion: Rolling Motion For blnced dik contrined to roll without liding, x rθ rα Rolling, no liding: F µ N rα Rolling, liding impending: F µ N rα Rotting nd liding: F µ N, rα independent k For the geometric center of n unblnced dik, O rα The ccelertion of the m center, G O + G O + + O ( ) ( ) G O t G O n 16-6
Smple Problem 16.6 m k m E E OB 4 kg 85 mm 3 kg The portion AOB of the mechnim i ctuted by ger D nd t the intnt hown h clockwie ngulr velocity of 8 rd/ nd counterclockwie ngulr ccelertion of 40 rd/. Determine: ) tngentil force exerted by ger D, nd b) component of the rection t hft O. SOLUTION: Drw the free-body-eqution for AOB, expreing the equivlence of the externl nd ective force. Evlute the externl force due to the weight of ger E nd rm OB nd the ective force ocited with the ngulr velocity nd ccelertion. Solve the three clr eqution derived from the free-body-eqution for the tngentil force t A nd the horizontl nd verticl component of rection t hft O. 16-7
Smple Problem 16.6 m m k E E OB 4 kg 85 mm 3 kg α 40 rd ω 8 rd/ SOLUTION: Drw the free-body-eqution for AOB. Evlute the externl force due to the weight of ger E nd rm OB nd the ective force. W W E OB I E m m I ( 4 kg) (9.81m ) 39. N ( 3kg) (9.81m ) 9.4 N E α m k α OB OB OB E 1.156 N m ( 4kg)( 0.085m) (40rd ( ) m ( rα) ( 3kg)( ) OB t OB 4.0 N ) 0.00m (40rd ( ) ( ) m rω ( 3kg)( 0.00 m)( 8rd ) OB n OB 38.4 N 1 1 ( m L ) ( 3kg)( 0.400m) α α 1 OB 1.600 N m 1 (40 rd ) ) 16-8
Smple Problem 16.6 Solve the three clr eqution derived from the freebody-eqution for the tngentil force t A nd the horizontl nd verticl component of rection t O. F M ( ) O M O ( 0.10 m) I Eα + mob ( OB ) t ( 0.00 m) + IOBα 1.156 N m + ( 4.0N)( 0.00 m) + 1.600 N m F 63.0 N F ( ) x F x W W E OB 39. N 9.4 N R x ( ) 4.0 N mob OB t R x 4.0 N I E α 1.156 N m ( ) 4.0 N m OB OB t ( ) 38.4 N m OB OB n I OB α 1.600N m R R y y F ( ) F y F y W E W OB OB ( ) OB 63.0 N 39. N 9.4 N 38.4 N m R y 4.0 N 16-9
Smple Problem 16.8 A phere of weight W i releed with no initil velocity nd roll without lipping on the incline. Determine: ) the minimum vlue of the coicient of friction, b) the velocity of G fter the phere h rolled 3 m nd c) the velocity of G if the phere were to move 3 m down frictionle incline. SOLUTION: Drw the free-body-eqution for the phere, expreing the equivlence of the externl nd ective force. With the liner nd ngulr ccelertion relted, olve the three clr eqution derived from the free-body-eqution for the ngulr ccelertion nd the norml nd tngentil rection t C. Clculte the friction coicient required for the indicted tngentil rection t C. Clculte the velocity fter 3 m of uniformly ccelerted motion. Auming no friction, clculte the liner ccelertion down the incline nd the correponding velocity fter 3 m. 16-30
Smple Problem 16.8 rα SOLUTION: Drw the free-body-eqution for the phere, expreing the equivlence of the externl nd ective force. With the liner nd ngulr ccelertion relted, olve the three clr eqution derived from the free-body-eqution for the ngulr ccelertion nd the norml nd tngentil rection t C. M C M C ( W in θ) r ( m ) ( mrα) r + ( mr ) W g ( ) 7 r + Iα rα r 5g in 30 rα 7 5 9.81m 5 + in 30 5 W g α r α 3.5m 5g inθ α 7r 16-31
Smple Problem 16.8 5g inθ α 7r rα 3.5m Solve the three clr eqution derived from the free-bodyeqution for the ngulr ccelertion nd the norml nd tngentil rection t C. F x ( F x ) mg inθ F m F ( ) y F y N mg N mg g F 7 coθ 0 mg co30 5g inθ 7 mg in 30 0.866mg 0.143mg Clculte the friction coicient required for the indicted tngentil rection t C. F µ µ N F N 0.143 0.866 mg mg µ 0.165 16-3
Smple Problem 16.8 Clculte the velocity fter 3 m of uniformly ccelerted motion. v 0 0 + 3.5m ( x ) v + x 0 ( ) ( 3m) v 4.6m 5g inθ α 7r rα 3.5m Auming no friction, clculte the liner ccelertion nd the correponding velocity fter 3 m. M 0 Iα α 0 G M G ( ) F x F x v 0 ( x ) v + x mg in 30 m ( ) ( 3m) 0 + 4.91m 0 ( 9.81m ) in 30 4.91m v 5.43m 16-33
Smple Problem 16.9 SOLUTION: Drw the free-body-eqution for the wheel, expreing the equivlence of the externl nd ective force. A cord i wrpped round the inner hub of wheel nd pulled horizontlly with force of 00 N. The wheel h m of 50 kg nd rdiu of gyrtion of 70 mm. Knowing µ 0.0 nd µ k 0.15, determine the ccelertion of G nd the ngulr ccelertion of the wheel. Auming rolling without lipping nd therefore, relted liner nd ngulr ccelertion, olve the clr eqution for the ccelertion nd the norml nd tngentil rection t the ground. Compre the required tngentil rection to the mximum poible friction force. If lipping occur, clculte the kinetic friction force nd then olve the clr eqution for the liner nd ngulr ccelertion. 16-34
Smple Problem 16.9 I mk 0.45kg m ( 50kg)( 0.70m) Aume rolling without lipping, rα ( 0.100m)α SOLUTION: Drw the free-body-eqution for the wheel,. Auming rolling without lipping, olve the clr eqution for the ccelertion nd ground rection. M C ( M C ) ( 00 N)( 0.040 m) m ( 0.100 m) 8.0 N m ( 50 kg)( 0.100 m) α + α + 10.74 rd (0.45 kg m ( 0.100 m) (10.74 rd ) 1.074 m F x F x F F F x F x N W 0 N + 00 N mg 146.3 N m + (50 kg) (1.074 m Iα ( 50kg) (1.074 m ) + 490.5 N ) ) α 16-35
Smple Problem 16.9 Compre the required tngentil rection to the mximum poible friction force. ( 490.5 N) 98.1N Fmx µ N 0.0 F > F mx, rolling without lipping i impoible. Without lipping, F 146.3 N N 490.5 N Clculte the friction force with lipping nd olve the clr eqution for liner nd ngulr ccelertion. F F N k µ k ( 490.5 N) 73.6 N 0.15 F x F x 00 N 73.6 N ( 50 kg).53 m M G M G ( 73.6 N)( 0.100 m) ( 00 N)( 0.0.060 m) α 18.94 rd (0.45 kg m ) α α 18.94 rd 16-36
Smple Problem 16.10 SOLUTION: Bed on the kinemtic of the contrined motion, expre the ccelertion of A, B, nd G in term of the ngulr ccelertion. The extremitie of 1. m rod weighing 5 kg cn move freely nd with no friction long two tright trck. The rod i releed with no velocity from the poition hown. Determine: ) the ngulr ccelertion of the rod, nd b) the rection t A nd B. Drw the free-body-eqution for the rod, expreing the equivlence of the externl nd ective force. Solve the three correponding clr eqution for the ngulr ccelertion nd the rection t A nd B. 16-37
Smple Problem 16.10 SOLUTION: Bed on the kinemtic of the contrined motion, expre the ccelertion of A, B, nd G in term of the ngulr ccelertion. Expre the ccelertion of B + B A B A With B A 4α, the correponding vector tringle nd the lw of ign yield A 5.46α 4. 90α B The ccelertion of G i now obtined from + where α G A G A G Reolving into x nd y component, x y 5.46α α in 60 α co 60 1.73α A 4.46α 16-38
Smple Problem 16.10 Drw the free-body-eqution for the rod, expreing the equivlence of the externl nd ective force. Iα m m I x y 1 1 3 kg.m 3α ml 5 1 5 1 1 1 ( 1.34α ) 5 kg 1 33.5α ( 1. m) ( 0.50α ) 13.0α Solve the three correponding clr eqution for the ngulr ccelertion nd the rection t A nd B. R R B B M E M E ( 5)( 0.50) ( 33.5α )( 1.34) + ( 13.0α )( 0.50) α +.33 rd F x F x in 45 R A 110 N F y F y + ( 33.5)(.33) 78.1 α R B.33 rd 110 ( 110 N) co 45 45 ( 13.0)(.33) R A 137 N N 16-39 + 3α 45 o