Integration of the equation of motion with respect to time rather than displacement leads to the equations of impulse and momentum.

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Dwain Lindsey
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2 Inegraion of he equaion of moion wih respec o ime raher han displacemen leads o he equaions of impulse and momenum. These equaions greal faciliae he soluion of man problems in which he applied forces ac during eremel shor periods of ime, as in impac problems, or oer specified inerals of ime.

3 Le s consider he general curilinear moion in space of a paricle of mass m, where he paricle is locaed b is posiion ecor r measured from a fied origin O. The eloci of he paricle is r and is angen o is pah. The resulan force of all forces on m is in he direcion of is a acceleraion. F

4 We ma wrie he basic equaion of moion for he paricle, as F ma m or F m d d d d m G G Where he produc of he mass and eloci is defined as he linear momenum G m of he paricle. This equaion saes ha he resulan of all forces acing on a paricle equals is ime rae of change of linear momenum.

5 In SI, he uni of linear momenum also equals N. s. m is kg. m/s, which Since he equaion of impulse and momenum is a ecor equaion, in addiion o he equali of he magniudes of and G, he direcion of he resulan force coincides wih he direcion of he rae of change in linear momenum, which is he direcion of he rae of change of eloci. Linear impulse momenum equaion is one of he mos useful and imporan relaionships in dnamics, and i is alid as long as mass m of he paricle is no changing wih ime. F

6 We now wrie he hree scalar componens of linear momenum equaion as F G F G Fz G z These equaions ma be applied independenl of one anoher.

7 The Linear Impulse-Momenum Principle All ha we hae done so far is o rewrie Newon s second law in an alernaie form in erms of momenum. Bu we ma describe he effec of he resulan force on he linear momenum of he paricle oer a finie period of ime simpl b inegraing he linear momenum equaion wih respec o ime. Mulipling he equaion b d gies Fd,,which we inegrae from ime o ime o obain dg Fd G dg G G G G F

8 ere he linear momenum a ime is G =m and he linear momenum a ime is G =m. The produc of force and ime is defined as he linear impulse of he force, and his equaion saes ha he oal linear impulse on m equals he corresponding change in linear momenum of m. Alernaiel, we ma wrie G Fd G I

9 which sas ha he iniial linear momenum of he bod plus he linear impulse applied o i equals is final linear momenum. m G m + = F d G m

10 The impulse inegral is a ecor which, in general, ma inole changes in boh magniude and direcion during he ime ineral. Under hese condiions, i will be necessar o epress F and in componen form and hen combine he G inegraed componens. The componens become he scalar equaions, which are independen of one anoher. Fd Fd m m G G G m m G G G Fzd mz mz Gz Gz Gz

11 In some cases, cerain forces are er large and of shor duraion. Such forces are called impulsie forces. An eample is a force of sharp impac. We frequenl assume ha impulsie forces are consan oer heir ime of duraion, so ha he can be brough ouside of he linear impulse momenum inegral. In addiion, we frequenl assume ha nonimpulsie forces can be negleced in comparison wih impulsie forces. An eample of a nonimpulsie force is he weigh of a baseball during is collision wih a ba he weigh of he ball, abou.45 N, is small compared wih he force eered on he ball b he ba, which is abou seeral housand Newons in magniude.

12 There are cases where a force acing on a paricle changes wih he ime in a manner deermined b eperimenal measuremens or b oher approimae means. In his case, a graphical or numerical inegraion mus be performed. If, for eample, a force F acing on a paricle in a gien direcion changes wih he ime as indicaed in he figure, he impulse, Fd of his force from o is he shaded area under he cure.

13 Conseraion of Linear Momenum If he resulan force on a paricle is zero during an ineral of ime, is linear momenum G remains consan. In his case, he linear momenum of he paricle is said o be consered. Linear momenum ma be consered in one direcion, such as, bu no necessaril in he - or z- direcions. G 0 G G m m This equaion epresses he principle of conseraion of linear momenum.

14 PROBLEMS. The 00-kg lunar lander is descending ono he moon s surface wih a eloci of 6 m/s when is rero-engine is fired. If he engine produces a hrus T for 4 s which aries wih he ime as shown and hen cus off, calculae he eloci of he lander when = 5 s, assuming ha i has no e landed. Graiaional acceleraion a he moon s surface is.6 m/s.

15 SOLUTION s m mg m m Fd s m g s s m kg m / (800) (800) (5)?, /.6, 5, / 6, 00 mg moion T

16 PROBLEMS. The 9-kg block is moing o he righ wih a eloci of 0.6 m/s on a horizonal surface when a force P is applied o i a ime = 0. Calculae he eloci of he block when = 0.4 s. The kineic coefficien of fricion is m k =0.3.

17 SOLUTION s m d d d m m Fd direcion in N N mg N F / (0.4) 36(0.) 7(0.) 0.6) 9( 0.3(88.3) (9.8) moion P W=mg N F f =m k N

18 PROBLEMS 3. A ennis plaer srikes he ennis ball wih her racke while he ball is sill rising. The ball speed before impac wih he racke is =5 m/s and afer impac is speed is = m/s, wih direcions as shown in he figure. If he 60-g ball is in conac wih he racke for 0.05 s, deermine he magniude of he aerage force R eered b he racke on he ball. Find he angle b made b R wih he horizonal. Commen on he reamen of he ball weigh during impac.

19 SOLUTION in direcion F 0 d R 0.05R 0.05R m m F 0 d R R in direcion R cos cos0 m m 0.06(9.8) 0.35 N b R R R R an b sin sin0 N N R R b 8.68 R R 0 0 R W=mg

PROBLEMS 4. The 40-kg bo has aken a running jump from he upper surface and lands on his 5-kg skaeboard wih a eloci of 5 m/s in he plane of he figure as shown.

20 PROBLEMS 4. The 40-kg bo has aken a running jump from he upper surface and lands on his 5-kg skaeboard wih a eloci of 5 m/s in he plane of he figure as shown. If his impac wih he skaeboard has a ime duraion of 0.05 s, deermine he final speed along he horizonal surface and he oal normal force N eered b he surface on he skaeboard wheels during he impac.

21 SOLUTION (m B +m S )g N m B Linear momenum is consered in -direcion; B mss mb ms 5 cos m / s 40 m B B 40 m S S N m m 5sin 30 0 N B N 440 N or N.44 kn S g d 0 0

22 In addiion o he equaions of linear impulse and linear momenum, here eiss a parallel se of equaions for angular impulse and angular momenum. Firs, we define he erm angular momenum. The figure shows a paricle P of mass m moing along a cure in space. The paricle is locaed b is posiion ecor r respec o a conenien origin O of fied coordinaes --z. wih

23 The eloci of he paricle is momenum is momenum ecor G m m he angular momenum, and is linear. The momen of he linear abou he origin O is defined as O of P abou O and is gien b he cross-produc relaion for he momen of a ecor. r o r m r G

24 The angular momenum hen is a ecor perpendicular o he plane A defined b r and. The sense of O is clearl defined b he righ-hand rule for cross producs.

25 The scalar componens of angular momenum ma be obained from he epansion z o z z o z k j i m or k m j z m i z m m r so ha z m z o z m z o m oz

26 Each of hese epressions for angular momenum ma be checked from he figure, which shows he hree linear momenum componens, b aking he momens of hese componens abou he respecie aes. In SI unis, angular momenum has he unis kg. m /s =N. m. s.

27 If F represens he resulan of all forces acing on he paricle M o P, he momen abou he origin O is he ecor cross produc M o r F r m We now differeniae o differeniaion of a cross produc and obain The erm o m d d r m wih ime, using he rule for he r m r m r m r mr 0 ma M o is zero since he cross produc of parallel ecors is zero.

28 Subsiuion ino he epression for momen abou O gies M o o This equaion saes ha he momen abou he fied poin O of all forces acing on m equals he ime rae of change of angular momenum of m abou O. This relaion, paricularl when eended o a ssem of paricles, rigid or nonrigid, proides one of he mos powerful ools of analsis in dnamics. The scalar componens of his equaion are M o o M o o M oz oz

29 The Angular Impulse-Momenum Principle To obain he effec of he momen on he angular momenum of he paricle oer a finie period of ime, we M o o inegrae from ime o. or M od d o o o o oal angularimpulse o o M od change in angular momenum r m r m o

30 The produc of momen and ime is defined as angular impulse and his equaion saes ha he oal angular impulse on m abou he fied poin O equals he corresponding change in angular momenum of m abou O. Alernaiel, we ma wrie M d o o o

31 Plane-Moion Applicaions Mos of he applicaions can be analzed as plane-moion problems where momens are aken abou a single ais normal o he plane moion. In his case, he angular momenum ma change magniude and sense, bu he direcion of he ecor remains unalered. M o d o o Fr sin d m d m d

32 Conseraion of Angular Momenum If he resulan momen abou a fied poin O of all forces acing on a paricle is zero during an ineral of ime, is angular momenum O remains consan. In his case, he angular momenum of he paricle is said o be consered. Angular momenum ma be consered abou one ais bu no abou anoher ais. o 0 O O This equaion epresses he principle of conseraion of angular momenum.

33 PROBLEM. The small sphere of mass m raeling wih speed srikes and becomes aached o he end of he saionar assembl ha pios freel abou a erical ais a O. Deermine he angular eloci w of he assembl afer impac and calculae he change E in he ssem energ. Neglec he mass of he rod compared wih m.

34 SOLUTION m E m m m m L ml m L m T T V V T E L L Lm Lm L Lm Lm k Lm i m Lj m r mass k Lm i m Lj m r mass Lmk i m i Lj m r d M g e O O O O O O O O O O L L z

35 PROBLEM. A pendulum consiss of wo 3. kg concenraed masses posiioned as shown on a ligh bu rigid bar. The pendulum is swinging hrough he erical posiion wih a clockwise angular eloci w = 6 rad/s when a 50-g bulle raeling wih eloci =300 m/s in he direcion shown srikes he lower mass and becomes embedded in i. Calculae he angular eloci w which he pendulum has immediael afer impac and find he maimum deflecion of he pendulum.

36 SOLUTION Angular momenum is consered during impac; M Od O 0, 0 O O O M O 0 r m r m r 0. j 3.(0.)(6) i 0.4 j 3.(0.4)(6) i 0.4 j 0.05(300 cos 0i 300sin 0 j) 0. cosj 0.sin i 3.( 0. cosi 0. sin j) 0.4 cosj 0.4sin i 3.5(0.4 cosi 0.4 sin j) w O w k 3.07 k k 0.8 cos.798k 0.5 cos k 0.5 sin k or in scalar form rad / s ( ccw) k 0.8 sin sin cos 0.5 sin cos k 0.64

37 SOLUTION Energ consideraions afer impac; T T T U T 0 T V g V e 0 T (maimum deflecion (Daum a O) ) O w.558 J w ΔV g 3.(9.8)(0. cos 0.) 3.5(9.8)(0.4 cos 0.4) 6.47( cos ) ( cos ) 0,cos o

We may write the basic equation of motion for the particle, as

We may write the basic equation of motion for the particle, as We ma wrie he basic equaion of moion for he paricle, as or F m dg F F linear impulse G dg G G G G change in linear F momenum dg The produc of force and ime is defined as he linear impulse of he force,