Ch. 9 Center of Mass Momentum. Question 6 Problems: 3, 19, 21, 27, 31, 35, 39, 49, 51, 55, 63, 69, 71, 77

Transcription

1 Ch. 9 Center of Mass Moentu Queston 6 Probles: 3, 9,, 7, 3, 35, 39, 49, 5, 55, 63, 69, 7, 77

2 Center of Mass Use center of ass when no longer dealng wth a pont partcle. The center of ass of a syste of partcles s the pont that oes as though: ) all of the syste s ass were concentrated there and ) all the external forces were appled there. Bascally, we treat a coplex object as a pont partcle that s located at the objects center of ass.

3 exaples: Throw a ball at an angle, t follows a parabolc trajectory. Flp an object such as a haer, baseball bat, or tenns racket. Dfferent ponts on the object wll follow dfferent paths. The center of ass of the object wll follow a parabolc path. Sple way to fnd center of ass. Balance an object on a sharp edge. The center of ass s aboe the balance pont.

4 Another way to fnd the center of ass. Let the object hang freely. Draw a plub lne and trace t on the object. Hang the object freely fro another pont and repeat. The pont where the plub lnes ntersect s the center of ass. Note: the center of ass does not hae to be where there s any ass. Exaples: doughnut, horseshoe, an epty water bottle

5 How to calculate the center of ass The center of ass s the su of the products of nddual segents of ass and ther respecte locatons, dded by the total ass. Assue we hae a syste of two dscrete partcles n -D. s at x s at x x x x M x co x You can see that f the two partcles hae the sae ass, the CoM s halfway n between the partcles.

6 Qute often there are ore than partcles. If partcles n the syste are n 3-D. M x x x x x x x n n n co co n co x M x (These are scalar equatons.) n co y M y n co z M z

7 Usng poston ector: r r co x ˆ x co y ˆ ˆj y z kˆ co ˆj z co kˆ We can replace the scalar equaton wth a ector equaton. r co M n r

8 Sold bodes (contnuous ass dstrbutons) When the ass s dstrbuted contnuously, we replace the suatons wth ntegrals. Break up object nto dfferental ass eleents each wth ass d. x co M xd y co M yd z co M zd

9 If the densty s not constant we can stll fnd the center of ass. If we know how t s dstrbuted, (x) we can use the ntegraton ethod to fnd the CoM. M M d = dv r co rd ( r) rdv

10 For the sake of splcty we wll only work wth objects of unfor densty. d dv M V Substtutng d = (M/V)d, we get: x co V xdv y co V ydv z co V zdv

11 Newton s nd Law for a syste of partcles. If we hae ultple nteractng partcles, we want to study the center of ass. Slde one hockey puck on a frctonless surface nto an dentcal puck. Before they collde the CoM s ong wth an ntal elocty. We wll see that after they collde the CoM wll antan ts elocty.

12 F M net a co ) F net the net external force actng on the syste. In the hockey puck exaple, snce the pucks were the syste, the forces they exert on each other are nternal. ) M s the total ass 3) a co s the acceleraton of the center of ass. If we want we can break ths equaton nto ts coponents. F net,x = M a co,x F net,y = M a co,y F net,z = M a co,z

13 Applcatons of studyng CoM ) Ballet dancer as dancer jups, rases ars and legs to rase the center of ass. The center of ass follows the parabolc trajectory. The oeent of the CoM decreases the heght that s attaned by the head and torso, producng the lluson that the dancer floats. See pctures on page 07. ) Hgh juper As the hgh juper goes oer the bar the body s arched. The center of ass doesn t hae to go oer the bar.

14 Lnear Moentu Lnear oentu s dfferent than angular oentu whch s n a later chapter. In ths chapter, wheneer oentu s entoned, t s lnear. Moentu s the product of the ass and elocty p = Moentu s a ector quantty (drecton s portant). Moentu s n the sae drecton as the object s elocty. The unts of oentu are: kg /s

15 Exaples A 000 kg truck ong wth elocty 30 /s p = (000 kg)(30 /s) = 3x0 4 kg /s A 3000 kg truck ong wth elocty 30 /s p = (3000 kg)(30 /s) = 9x0 4 kg /s A 0.4 kg baseball wth elocty 40 /s p = (0.4 kg)(40 /s) = 5.6 kg /s A 6 kg bowlng ball wth elocty 5 /s p = (6 kg)(4 /s) = 4 kg /s

16 Moentu F net = a and: a = / t F net = / t Use oentu p = If the ass s constant: So: F net p dp dt dt dp dt d dt The lnear oentu of a syste of partcles can only change f there s a net external force.

17 F net p dp dt dt Collson and pulse If force s te dependant: d p F( t) dt t t f p dp f t p t f F( t) dt p J t t f F( t) dt J s referred to as the pulse. (change n oentu)

18 Force s. te graph The are under a Force s. te graph tells us the pulse. F J t t f t If we don t know how the force ares but we know the aerage force we can fnd the pulse. J = F ae t F t J t f t

19 Newton s 3 rd Law Snce forces occur n equal and opposte pars so do pulses. When a bat hts a ball. The bat exerts an pulse on the ball. The ball exerts an pulse wth the sae agntude but opposte drecton on the bat.

20 Moentu s a ector and can be splt nto coponents. For an solated syste, f a coponent of the oentu s zero, then the coponent of p n that drecton wll also be zero. If a syste s not solated, p would be nonzero. The key s that deternng what akes up the syste s portant.

21 Conseraton of Moentu When there s no net external force actng on a syste, the total oentu of the syste reans constant wth te. When a collson occurs n an solated syste, the nddual oentus of objects y change, but the total oentu ector of the whole syste s a constant. For a syste wth two partcles: + = f + f

22 Archer exaple An archer stands on frctonless ce and shoots a 0.5kg arrow horzontally at 50 /s. The cobned ass of the archer and the bow s 60 kg. Wth what elocty wll the archer oe after shootng the arrow? Snce the syste s the archer and the arrows, t s solated. The forces that the bow and arrow exert on each other are nternal forces. Use conseraton of oentu + = f + f

23 Archer Intally the total oentu s zero. The ntal eloctes of the archer and the arrow are both zero. So after the arrow s shot, the total oentu ust stll be zero.

24 Archer (labeled the archer as and the arrow as ) + = f + f (60kg)0/s + (0.5kg)0/s = (60kg) f + (0.5kg)(50/s) 0 = (60kg) f + (0.5kg)(50/s) (60kg) f = - (0.5kg)(50/s) f = /s If the arrow s shot to the rght, the archer oes to the left.

25 Collsons When objects collde, oentu s always consered. Knetc energy s not always consered. Soe of the knetc energy s turned nto sound, work needed to defor the object, heat, nternal energy.

26 Collsons We wll be lookng at two types of collsons. Perfectly elastc collsons oentu and knetc energy s consered the objects leae each other wth the sae speed as they ht each other there s no deforaton of the objects durng the collson exaples: Pool balls (approxately) Proton-proton collson

27 Collsons Perfectly Inelastc collsons oentu s consered, but knetc energy s lost the objects stck together after the collson. (they oe wth the sae elocty) exaples: Two cars httng each other and the bupers lock up Football player beng tackled by another player Person catchng a ball Shootng a ball nto a ballstc pendulu

28 Collsons Nearly all collsons n the real world fall soewhere n between beng perfectly elastc and perfectly nelastc. When you ht a baseball the ball doesn t stck to the bat, so the collson sn t perfectly nelastc. Soe of the knetc energy s conerted to sound and s used to oentarly defor the ball and bat, so the collson sn t perfectly elastc ether.

29 Perfectly Inelastc Collsons Snce the two objects hae the sae elocty after the collson, f = f = f we can rewrte the conseraton of energy as: + = ( + ) f Solng for the fnal elocty ges us: f Reeber to treat and as ectors. Ther drecton atters.

30 Inelastc Collson Shoot a 0.05 kg bullet at 300 /s nto a 5kg wooden block. The bullet gets stuck n the block, so they oe together afterwards. Fnd the fnal elocty of the bullet/block. Cons. of Mo. says: + = ( + ) f Let the bullet be ass and the block be ass. f (0.05kg)(300 / s) 0.05kg (5kg)(0 5kg / s).97 / s

31 Perfectly Elastc Collson Moentu s consered + = f + f Knetc Energy s consered ½ + ½ = ½ f + ½ f Cobnng these two equatons, f I gae you the ntal eloctes of the objects you could sole for the two fnal eloctes sultaneously.

32 Elastc Collsons Do a bunch of algebra and you wll get: and ) ( f ) ( f

33 Specal cases for elastc collsons If the target ( ) s statonary, = 0 ) ( f ) ( f ) ( f f

34 f ( ) If the two asses are equal we get: V f = and f = f ( ) If the target s statonary and the two asses are equal we get: V f = 0 and f = If the target s statonary and the target s asse ( >> ) we get: and f f If the target s statonary and the projectle s asse ( >> ) we get: and f f

35 Soe sual exaples of elastc collsons Do saple proble 9-

36 Collsons n ore than -Denson Reeber that oentu s a ector. We can expand our work nto densons. All you need to realze s that the oentu n the x-drecton s consered. In addton, the oentu n the y-drecton s consered. The next exaple s a projectle httng a target wth an elastc, glancng collson (not head on) wth a target.

37 y f X-drecton: = f cos + f cos Y-drecton: 0 = - f sn + f sn In addton we wrte fro conseraton of energy: There are 7 arables:,,, f, f,,. If we know 4 of the we can fnd the other 3. f f f x

38 Ballstc Pendulu A 0.05 bullet wth elocty 50 /s s shot nto a 3 kg ballstc pendulu. Fnd how hgh the pendulu rses after the bullet gets stuck nsde.

39 Ballstc Pendulu Frst use conseraton of oentu to fnd the fnal elocty of the bullet/pendulu. Ths s an nelastc collson. + = ( + ) f (0.05kg)(50/s)+(3kg)(0/s) = (3.05kg) f f =.5 /s

40 Ballstc Pendulu Now that we know the ballstc pendulu wth the bullet n t begns to swng wth a speed of.5 /s, we use conseraton of energy to fnd how hgh t swngs. ½ = g y ½ (.5/s) = g y y = 0.3

41 Systes wth aryng ass Exaple where the ass ares s rocket oton. As the rocket burns fuel, ts ass s decreased. Gases are expelled at a hgh elocty out the botto of the rocket. Snce the rocket and the expelled gas exert equal and opposte forces on each other (nternal forces), the total oentu s consered. To consere the oentu, the rocket goes n the drecton opposte to the relate elocty of the gas.

42 ) Frst rocket equaton: R rel = Ma R nstantaneous rate of fuel consupton V rel relate speed of gas product to rocket M nstantaneous ass a nstantaneous acceleraton Left hand sde of equaton has unts of force R rel s the thrust (T) of the engne.

43 The rocket acceleratng so ts elocty s non-constant. By equatng the oentu carred away by the gas to the oentu of the rocket we get: dm dt rel M d dt d f f d rel rel dm M rel M M f ln dm M M M f nd Rocket equaton An deal rocket would reach the destnaton, wth only the payload and ths would be the ost effcent case. (Maxze M /M f )

44 Probles: 4, 0, 36, 4, 54, 78