12.1 Introduction 12-1

Transcription

1 12.1 Intoduction So fa we have esticted ouselves to consideing systems consisting of discete objects o point-like objects that have fixed amounts of mass. We shall now conside systems in which mateial flows between the objects in the system, fo example we shall conside coal falling fom a hoppe into a moving aiload ca, sand leaking fom aiload ca fuel, gain moving fowad into a aiload ca, and fuel ejected fom the back of a ocket, In each of these examples mateial is continuously flows into o out of an object. We have aleady shown that the total extenal foce causes the momentum of a system to change, 12-1

2 d p total system F =. (12.2.1) ext dt We shall analyze how the momentum of the constituent elements ou system change ove a time inteval [t,t + Δt], and then conside the limit as Δt 0. We can then explicit calculate the deivative on the ight hand side of Eq. (12.2.1) and Eq. (12.2.1) becomes dp Δ p p (t + Δt) p system system system system F total = = lim = lim. (12.2.2) ext dt Δt 0 Δt Δt 0 Δt We need to be vey caeful how we apply this genealized vesion of Newton s Second Law to systems in which mass flows between constituent objects. In paticula, when we isolate elements as pat of ou system we must be caeful to identify the mass Δm of the mateial that continuous flows in o out of an object that is pat of ou system duing the time inteval Δt unde consideation. We shall conside fou categoies of mass flow poblems that ae chaacteized by the momentum tansfe of the mateial of mass Δm Tansfe of Mateial into an Object, but no Tansfe of Momentum Conside fo example ain falling vetically downwad with speed u into ca of mass m moving fowad with speed v. small amount of falling ain Δm has no component of momentum in the diection of motion of the ca. Thee is a tansfe of ain into the ca but no tansfe of momentum in the diection of motion of the ca (Figue 12.1). fictionless u ain v fictionless m u v Figue 12.1 Tansfe of ain mass into the ca but no tansfe of momentum in diection of motion Tansfe of Mateial Out of an Object, but no Tansfe of Momentum The mateial continually leaves the object but it does not tanspot any momentum away fom the object in the diection of motion of the object (Figue 12.2). Conside an ice skate gliding on ice at speed v holding a bag of sand that is leaking staight down with espect to the moving skate. The sand continually leaves the bag but it does not tanspot any momentum away fom the bag in the diection of motion of the object. In Figue 12.2, sand of mass Δm s leaves the bag. 12-2

3 Figue 12.2 Tansfe of mass out of object but no tansfe of momentum in diection of motion Tansfe of Mateial Impulses Object Via Tansfe of Momentum Suppose a fie hose is used to put out a fie on a boat of mass m b. ssume the column of wate moves hoizontally with speed u. The incoming wate continually hits the boat popelling it fowad. Duing the time inteval Δt, a column of wate of mass Δm s will hit the boat that is moving fowad with speed v inceasing it s speed (Figue 12.3). Figue 12.3 Tansfe of mass of wate inceases speed of boat Mateial Continually Ejected Fom Object esults in Recoil of Object When fuel of mass Δm f is ejected fom the back of a ocket with speed u elative to the ocket, the ocket of mass m ecoils fowad. Figue 12.4a shows the ecoil of the ocket in the efeence fame of the ocket. The ocket ecoils fowad with speed Δv. In a efeence fame in which the ocket is moving fowad with speed v, then the speed afte ecoil is v + Δv. The speed of the backwadly ejected fuel is u v (Figue 12.4b). (a) (b) 12-3

4 Figue 12.4 Tansfe of mass out of ocket povides impulse on ocket in (a) efeence fame of ocket, (b) efeence fame in which ocket moves with speed v We must caefully identify the momentum of the object and the mateial tansfeed at time t in ode to detemine p. We must also identify the momentum of the object system and the mateial tansfeed at time t + Δt in ode to detemine p (t + Δt) as well. system Recall that when we defined the momentum of a system, we assumed that the mass of the system emain constant. Theefoe we cannot ignoe the momentum of the tansfeed mateial at time t + Δt even though it may have left the object; it is still pat of ou system (o at time t even though it has not flowed into the object yet) Woked Examples Example 12.1 Filling a Coal Ca n empty coal ca of mass m 0 stats fom est unde an applied foce of magnitude F. t the same time coal begins to un into the ca at a steady ate b fom a coal hoppe at est along the tack (Figue 12.5). Find the speed when a mass m c of coal has been tansfeed. Figue 12.5 Filling a coal ca Solution: We shall analyze the momentum changes in the hoizontal diection, which we call the x -diection. Because the falling coal does not have any hoizontal velocity, the falling coal is not tansfeing any momentum in the x -diection to the coal ca. So we shall take as ou system the empty coal ca and a mass m c of coal that has been tansfeed. Ou initial state at t = 0 is when the coal ca is empty and at est befoe any coal has been tansfeed. The x -component of the momentum of this initial state is zeo, p x (0) = 0. (12.3.1) 12-4

5 Ou final state at t = t is when all the coal of mass m = bt has been tansfeed into the f c f ca that is now moving at speed v f. The x -component of the momentum of this final state is p ) = (m 0 + m + bt f. (12.3.2) x (t f c )v f = (m 0 )v f Thee is an extenal constant foce F x = F applied though the tansfe. The momentum pinciple applied to the x -diection is t f F dt = Δp = p (t f ) p (0). (12.3.3) x x x x 0 Because the foce is constant, the integal is simple and the momentum pinciple becomes So the final speed is Ft f = (m 0 + bt f )v f. (12.3.4) Example 12.2 Emptying a Feight Ca Ft f v f =. (12.3.5) (m 0 + bt f ) feight ca of mass m c contains sand of mass m s. t t = 0 a constant hoizontal foce of magnitude F is applied in the diection of olling and at the same time a pot in the bottom is opened to let the sand flow out at the constant ate b = dm s / dt. Find the speed of the feight ca when all the sand is gone (Figue 12.6). ssume that the feight ca is at est at t = 0. Figue 12.6 Emptying a feight ca Solution: Choose the positive x -diection to point in the diection that the ca is moving. Choose fo the system the amount of sand in the fight ca at time t, m c. t time t, 12-5

6 the ca is moving with velocity v = v î. The momentum diagam fo the system at c c time t is shown in the diagam on the left in Figue v c v c + v c m c m c + m c v c + v c time t m s time t + t Figue 12.7 Momentum diagam at time t and at time t + Δt The momentum of the system at time t is given by p sys = m c v c. (12.3.6) Duing the time inteval [ t, t + Δ t], an amount of sand of mass Δm s leaves the feight ca and the mass of the feight ca changes by m (t + Δt) = m + Δm, whee Δm = Δm. c c c c s t the end of the inteval the ca is + Δ v moving with velocity v (t + Δt) = v = (v c + Δv )î. The momentum diagam fo the system at time c c c c t + Δt is shown in the diagam on the ight in Figue The momentum of the system at time t + Δt is given by p sys (t + Δt) = (Δm + m + Δm )( v + Δ v ) = m ( v + Δ v ).(12.3.7) s c c c c c c c Note that the sand that leaves the ca is shown with velocity v c + Δv. This implies c that all the sand leaves the ca with the velocity of the ca at the end of the inteval. This is an appoximation. Because the sand leaves continuous, the velocity will vay fom v to v + Δv but so does the change in mass of the ca and these two c c c contibutions to the system s moment exactly cancel. The change in momentum of the system is then Δ p = p (t + Δt) p = m c ( v + Δ v ) m c v = m c Δv.(12.3.8) sys sys sys c c c c Thoughout the inteval a constant foce F = Fî momentum pinciple becomes is applied to the system so the 12-6

7 F p (t + Δt) p Δ v sys sys c = lim = lim m c = m Δt 0 Δt Δt 0 Δt c d v c. dt (12.3.9) Because the motion is one-dimensional, Eq. (12.3.9) witten in tems of x -components becomes dv c F = mc. ( ) dt Denote by initial mass of the ca by m c,0 = m c + m s whee m c is the mass of the ca and m s is the mass of the sand in the ca at t = 0. The mass of the sand that has left the ca at time t is given by t t dm s m = dt = bdt = bt. ( ) Thus Theefoe Eq. ( ) becomes s 0 dt 0 m = m bt = m + m bt. ( ) c c,0 c s dv F = (m + m bt) c c s. ( ) dt This equation can be solved fo the x -component of the velocity at time t, v c, (which in this case is the speed) by the method of sepaation of vaiables. Rewite Eq. ( ) as Fdt dv =. ( ) c (m c + m s bt) Then integate both sides of Eq. ( ) with the limits as shown v c=v c (t ) t =t Fdt d v c =. ( ) v =0 t =0 m + m bt c c s Integation yields the speed of the ca as a function of time F t = t F m + m bt F m + m c s c s v c = ln(m + m s bt ) = ln ln. ( ) c b t = 0 b m + m = b m + m bt c s c s 12-7

8 In witing Eq. ( ), we used the popety that ln(a) ln(b) = ln(a / b) and theefoe ln(a / b) = ln(b / a). Note that m + m m + m bt, c s c s so the tem m + m c s ln 0, and the speed of the ca inceases as we expect. m + m bt c s Example 12.3 Filling a Feight Ca Gain is blown into ca fom ca B at a ate of b kilogams pe second. The gain leaves the chute vetically downwad, so that it has the same hoizontal velocity, u as ca B, (Figue 12.8). Ca is initially at est befoe any gain is tansfeed in and has mass m,0. t the moment of inteest, ca has mass m and speed v. Detemine an expession fo the speed ca as a function of time t. Figue 12.8 Filling a feight ca Solution: Choose positive x -diection to the ight in the diection the cas ae moving. Define the system at time t to be the ca and gain that is aleady in it, which togethe has mass, and the small amount of mateial of mass Δm that is blown into ca m duing the time inteval [t,t + Δt]. t time that is moving with x -component of the velocity v. t time t, ca is moving with velocity v = v î, and the mateial blown into ca is moving with velocity u = uî t time t + Δt, ca is moving with velocity v + Δv = (v + Δv )î, and the mass of ca is m (t + Δt) = m + Δm, whee Δm = Δm g. The momentum diagam fo times t and fo t + Δt is shown in Figue u î g v m g v + v m m + m time t time t + t Figue 12.9 Momentum diagam at times t and t + Δt 12-8

9 The momentum at time t is The momentum at time t + Δt is P sys P = m v + Δm g u. ( ) sys (t + Δt) = (m + Δm )( v + Δv ). ( ) Thee ae no extenal foces acting on the system in the x -diection and the extenal foces acting on the system pependicula to the motion sum to zeo, so the momentum pinciple becomes P (t + Δt) P sys sys 0 = lim. ( ) Δt 0 Δt Using the esults above (Eqs. ( ) and ( ), the momentum pinciple becomes (m v v v u 0 = lim + Δm )( + Δ ) (m + Δm ) g. ( ) Δt 0 Δt which afte using the condition that Δm = Δm g and some eaangement becomes m Δv Δm ( v u ) Δm v 0 Δ = lim + lim + lim. ( ) Δt 0 Δt Δt 0 Δt Δt 0 Δt In the limit as, the poduct Δm Δ v is a second ode diffeential (the poduct of two fist ode diffeentials) and the tem Δm Δ v / Δt appoaches zeo, theefoe the momentum pinciple yields the diffeential equation 0 = m The x -component of Eq. ( ) is then v dt + dm dt ( v u). ( ) d 0 = m dv dt + dm dt (v u). ( ) Reaanging tems and using the fact that the mateial is blown into ca at a constant ate b dm / dt, we have that the ate of change of the x -component of the velocity of ca is given by dv b(u v ) =. ( ) dt m 12-9

10 We cannot diectly integate Eq. ( ) with espect to dt because the mass of the ca is a function of time. In ode to find the x -component of the velocity of ca we need to know the elationship between the mass of ca and the x -component of the velocity of the ca. Thee ae two appoaches. In the fist appoach we sepaate vaiables in Eq. ( ) whee we have suppessed the dependence on t in the expessions fo m and v yielding which becomes the integal equation dv u v dm =, ( ) m v = v (t ) m = m (t ) dv dm =, ( ) =0 u v = m v m m,0 whee m,0 is the mass of the ca befoe any mateial has been blown in. fte integation we have that Exponentiate both side yields u m ln = ln. ( ) u v m,0 u m =. ( ) u v m,0 We can solve this equation fo the x -component of the velocity of the ca m m,0 v = u. ( ) m Because the mateial is blown into the ca at a constant ate b dm / dt, the mass of the ca as a function of time is given by m = m,0 + bt. ( ) Theefoe substituting Eq. ( ) into Eq. ( ) yields the x -component of the velocity of the ca as a function of time bt v = u. ( ) m,0 + bt In a second appoach, we substitute Eq. ( ) into Eq. ( ) yielding 12-10

11 Sepaate vaiables in Eq. ( ): which then becomes the integal equation dv b(u v ) =. ( ) dt m + bt,0 dv bdt =, ( ) u v m + bt,0 Integation yields dv dt =. ( ) v =0 u v t =0 m,0 + bt v =v (t ) t =t u m + bt,0 ln = ln. ( ) u v m,0 gain exponentiate both sides esulting in u m,0 + bt =. ( ) u v m,0 fte some algebaic manipulation we can find the speed of the ca as a function of time bt v = m,0 + bt u. ( ) in ageement with Eq. ( ). Check esult: We can ewite Eq. ( ) as (m,0 + bt)v = btu, ( ) which illustates the point that the momentum of the system at time t is equal to the momentum of the gain that has been tansfeed to the system duing the inteval [0,t]. Example 12.4 Boat and Fie Hose buning boat of mass m 0 is initially at est. fie fighte stands on a bidge and spays wate onto the boat. The wate leaves the fie hose with a speed u at a ate α (measued in kg s -1 ). ssume that the motion of the boat and the wate jet ae hoizontal, that gavity does not play any ole, and that the ive can be teated as a fictionless suface. lso assume that the change in the mass of the boat is only due to the wate jet and that all the wate fom the jet is added to the boat, (Figue 12.10)

12 Figue Example 12.4 a) In a time inteval [ t, t + Δt], an amount of wate Δm hits the boat. Choose a system. Is the total momentum constant in you system? Wite down a diffeential equation that esults fom the analysis of the momentum changes inside you system. b) Integate the diffeential equation you found in pat a), to find the velocity v( m ) as a function of the inceasing mass m of the boat, m 0, and u. Solution: Let s take as ou system the boat, the amount of wate of mass Δm that entes w the boat duing the time inteval [ t, t + Δt] and whateve wate is in the boat at time t. The wate fom the fie hose has a speed u. Denote the mass of the boat (including some wate) at time t by m b m b, and the speed of the boat by v v b. t time t + Δt the speed of the boat is v + Δv. Choose the positive x - diection in the diection that the boat is moving. Then the x -components of the momentum of the system at time t and t + Δt ae shown in Figue u v t m w m b v + v m w t + t m b Figue Momentum diagams fo buning boat Because we ae assuming that the buning boat slides with negligible esistance and that gavity has a negligible effect on the ac of the wate jet, thee ae no extenal foces acting on the system in the x -diection. Theefoe the x -component of the momentum of the system is constant duing the inteval [ tt+, Δt] and so 12-12

13 p ( t+ Δt) p ( x x t 0 = lim ). ( ) Δ t 0 Δt Using the infomation fom the figue above, Eq. ( ) becomes (m b + Δm w )( v + Δ v) ( Δ mwu + mbv) 0 = lim. ( ) Δ t 0 Δt Eq. ( ) simplifies to Δv Δm w Δm w Δv Δm 0 = lim w mb + lim v + lim lim u. ( ) Δ 0 t Δ t Δ t 0 Δ t Δ t 0 Δt Δ t 0 Δt The thid tem vanishes when we take the limit Δt 0 because it is of second ode in the infinitesimal quantities (in this case Δm Δ v) and so when dividing by Δt the w quantity is of fist ode and hence vanishes since both Δm w 0 and Δv 0. Eq. ( ) becomes Δv Δm w Δm 0 = lim w mb + lim v lim u. ( ) Δ t 0 Δt Δ t 0 Δ t Δ t 0 Δ t We now use the definition of the deivatives: Δv dv Δm w dmw lim = ; lim =. ( ) Δ 0 t Δt dt Δ t 0 Δt dt in Eq. ( ) to fund the diffeential equation descibing the elation between the acceleation of the boat and the time ate of change of the mass of wate enteing the boat dv dm 0 = w mb + (v u). ( ) dt dt The mass of the boat is inceasing due to the addition of the wate. Let m () t denote the w mass of the wate that is in the boat at time t. Then the mass of the boat can be witten as m b = m 0 + m w, ( ) whee m 0 is the mass of the boat befoe any wate enteed. Note we ae neglecting the effect of the fie on the mass of the boat. Diffeentiating Eq. ( ) with espect to time yields dm b dm = w, ( ) dt dt Then Eq. ( ) becomes dv dm 0 = b mb + (v u). ( ) dt dt 12-13

14 (b) We can integate this equation though the sepaation of vaiable technique. Rewite Eq. ( ) as (cancel the common facto dt ) dv dm = b. ( ) v u m We can then integate both sides of Eq. ( ) with the limits as shown b Integation yields v(t ) m b (t ) dv dm = b ( ) v u v=0 m 0 m b v u m b ln = ln ( ) u m 0 Recall that ln( a/ b ) = ln( b/ a) so Eq. ( ) becomes v u m 0 ln = ln ( ) u m b lso ecall that exp(ln( a/ b )) = a/ b and so exponentiating both sides of Eq. ( ) yields v u m = 0 ( ) u m b So the speed of the boat at time t can be expessed as Check esult: m 0 v = u 1 ( ) m b We can ewite Eq. ( ) as m b (v u) = m 0 u m b v = (m b m 0 )u. ( ) Recall that the mass of the wate that entes the ca duing the inteval [0,t] is m w = m b m 0. Theefoe Eq. ( ) becomes m b v = m w u. ( ) 12-14

15 Duing the inteaction between the jet of wate and the boat, the wate tansfes an amount of momentum m u to the boat and ca poducing a momentum m v. w b Because all the wate that collides with the boat ends up in the boat, all the inteaction foces between the jet of wate and the boat ae intenal foces. The boat ecoils fowad and the wate ecoils backwad and though collisions with the boat stays in the boat. Theefoe if we choose as ou system, all of the wate that eventually ends up in the boat and the boat then the momentum pinciple states p = p (0), ( ) sys sys whee p boat. sys (0) = m u is the momentum of all of the wate that eventually ends up in the w Note that the poblem didn t ask to find the speed of the boat as a function t. We shall now show how to find that. We begin by obseving that dm b dt dm = w α ( ) dt whee the constant α is measued in kg s -1 and is specified as a given constant accoding to the infomation in the poblem statement. The eason is that α is the ate that the wate is ejected fom the hose but not the ate that the wate entes the boat. u t m = u t Figue Mass pe unit length of wate jet Conside a small amount of wate that is moving with speed u that, in a time inteval Δt, flows though a coss sectional aea oiented pependicula to the flow (see Figue 12.12). The aea is lage than the coss sectional aea of the jet of wate. The amount of wate that floes though the aea element Δm = λuδt, whee λ is the mass pe unit length of the jet and uδt is the length of the jet that flows though the aea in the inteval Δt. The mass ate of wate that flows though the coss sectional aea element is then Δm α = = λu. ( ) Δt 12-15

16 In the Figue we conside a small length uδt of the wate jet that is just behind the boat at time t. Duing the time inteval [ tt+, Δt], the boat moves a distance vδ t. u t v t v t (u v) t t + t Figue mount of wate that ente boat in time inteval [t,t + Δt] Only a faction of the length uδt of wate entes the boat and is given by α Δm w = λ(u v)δt = (u v)δt ( ) u Dividing Eq. ( ) though by Δt and taking limits we have that dm w Δm w α = lim = (u v) = α(1 v ). ( ) dt Δt 0 Δt u u Substituting Eq. ( ) and Eq. ( ) into Eq. ( ) yields dm b m 0 = α(1 v ) = α. ( ) dt u m b We can integate this equation by sepaating vaiables to find an integal expession fo the mass of the boat as a function of time m b (t ) t m dm b = α m 0 dt. ( ) b m 0 t=0 We can easily integate both sides of Eq. ( ) yielding 12-16

17 The mass of the boat as a function of time is then 1 (m b 2 m 0 2 ) = α m b,0 t. ( ) 2 αt m b = m ( ) m 0 We now substitute Eq. ( ) into Eq. ( )yielding the speed of the buning boat as a function of time 1 v( t ) = u 1 ( ) αt 1+ 2 m b, Rocket Populsion ocket at time t = t i is moving with velocity v,i with espect to a fixed efeence fame. Duing the time inteval [t i,t f ] the ocket continuously buns fuel that is continuously ejected backwads with velocity u elative to the ocket. This exhaust velocity is independent of the velocity of the ocket. The ocket must exet a foce to acceleate the ejected fuel backwads and theefoe by Newton s Thid law, the fuel exets a foce that is equal in magnitude but opposite in diection acceleating the ocket fowad. The ocket velocity is a function of time, v. Because fuel is leaving the ocket, the mass of the ocket is also a function of time, m, and is deceasing at a ate dm / dt. Let F ext denote the total extenal foce acting on the ocket. We shall use the momentum pinciple, to detemine a diffeential equation that elates dv / dt, dm / dt, u, v, and F, an equation known as the ocket equation. ext We shall apply the momentum pinciple duing the time inteval [t,t + Δt] with Δt taken to be a small inteval (we shall eventually conside the limit that Δt 0 ), and t i < t < t f. Duing this inteval, choose as ou system the mass of the ocket at time t, m = m = m + m f, ( ) sys,d whee m,d is the dy mass of the ocket and m f is the mass of the fuel in the ocket at time t. Duing the time inteval [t,t + Δt], a small amount of fuel of mass Δm f (in the 12-17

18 limit that Δt 0, Δm f 0 ) is ejected backwads with velocity u to the ocket. Befoe the fuel is ejected, it is taveling at the velocity of the ocket and so duing the time inteval [t,t + Δt], the elected fuel undegoes a change in momentum and the ocket ecoils fowad. t time t + Δt the ocket has velocity v (t + Δt). lthough the ejected fuel continually changes its velocity, we shall assume that the fuel is all ejected at the instant t + Δt and then conside the limit as Δt 0. Theefoe the velocity of the ejected fuel with espect to the fixed efeence fame is the vecto sum of the elative velocity of the fuel with espect to the ocket and the velocity of the ocket, u + v (t + Δt). Figue epesents momentum diagams fo ou system at time t and t + Δt elative to a fixed inetial efeence fame in which velocity of the ocket at time t is v. m time t v u + v (t + t) m f m + m v (t + t) time t + t Figue Momentum diagams fo system at time t and t + Δt The momentum of the system at time t is p sys = m v. ( ) Note that the mass of the system at time t is m sys = m. ( ) The momentum of the system at time t + Δt is p sys (t + Δt) = m (t + Δt) v (t + Δt) + Δm f ( u + v (t + Δt)), ( ) whee m (t + Δt) = m + Δm. With this notation the mass of the system at time t + Δt is given by m sys = m (t + Δt) + Δm f = m + Δm + Δm f. ( ) Because the mass of the system is constant, setting Eq. ( ) equal to Eq. ( ) equies that 12-18

19 Δm = Δm f. ( ) The momentum of the system at time t + Δt (Eq. ( )) can be ewitten as p (t + Δt) = (m + Δm ) v (t + Δt) Δm ( u + sys v (t + Δt)), ( ) p sys (t + Δt) = m v (t + Δt) Δm u We can now apply Newton s Second Law in the fom of the momentum pinciple, F ext = m = lim Δt 0 lim Δt 0 (m v (t + Δt) Δm u) m v v Δt (t + Δt) v Δm lim u Δt Δt 0 Δt.. ( ) We now take the limit as dv dm F ext = m u. ( ) dt dt Eq. ( ) is known as the ocket equation. Suppose the ocket is moving in the positive x -diection with an extenal foce given by F = F î Then u = u î, whee u > 0 is the elative speed of the fuel and it is moving ext ext,x in the negative x -diection, v = v î. Then the ocket equation (Eq. ( )) becomes,x dv,x dm F = m + u. ( ) ext,x dt dt Note that the ate of decease of the mass of the ocket, dm / dt, is equal to the negative of the ate of incease of the exhaust fuel dm dm f =. ( ) dt dt We can ewite Eq. ( ) as dm dv,x F u = m. ( ) ext,x dt dt The second tem on the left-hand-side of Eq. ( ) is called the thust 12-19

20 dm dm = u = f Fthust,x u. ( ) dt dt Note that this is not an exta foce but the esult of the fowad ecoil due to the ejection of the fuel. Because we ae buning fuel at a positive ate dm f / dt > 0 and the speed u > 0, the diection of the thust is in the positive x -diection Rocket Equation in Gavity-fee Space We shall fist conside the case in which thee ae no extenal foces acting on the system, then Eq. ( ) becomes dm dv,x u = m. ( ) dt dt In ode to solve this equation, we sepaate the vaiable quantities v,x and m and multiply both sides by dt yielding dm dv = u,x. ( ) m We now integate both sides of Eq. ( ) with limits coesponding to the values of the x -component of the velocity and mass of the ocket at times t i when the ejection of the buned fuel began and the time t f when the pocess stopped, v = v m = m,x,x, f, f u dv,x = m dm. ( ) v = v m = m,x,x,i,i Pefoming the integation and substituting in the values at the endpoints yields m, f v v = u ln. ( ),x, f,x,i m,i Because the ocket is losing fuel, m < m, we can ewite Eq. ( ) as, f,i m,i v v = u ln. ( ),x, f,x,i m, f 12-20

21 We note ln(m / m ) > 1. Theefoe v > v, as we expect. fte a slight,i, f,x, f,x,i eaangement of Eq. ( ), we have an expession fo the x -component of the velocity of the ocket as a function of the mass m of the ocket v,x, f = v,x,i m,i + u ln. ( ) m, f Let s examine ou esult. Fist, let s suppose that all the fuel was buned and ejected. Then m m is the final dy mass of the ocket (empty of fuel). The atio, f,d R = m,i m,d ( ) is the atio of the initial mass of the ocket (including the mass of the fuel) to the final dy mass of the ocket (empty of fuel). The final velocity of the ocket is then v = v + u ln R. ( ),x, f,x,i This is why multistage ockets ae used. You need a big containe to stoe the fuel. Once all the fuel is buned in the fist stage, the stage is disconnected fom the ocket. Duing the next stage the dy mass of the ocket is much less and so R is lage than the single stage, so the next bun stage will poduce a lage final speed then if the same amount of fuel wee buned with just one stage (moe dy mass of the ocket). In geneal ockets do not bun fuel at a constant ate but if we assume that the buning ate is constant whee then we can integate Eq. ( ) dm f dm b = = ( ) dt dt m = m (t ) t =t dm = b dt ( ) m = m,i t =t i and find an equation that descibes how the mass of the ocket changes in time m = m b(t t i ). ( ),i Fo this special case, if we set t f = t in Eq. ( ), then the velocity of the ocket as a function of time is given by m,i v = v + u ln. ( ),x, f,x,i m bt,i 12-21

22 Example 12.4 Single-Stage Rocket Befoe a ocket begins to bun fuel, the ocket has a mass of m,i = kg, of which the mass of the fuel is m f,i = kg. The fuel is buned at a constant ate with total bun time is 510 s and ejected at a speed u = 3000 m/s elative to the ocket. If the ocket stats fom est in empty space, what is the final speed of the ocket afte all the fuel has been buned? Solution: The dy mass of the ocket is m m m = kg, hence,d,i f,i R = m / m,i,d = The final speed of the ocket afte all the fuel has buned is Example 12.5 Two-Stage Rocket v = Δv = u ln R = 6250 m/s. ( ), f Now suppose that the same ocket in Example 12.4 buns the fuel in two stages ejecting the fuel in each stage at the same elative speed. In stage one, the available fuel to bun is m = kg with bun time 150 s. Then the empty fuel tank and accessoies f,1,i fom stage one ae disconnected fom the est of the ocket. These disconnected pats have a mass m = kg. ll the emaining fuel with mass is buned duing the second stage with bun time of 360 s. What is the final speed of the ocket afte all the fuel has been buned? Solution: The mass of the ocket afte all the fuel in the fist stage is buned is m = m m f,1,i = kg and R 1 = m / m,1,d = The change in speed,1,d,1,i,1,i afte the fist stage is complete is Δv,1 = u ln R 1 = 3840 m/s. ( ) fte the empty fuel tank and accessoies fom stage one ae disconnected fom the est of the ocket, the emaining mass of the ocket is m = kg. The emaining fuel has mass m f,2,i,2,d = kg. The mass of the ocket plus the unbuned fuel at the beginning of the second stage is m = kg. Then = m / m = 3.05.,2,i R 2,2,i,2,d Theefoe the ocket inceases its speed duing the second stage by an amount Δv,2 = u ln R 2 = 3340 m/s. ( ) The final speed of the ocket is the sum of the change in speeds due to each stage, 12-22

23 v f = Δv = u ln R 1 + u ln R 2 = u ln(r 1 R 2 ) = 7190 m/s, ( ) which is geate than if the fuel wee buned in one stage. Plots of the speed of the ocket as a function time fo both one-stage and two-stage buns ae shown Figue Figue Plots of speed of ocket fo both one-stage bun and two-stage bun Rocket in a Constant Gavitational Field: Now suppose that the ocket takes off fom est at time t = 0 in a constant gavitational field then the extenal foce is total F = m g. ( ) ext Choose the positive x -axis in the upwad diection then F ext,x = m g. Then the ocket equation (Eq. ( ) becomes dm dv,x m g u = m. ( ) dt dt Multiply both sides of Eq. ( ) by dt, and divide both sides by m. Then Eq. ( ) can be witten as dm dv,x = gdt u. ( ) m We now integate both sides 12-23

24 v (t ) m (t ),x t dm dv = u g dt, ( ),x m v 0 m 0,x,i =,i whee m,i is the initial mass of the ocket and the fuel. Integation yields fte all the fuel is buned at t m, f = m,d and so m m,i v,x = u ln gt = u ln gt. ( ) m,i m = t f, the mass of the ocket is equal to the dy mass v,x (t f ) = u ln R gt f. ( ) The fist tem on the ight hand side is independent of the bun time. Howeve the second tem depends on the bun time. The shote the bun time, the smalle the negative contibution fom the thid tun, and hence the ocket ends up with a lage final speed. So the ocket engine should bun the fuel as fast as possible in ode to obtain the maximum possible speed

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