Math 1302, Week 7: Variable mass systems - UCL

1 math 1302, week 7: Variable mass systemsExample: Coupling of two moving carriagesConsider two train carriages of massm1,m2moving on the same track with speedsU1andU2,whereU1> U2(see figure 1). When they catch each other up they couple together to make asingle coupled pair of carriages that moves with speedV. : We have conservation of momentum:m1m2m1+m2U1U2 VFigure 1: Two carriages coupling together into a singlemassmomentum before = momentum afterm1U1+m2U2= (m1+m2)VSo thatV=m1U1+m2U2m1+ the coupled carriages move with the speed of the centre of mass of the uncoupled about energy changes? Before colliding the total kinetic energy isTb=12m1U21+12m2U22,and after colliding and joining the kinetic energy isTa=12(m1+m2)(m1U1+m2U2m1+m2)21 Hence the change in kinetic energy T=Ta Tbis given by T=12(m1+m2)(m1U1+m2U2m1+m2)2 12m1U21 12m2U22=12m21U21+m22U22+ 2m1m2U1U2m1+m2 12m1U21 12m2U22=12(m1+m2)[m21U21+m22U22+ 2m1m2U1U2) (m21+m1m2)U21 (m1m2+m22)U22]=12(m1+m2)(2m1m2U1U2) (m1m2)U21 (m1m2)U22)= m1m2(U1 U2)22(m1+m2)< energy is lost in the collision.

2 (Where does it go?) Notice that this is just the same as twobodies colliding, but with coefficient of restitutione= mass systemsWhen the mass of a system is changing with time, we will need a generalisation of Newton ssecond law. We will use thatforce = rate of change of reduces to Newton s 2nd law when the mass is constant with of particle through cloud of stationary dustm,vm + m, v + v, MM - m tFigure 2: Particle moving through a cloud of stationarydust particles2We consider a particle moving through a cloud of stationary dust (see figure 2). The particlehas massm(t) at timet, and the total mass of the dust cloud isM(t). Since the cloud dust isstationary, it has no momentum. Let the velocity of the particle at timetbev(t) and at timet+ tlet it bev+ v. Thentotal momentum att=mv momentum of particle+M0 momentum of dust= momentum att+ t= (m+ m)(v+ v) mtm of particle+ (M m)0 zero mtm of dust=mv+m v+ mv+ m in momentum t=1 t{(mv+m v+ mv+ m v) (mv)}Now let t 0 to obtainrate of change of momentum = lim t 01 t(m v+ mv+ m v) =mdvdt+dmdtv=ddt(mv).

3 When there is an external forceFacting on the system we obtainF= rate of change of momentum =ddt(mv).(1)Example: falling raindropSuppose that a raindrop falls through a cloud and accumulates mass at a ratekmvwherek >0is a constant,mis the mass of the raindrop, andvits velocity. What is the speed of the raindropat a given time if it starts from rest, and what is its mass ?Solution: We are takingxas distance fallen andv= x. Then the external force is its weightmgand so from (1) we havemg=ddt(mv) =mdvdt+vdmdt=mdvdt+kmv2,since we are told thatdm/dt=kmv. Cancelling the mass and rearrangingdvdt=g kv2,so that v0dvg kv2= t0dt= setV2=g/kand use partial fractions to gett= v0dvg kv2=12kV v01V+v+1V vdv=12kVlog(V+vV v)soV+v= (V v)e2kVt, (e2kVt 1e2kVt+1)=Vtanh(V kt), so thatv= gktanh( kgt).3 Now we may find the mass : We havedmdt=kmv=km gktanh( kgt) =m kgtanh( kgt).

4 Thus t01mdmdtdt= t0 kgtanh( kgt)dt mm0dmm= t0 kgtanh( kgt)dtlogm logm0= log cosh( kgt)which givesm=m0cosh( kgt).Example: Raindrop falling through a cloud accumulating mass a given rateA raindrop falls through a cloud while accumulating mass at a rate r2whereris its radius(assume that the raindrop remains spherical) and >0. Find its velocity at timetif it startsfrom rest with : We have thatdmdt= r2. Butm=43 r3 where is the density. So r2=dmdt=ddt(43 r3 )= 4 givesdr/dt= where = 4 , which givesr= t+C,whereCis a constant. Usingr=aatt= 0 we obtainC=aand hencer= t+a. Now wehave1mdmdt=34 r3 r2=3 4 r=3 ( t+a).Thus fromddt(mv) =mgwe obtain vm+v m=mgso that v+ mmv=gand hencedvdt+3 ( t+a)v= can be solved using the integrating factorI= exp( 3 ( t+a)dt) = exp(3 log(a+ t)) =(a+ t)3. This givesddt[v(a+ t)3] =g(a+ t) (a+ t)3=C +g4 (a+ t)4.

5 Usingv(0) = 0 we getC = ga4/4 and sov(t) =g4 ((a+ t) a4(a+ t) 3).4mvv + v m + m m-v + v tFigure 3: Particle moving and gaining m >0 or losing m <0 mass from mass moving at zero relative velocityMass lost or gained at zero relative velocityThe example to keep in mind here is a hot air balloon containing a bag of sand. The sand isreleased to control the height of the balloon. As the sand trickles out it is (effectively) stationaryrelative to the balloon. Consider, then, figure 3. Particle has massmand is moving with velocityvat timet. At a later timet+ tit has massm+ mand velocityv+ v. The mass movingwith the particle with zero relative velocity has mass mand velocityv+ v. Hencemomentum att=mv,andmomentum att+ t= (m+ m)(v+ v) + ( m)(v+ v)rate of change of momentum = lim t 01 t{(m+ m)(v+ v) + ( m)(v+ v) mv}= lim t 01 t{mv+ mv+m v+ m v mv m v mv}= for an external force ofFwe getF=mdvdt.

6 (2)Balloon risingSuppose a balloon constant massMcontains a bag of sand massm0experiences a constantupward thrust ofC. Initially it is in equilibrium, and then the sand is released at a constantrate so that it is all released in timet0. Find the height of the balloon and its velocity when allthe sand has been : As the sand leaves the bag its speed relative to the ground is the same as that of theballoon, its speed relative to the balloon is zero, so we may use (2):(M+m)dvdt=C (M+m) +m the sand empties in timet0at a constant rate, say . Thusm(t) =m0 t(solvedm/dt= ). If at timet0we havem= 0 then we must have =m0/t0. Hencedvdt=CM+m0 t overt:v(t) v(0) = gt C [log(M+m0 t)]t0= gt C log(M+m0 tM+m0).Now if the balloon is initially in equilibrium the upward thrust equals its weight:C= (M+m0) using thatv(0) = 0 and =m0/t0we arrive at:v(t) = gt (M+m0)gt0m0log(1 m0t(M+m0)t0).

7 To find the height, lets first set =m0/(t0(M+m0)). Then we havedxdt=v= gt g log(1 t).Hencex(t) = gt22 g log(1 t)dt+ clearly need the integral logz dz. To do this we can either argue thatddz(zlogz) =logz+z/z= logz+ 1, so thatddz(zlogz z) = logzand hence logz dz= ddz(zlogz z)dz=zlogz z+ constant,or we may use integration by parts after splitting logz= logz 1 and useu= logz, dv/dz= 1,etc. The end result is that using the substitutionz= 1 t log(1 t)dt= logz( 1 )dz= 1 z(logz 1)+K = 1 (1 t)[log(1 t) 1]+K .This yieldsx(t) =K gt22+g 2(1 t){log(1 t) 1},whereK is found from initial conditions:x(t) =x(0) +g 2 gt22+g 2(1 t){log(1 t) 1}6mvv + v m + m tMuu + u M + MFigure 4: Particle moving with velocityvand gaining m >0 or losing m <0 mass from massMmoving Variable mass problemsConsider figure 4. We havemomentum before =mv+M u,andmomentum after = (m+ m)(v+ v) + (M+ M)(u+ u).

8 Thusrate of change in momentum = lim t 01 t{(m+ m)(v+ v) + (M+ M)(u+ u) (mv+M u)}= lim t 01 t{ mv+m v+ m v+ M u+M u+ M u}=vdmdt+mdvdt+udMdt+ , we may also use that m= M( mass conservation), so thatdm/dt= dM/dtandrate of change in momentum =ddt(mv) udmdt+ is a general result. Note that whenu= 0, we obtain rate of change in momentum =ddt(mv)as in equation (1). Also, whenu=vwe have rate of change in momentum = (m+M)dvdtas inequation (2).Example: Rocket motion (I)A rocket of massmemits mass backwards at speedurelative to the rocket at a constant ratek. Ignoring gravity and air resistance find its speedvat timetif att= 0 it has speedv0and7mm + m vv + v ( m)v + v - u Figure 5: Rocket burning fuel that is ejected at velocityurelative to the +m0, wherem0is the amount of fuel for : Letmbe the mass of the rocket and fuel at timet. We refer to figure of change in momentum = lim t 01 t{(m+ m)(v+ v) + ( m)(v+ v u) mv}= lim t 01 t{ mv+m v+ m v mv+u m m v}=mdvdt+ is no external force acting so the rate of change of momentum is zero:mdvdt= kis the constant loss of fuel mass .

9 Thusm(t) =M0 kt=M+m0 ktfort < m0/k(after which there is no more fuel to burn).Thusdvdt=kuM0 kt,which integrates to givev(t) = ulog (M0 kt) + 0 we havev=v0we obtainv(t) =v0 ulog(1 kM0t),(t m0/k).Example: Rocket motion (II)The rocket now moves under the pull of gravity and air resistancekmv2and the fuel is burnedso thatm=m0e btwhereb > g/uis a haveF= mg kmv2for the external force. This equals the rate of change of momentum:F=mdvdt+ +udmdt= mg and dividing bym,dvdt+umdmdt= g we are told thatm=m0e bt, so that1mdmdt= b. This givesdvdt bu= g kv2,which we rewrite asdvdt= 2 kv2,where 2=bu g > use partial fractions to obtain 1 kv+1 + kvdv= 2 t+C,whereCis a constant. Thuslog( + kv kv)= 2 kt+C .Using thatv(0) = 0 we getC = 0. Tidying up we finally obtainv(t) = ktanh( kt).Ast we havev limt ktanh( kt)= limt k(1 e 2 kt1 +e 2 kt)= the rocket reaches a limiting velocity k= bu (2004 Exam question 5)The mass of a spacecraft at timetism(t) and its velocity isV(t).

10 Fort <0,m(t) =MandV=U iwhereMandUare constants andiis a constant unit vector. For 0< t < T, thecraft encounters a stream of particles which have velocityw(cos i+ sin j) wherewand areconstants andjis a unit vector orthogonal constant mass of the particles enter the craft per unit time and these particles are thereafterstationary relative to the (a) IfV=ui+vj, show thatmdudt+dmdtu= wcos , mdvdt+dmdtv= wsin ,dmdt= .(b) Solve these equations to show that at timeTthe direction of motion of the craft has beenturned through an angle , wheretan = wTsin M U+ wTcos Solution: Consider the before and after picture shown in figure 6 for the momentum changeFigure 6: Rocket in a stream of particlesin time t. Since the particles pass through the box at a constant rate, in the absence of thespaceship the mass inside a fixed virtual box (we need only consider 2D) is some the spaceship is passing through the box and is there att(before figure) andt+ t(after figure).