Dynamics of Systems of Variable Mass

Chapter 8

Dynamics of Systems of Variable Mass

8.1 Introduction So far we have considered DMSs and CMSs in which masses of particles mn and their number have not changed. In nature and technology, however, phenomena are commonly known where the number of particles of a system or their mass change over time. If floating icebergs are heated by the Sun’s rays, then the ice melts and their mass decreases. If the falling snow becomes frozen to the floating icebergs, then their mass increases. Earth’s mass increases when meteorites fall on its surface. In turn, the mass of the meteorites before they reach Earth’s surface decreases as a result of burning in Earth’s atmosphere. The mass of rockets decreases as the fuel they contain burns. The mass of elements transported on a conveyor belt changes as a result of their loading and unloading.

8.2 Change in Quantity of Motion and Angular Momentum Let the mass of a mechanical system m.t/ be changing in time according to the equation m .t/ D m0  m1 .t/ C m2 .t/ ; (8.1) where m.t/ D m.t0 /, m1 .t/  0, (m2 .t/  0) denotes the mass of particles leaving (entering) the system (Fig. 8.1). Let us choose a time instant t during motion of the system, and let for this instant the momentum p of the considered system of particles increase by p during time t. Then, by p let us denote the momentum of analogous system, but of a constant mass. At the instant t C t the quantity of motion of a system of variable mass is equal to p C p D p C p  p1 C p2 : (8.2) J. Awrejcewicz, Classical Mechanics: Dynamics, Advances in Mechanics and Mathematics 29, DOI 10.1007/978-1-4614-3740-6 8, © Springer Science+Business Media New York 2012

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Fig. 8.1 Motion of a body of variable mass with respect to the inertial coordinate system O 0 X10 X20 X30

This means that the increment of momentum of the investigated system follows from the increment of momentum of a system of constant mass and the additional quantity of motion delivered .p2 / and removed .p1 / to/from the system during time t. From the preceding equation we obtain p D p  p1 C p2 because at the instant t we have

p D p :

(8.3)

(8.4)

Dividing by t and on the assumption that t ! 0 we get p p p1 p2 D lim  lim C lim ; t !0 t t !0 t t !0 t t !0 t

(8.5)

dp R D F C FR 1 C F2 ; dt

(8.6)

lim

hence

where dp p D ; t !0 t dt p1 p2 FR ; FR ; 1 D  lim 2 D lim t !0 t t !0 t F D lim

(8.7)

and F is a main vector of a system of external forces acting at the time instant t. Equation (8.6) extends the well-known theorem concerning the change in the quantity of motion (momentum) of a system. On its right-hand side additionally R appear the so-called thrust forces, FR 1 and F2 .

8.3 Motion of a Particle of a Variable Mass System

343

In a similar way one can generalize the theorem regarding the change in angular momentum (moment of momentum) of a system. Applying an argument analogous to the previous one, we obtain K C K D K C K  K1 C K2 ;

(8.8)

where K is the moment of momentum of the system with respect to a certain arbitrary chosen fixed pole in the coordinate system O 0 X10 X20 X30 , and K1.2/ denotes the sum of moments of a quantity of motion for those particles that left (entered) the considered system of variable mass during the time interval t. Dividing the preceding equation by t and proceeding to the limit as t ! 0 we have dK R D M C MR 1 C M2 ; dt

(8.9)

where K M D ; t !0 t dt K1 K2 MR ; MR : 1 D  lim 2 D lim t !0 t t !0 t M D lim

(8.10)

Equation (8.9) is a generalization of a theorem concerning changes in the angular momentum of a mechanical system. On its right-hand side additionally appear R moments of a thrust force, MR 1 and M2 .

8.3 Motion of a Particle of a Variable Mass System Let us consider a particle A belonging to the investigated system of variable mass, and let the mass of this particle be described by (8.11) in the form mA .t/ D mA .t0 /  mA1 .t/ C mA2 .t/ :

(8.11)

The kinematics of a particle of variable mass is presented in Fig. 8.2. In Fig. 8.2 the absolute velocity of a piece of mass m2 is denoted by u2 , whereas the absolute velocity of a piece of mass m1 is denoted by u1 . We will assume that mA1  mA .t0 / and mA2  mA .t0 /. In order to derive the differential equation of the motion of a particle of variable mass m.t/, we will make use of (8.6). A quantity of motion (momentum) of particle A at an arbitrary time instant t reads p .t/ D m .t/ v .t/ ;

(8.12)

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8 Dynamics of Systems of Variable Mass

Fig. 8.2 Particle of mass m0 in absolute system OX1 X2 X3 at time instant t and piece of mass m1 expelled (absorbed m2 ) from (by) particle A

and the changes in momentum that follow from absorbing mass m2 and expelling mass m1 by particle A during the time interval t are respectively equal to pi D mi ui ;

i D 1; 2:

(8.13)

According to (8.7) we have FR 1 D  lim

t !0

FR 2 D lim

t !0

p1 m1 u1 dm1 D  lim D u1 ; t !0 t t dt

p2 m2 u2 dm2 D lim D u2 : t !0 t t dt

(8.14)

Substituting (8.12) and (8.14) into (8.6) we obtain d dm1 dm2 Œm .t/ v .t/ D F  u1 C u2 ; dt dt dt

(8.15)

and following the transformations we have mRrA D F  m P 1 .u1  v/ C m P 2 .u2  v/ :

(8.16)

The obtained (8.16) is called a generalized Meshcherskiy1 equation, and it describes the motion of a particle of variable mass. If the mass of particle A does not change, then m P1 D m P 2 D 0, and from (8.16) we obtain Newton’s second law on the motion of particle A of constant mass m. 1

Ivan Meshcherskiy (1859–1935), professor working mainly in Saint Petersburg.

8.3 Motion of a Particle of a Variable Mass System

345

In previous calculations the dynamics of a particle of variable mass was presented descriptively during the derivation of (8.6) and (8.9). Presently we will proceed in a different way (see [1, 2]) by taking into account only the change in momentum of particle A. Let an elementary mass dm2 of velocity u2 .t/ be added, and an elementary mass dm1 of velocity u1 .t/ be removed, to/from particle A of mass m.t/ and velocity v.t/. The momenta at the time instants t and t C dt are equal to p .t/ D .m C dm1 / v C dm2 u2 ; p .t C dt/ D .m C dm2 / .v C dv/ C dm1 u1 : The increment of momentum is equal to p .t C dt/  p .t/ D vdm2 C dvdm2 C mdv C dm1 u1  vdm1  dm2 u2 ; and neglecting differentials of the second order and dividing by dt we obtain the generalized Meshcherskiy equation (8.16), where F D dp=dt. Following the introduction of relative velocities, wi D ui  v;

i D 1; 2:

(8.17)

Respectively expelling and absorbing the mass by particle A (8.16) takes the form mRrA D F  m P 1 w1 C m P 2 w2 : (8.18) Taking into account relation (8.17), (8.18) is identical to (8.16). If the case of separation of mass from particle A is considered alone, then from (8.11) for mA2  0 we obtain m.t/ D m.t0 /  m1 .t/; (8.19) hence m.t/ P D m P 1 .t/:

(8.20)

Substituting (8.20) into (8.18) we get mRrA D F C FR 1:

(8.21)

The preceding equation is called a Meshcherskiy equation. From (8.21) it follows that the effect of separation of mass is equivalent to the action of an additional force FR P 1 on particle A, called a thrust force. The thrust force FR 1 D mw 1 (removal of mass) has a sense opposite to the sense of velocity w1 , whereas the thrust force FR 2 (addition of mass) has the same sense as the sense of the relative velocity w2 . The quantity m P 1 .m P 2 / is called the mass removal (addition) per second. In a special case, where the absolute velocity of the mass that separates is u1 D 0, (8.21) takes the form m

dv dm DF v dt dt

(8.22)

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8 Dynamics of Systems of Variable Mass

or

d.mv/ D F: (8.23) dt We have shown that if the absolute velocity of the mass that separates is equal to zero, then the derivative of momentum of particle A balances the external forces acting on this particle. If, in turn, the relative velocity of the mass that separates is w1 D u1  v D 0, then from (8.21) we obtain m.t/

dv D F: dt

(8.24)

In this case we obtained an equation that is formally consistent with Newton’s second law on the motion of a particle of constant mass.

8.4 Motion of a Rocket (Two Problems of Tsiolkovsky) Let us now consider two problems of Tsiolkovsky.2

8.4.1 First Tsiolkovsky Problem Let a rocket, treated further as a particle, be moving in space, and let the action of external forces on it be negligibly small. The initial conditions of motion are as follows: v.0/ D v0 , m.t/ D m0 C m1 .t/, where m0 is the mass of the rocket and m1 .t/ is the mass of fuel .m1 .0/ D m10 /. In the considered case, the Meshcherskiy equation, (8.21), takes the form m

dv dm D w1 : dt dt

(8.25)

Let us assume that the relative velocity of combustion gases w1 D u1 v D const and its sense are opposite to those of velocity vector v. It follows that a rocket moves along a straight line according to the sense of vector v (Fig. 8.3). Following the projection (multiplication by E1 ) of (8.25) onto the axis OX we obtain m

2

dm dv D w1 dt dt

Konstantin Tsiolkowsky (1857–1935), Russian teacher of mathematics and physics of Polish origin; precursor to the theory of rocket flight.

8.4 Motion of a Rocket (Two Problems of Tsiolkovsky)

347

Fig. 8.3 Motion of rocket in a force-free field

or dv D w1

dm : m

(8.26)

Integrating (8.26) we have v.t/ D w1 ln m C C;

(8.27)

where C is the constant of integration equal to C D v0 C w1 ln.m0 C m10 /. Finally, the time change in the velocity of a rocket is described by the scalar equation   m0 C m10 dx :  v.t/ D v0 C w1 ln dt m.t/

(8.28)

The maximum velocity is reached by the rocket after the fuel is completely spent, that is, when m.t / D m0 , and it is equal to   m10 : v.t / D v0 C w1 ln 1 C m0

(8.29)

The obtained equation is called a rocket equation. The maximum velocity of a rocket does not depend on the process of fuel combustion, that is, whether combustion proceeds slowly or quickly. The constant quantity m10 =m0 is also known as a Tsiolkovsky constant. In contrast, the trajectory of motion of a rocket does depend on the process of fuel combustion. Integrating (8.28), for the initial condition x.0/ D 0, we have Zt x.t/ D v0 t C w1

ln 0

where m00 D m0 C m10 .

m00 d; m./

(8.30)

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8 Dynamics of Systems of Variable Mass

Fig. 8.4 Vertical motion of rocket in Earth’s gravitational field

8.4.2 Second Tsiolkovsky Problem Let a rocket, treated further as a particle, move vertically upward in a uniform gravitational field of Earth, its resistance to motion being neglected. The relative velocity of ejection of fuel combustion products is constant and directed vertically downward (Fig. 8.4). In this case after projection of the Meshcherskiy equation (8.21) onto the axis OX3 we get dv dm m D mg  w1 ; (8.31) dt dt or, separating the variables, d.v C gt/ D w

dm : m

(8.32)

Integrating the preceding equation we have v C gt D w ln m C C:

(8.33)

C D v0 C w ln m00 :

(8.34)

The constant C is equal to

8.4 Motion of a Rocket (Two Problems of Tsiolkovsky)

349

Substituting the obtained value of C into (8.33) we have  0  dx3 m0  v.t/ D v0  gt C w ln : dt m.t/

(8.35)

If we assume the initial conditions to be x3 .0/ D 0, v0 D 0, then following the integration of (8.35) we get Zt x3 .t/ D w

 1 m00 d  gt 2 : ln m.t/ 2 

(8.36)

0

Let the fuel combustion take place according to the following process: m.t/ D m00 e˛t ;

(8.37)

where ˛ is a constant coefficient characterizing the speed of fuel combustion. The mass of combustion products m1 .t/ can be calculated from m.t/ C m1 .t/ D m0 C m10  m00 and is equal to   m1 .t/ D m0 C m10  .m0 C m10 / e˛t D m00 1  e˛t :

(8.38)

The thrust force is equal to P 1 w1 D m00 w1 ˛e˛t D m.t/w1 ˛; F1R D m

(8.39)

where ˛w1 is the acceleration imposed on the rocket due to fuel combustion. Because we assumed certain combustion process described by (8.37), from (8.35) we have  0  m0 ; v.t/ D v0  gt C w ln ˛t e and for v0 D 0 we obtain

v.t/ D .˛w  g/ t:

(8.40)

In turn, from (8.36) (or by integrating (8.40)) we have x3 .t/ D .˛w  g/

t2 : 2

(8.41)

From the last equation it follows that the launch of the rocket is possible if ˛w > g, that is, the acceleration coming from a thrust force F1R should exceed the acceleration of gravity.

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8 Dynamics of Systems of Variable Mass

If the fuel is burned completely at the time instant t D tf , then according to (8.37) we have m.tf / D m0 C m1 .tf / D m00 e˛tf ; that is,

m0 D m00 e˛tf ;

(8.42)

because at the instant tk we have no more fuel, that is, m1 .tf / D 0. From (8.42) we can determine the time required for complete combustion of fuel by a rocket, which is equal to  (8.43) tf D ; ˛ where   m10  D ln 1 C : m0 From (8.40) and (8.41) one can determine the velocity and ceiling height of a rocket corresponding to the time instant when the fuel is spent: vf D

 .˛w  g/ ; ˛

(8.44)

x3f D

 2 .˛w  g/ : 2˛ 2

(8.45)

Because at the instant when the fuel has run out t D tf and vf D v.tf /, for such initial conditions a rocket of mass m.tf / D m0 additionally climbs in Earth’s gravitational field at the height hd D

v2f 2g

D

2 .˛w  g/2 : 2˛ 2 g

(8.46)

We obtain the maximum height h of the rocket using (8.45) and (8.46): h D hd C x3f

 2w D 2



w 1  g ˛

 :

(8.47)

The height reached by a rocket depends on the coefficient of the fuel combustion rate ˛. For example, at a rapid (explosive) rate of fuel combustion the height attained is equal to  2 w2 hmax D : (8.48) 2g

8.5 Equations of Motion of a Body with Variable Mass

351

8.5 Equations of Motion of a Body with Variable Mass A group of particles n D 1; : : : ; N , between which the mutual distances do not change and at least one of which is a particle with variable mass, is called a rigid body of variable mass [3]. According to the previous calculations, let the particles of the body (the material system) change their mass according to (8.1), that is, mn .t/ D m0n  m1n .t/ C m2n .t/;

n D 1; : : : ; N;

(8.49)

where m1n .t/ is the total mass lost by particle n at time t, and m2n .t/ is the total mass gained by the particle at time t. Let us further consider the case of motion of a rigid body with variable mass about a certain fixed point O (motion about a point of a system with variable mass). The angular momentum KO of the system about point O is equal to (in the system rigidly connected to the body OX100 X200 X300 ) dKO R C !  KO D MZ O C MO ; dt

(8.50)

where MZ O is the main moment of external forces acting on the system with respect to point O, and MR O is the additional moment of a thrust force that needs to be determined. According to relation (8.8) we have dK1O D

N X

dm1n rn  u1n ;

nD1

dK2O D

N X

dm2n rn  u2n ;

(8.51)

nD1

where rn is a radius vector of particle n, and on that basis a moment of thrust forces is equal to R R MR O D M1O C M2O ;

where MR 1O D 

N X dm1n nD1

MR 2O D 

dt

N X dm2n nD1

dt

rn  u1n ;

rn  u2n :

(8.52)

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8 Dynamics of Systems of Variable Mass

Introducing the notion of relative velocity wn according to equations u1n D vn C w1n ; u2n D vn C w2n ;

(8.53)

we have MR O D

N X dm1n nD1

D

N X

dt 

rn 

nD1

C

D

rn  .vn C w1n / C

N X dm2n nD1

dm2n dm1n  dt dt

dt

rn  .vn C w2n /

  vn

  dm1n dm2n w1n C w2n rn   dt dt nD1

N X

N X

 rn 

nD1

 X N dm2n dm1n dmn w2n  w1n C vn ; rn  dt dt dt nD1

(8.54)

where (8.49) was used. Eventually we obtain W MR O D MO C

dI !; dt

(8.55)

where MW O D

N X nD1

 rn 

 dm2n dm1n w2n  w1n ; dt dt

N X dI dmn !D .!  rn / ; rn  dt dt nD1

(8.56)

and I is the matrix of the inertia tensor of a body for point O, and in this case the matrix depends on time. Because KO D I!, from (8.50) and taking into account (8.55) we obtain d! dI dI W !CI C !  I! D MZ !; O C MO C dt dt dt hence I

d! W C !  I! D MZ O C MO : dt

(8.57)

8.5 Equations of Motion of a Body with Variable Mass

353

If the axes of a coordinate system during the process of gaining and losing mass remain the principal axes of inertia, then (8.57) has the following scalar representation: d!1 C .I3 .t/  I2 .t// !2 !3 D M1 C M1W ; dt d!2 C .I1 .t/  I3 .t// !1 !3 D M2 C M2W ; I2 .t/ dt d!3 C .I2 .t/  I1 .t// !1 !2 D M3 C M3W ; I3 .t/ dt

I1 .t/

(8.58)

where Ii .t/ are the moments of inertia of the body with respect to the axes OXi , Mi are the projections of a main vector of external forces onto these axes, and ! D !1 E1 C !2 E2 C !3 E3 . In the case of rotation of the body about a fixed axis (let it be the axis OX3 ), we have ! D !3 E3 , and from the last equation of (8.58) we obtain I3 .t/

d!3 D M3 C M3W : dt

(8.59)

As distinct from the previously considered case of the rotation of a rigid body about a fixed axis, on the right-hand side additionally appeared the moment of a thrust force, and on the left-hand side the mass moment of inertia of a body changing in time. Example 8.1. Figure 8.5 shows a drum having moment of inertia I0 with respect to the axis OX3 perpendicular to the plane of the drawing and passing through point O, onto which a rope of length S and mass m is wound. Determine the angular velocity of the drum on the assumption that the rope started to reel out from the drum at an initial velocity of zero and the drum axis was horizontal. For the solution of the problem we make use of (8.59). In this case M3W D .!1  !/

dI ; dt

where !1 is the angular velocity of an elementary moment of inertia dI of a rope separating from a drum that is rotating with angular velocity !. The element of the rope leaving the drum has a velocity equal to the peripheral speed of the drum, that is, r!1 D r!, i.e., M3W D 0. The equation of motion of the investigated system is analogous to (8.24) for the rotational motion. The problem reduces to the analysis of equation I.'.t//

d! D M Z; dt

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8 Dynamics of Systems of Variable Mass

Fig. 8.5 Rope reeling out of a drum

where I.'.t// D I0 C mr 2 

m .r'/r 2 : S

In turn, the moment M Z follows from the action of the force coming from the rope reeling out from the drum and is equal to M Z .'.t// D

m .r'/gr: S

Because d! d! d' d! D D! ; dt d' dt d' from the equation of motion we have 

I0 C mr 2 

m m 3  d! r ' ! D r 2 g'; S d' l

and separating the variables we get !d! D

mr 2 g ' S I0 C mr 2 

m 3 d': Sr '

8.5 Equations of Motion of a Body with Variable Mass

355

Setting I0 D 3mr 2 we have !2 g D 2 S

Z

 ' d' C C : 4  Sr '

The obtained indefinite integral is calculated by substitution: t D4

r '; S

hence S d' D  dt; r

'D

S .4  t/ : r

We have then Z

that is,

Z

4t S2 dt D 2 .4 ln jtj C t/ t r 4S 2 ˇˇ S r ˇˇ S D 2 ln ˇ4  ' ˇ  ' C C ; r S r r

' D 4  Sr '

  r ˇˇ g 4S ˇˇ !2  D ln ˇ4  ' ˇ  ' C C : 2 r r S

The integration constant is determined from the initial condition !.0/ D 0 and is equal to C D

4S ln 4: r

The desired function 

  12 2g 4S h r i 4S !  !Œ'.t/ D  ln 4  '  ' C ln 4 : r r S r The maximum angular velocity ! D !max is obtained after the rope has been completely unwound, that is, substituting '  'max D Sr into the preceding formula.  Example 8.2. A body of mass m is thrown upward with initial speed v0 , and there is a chain of unit mass  stacked on a horizontal plane and attached to the body. Determine the maximum height attained by the chain (Fig. 8.6). During the motion of the chain its links are successively lifted from the stationary stack, that is, their absolute velocity is equal to zero. The problem is therefore described by (8.23), which in our case takes the form d Œ.m C x/ x P D mg: dt

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8 Dynamics of Systems of Variable Mass

Fig. 8.6 Projection of ball of mass m with attached chain

The first integral of the preceding equation reads mxP C x xP D mgt C C1 or

 x 2 d mx C D mgt C C1 : dt 2 The second integral is equal to mx C

t2 x 2 D mg C C1 t C C2 : 2 2

Let x.0/ D x0 , x.0/ P D v0 . The constant C1 is found from the equation C1 D .m C x0 / v0 ; and the constant C2 reads

 x0  x0 : C2 D m C 2

References

357

The maximum height is attained for a velocity of mass equal to xP  v.t / D 0, that is, for the time instant t D

.m C x0 /v0 C1 D : mg mg

The desired quantity x.t / D x is determined from the equation mx C

.m C x0 /2 v20 .m C x0 /2 v20  x0  x2 D C C mC x0 2 2mg mg 2

or, following transformation, x2

   2m 2m .m C x0 /2 v20 x  C C x0 x0 D 0: C  mg 

For x0 D 0 we have x2 C

mv20 2m x  D 0:  g

Solving the preceding quadratic equation and rejecting the negative root we obtain s   v20 m 1 4m m x D  C :  C  2   g Finally, this chapter can be supplemented by the classic works [4–9].

References 1. J. Nizioł, Methods of Solving Mechanical Tasks (WNT, Warsaw, 2002) (in Polish) 2. W. Kurnik, Lectures on General Mechanics (Warsaw Technological University Press, Warsaw, 2005) (in Polish) 3. A.D. Markeev, Theoretical Mechanics (Nauka, Moscow, 1990) (in Russian) 4. S.T. Thornton, J.B. Marion, Classical Dynamics of Particles and Systems (Saunders College Publishers, New York, 1995) 5. A.P. Arya, Introduction to Classical Mechanics (Addison-Wesley, San Francisco, 1998) 6. J.V. Jos´e, E.J. Saletan, Classical Dynamics: A Contemporary Approach (Cambridge University Press, Cambridge, 1998) 7. H. Goldstein, C. Poole, J. Safko, Classical Mechanics, 3rd edn. (Addison-Wesley, San Francisco, 2002) 8. T.W.B. Kibble, F.H. Berkshire, Classical Mechanics, 5th edn. (Imperial College Press, London, 2004) 9. R.D. Gregory, Classical Mechanics (Cambridge University Press, Cambridge, 2011)