Download 6 Lecture 6: Momentum and variable

Mathematics for Physics 3: Dynamics and Differential Equations
6
45
Lecture 6: Momentum and variable-mass problems
So far we have written the second Newton law as
m
dv
=F
dt
(114)
considering a single body with a fixed mass m. A more general formulation, which allows to consider systems
composed of with several bodies, or, equivalently a body with a varying mass, is the following:
• Define the momentum of particle i as pi = mi v i . The momentum is a vector. A system composed of several
particles has the momentum which is the vector sum of all the constituent particle momenta:
�
�
P =
pi =
mi v i .
(115)
i
i
• Distinguish between internal forces and external forces. An isolated system is a system on which no
external forces act, F ext = 0; in such a system, the total momentum is conserved. Internal forces are forces
acted by one part of the system on another, these are irrelevant for the motion of the system as a whole: they
cannot change the total momentum of the system. Assume for example an isolated system composed of two
particles of masses m1 and m2 respectively. According to the II Newton law:
m1
dv 1
= F 2→1 ;
dt
m2
dv 2
= F 1→2 .
dt
However, by the III Newton law the force acted by particle 1 on 2 is equal in size and opposite in direction to
the one acted by particle 2 on 1. This means that
m1
dv 1
dv
= −m2 2
dt
dt
or
� dP
d �
m1 v 1 + m2 v 2 =
= 0,
dt
dt
namely the total momentum of an isolated system is conserved.
• Assume now a non-vanishing net external force F ext on the system. The II Newton law then states:
dP
= F ext
dt
(116)
namely, it states that the rate of change of the total momentum P of the system is equal to the external force.
Eq. (116) is a general formulation of Newtonian dynamics. Momentum conservation of an isolated system can
of course be obtained in the special case of a vanishing external force. Clearly, another special case is the one
of a body of a fixed mass m, in which case the mass can be pulled out of the time derivarive on the l.h.s. of
(116), returning to our previous formulation of (114).
The II Newton law in (116) is most conveniently used in its integrated form. Consider for example the effect of
the force between time t and time t + ∆t. Integrating (116) we get:
�
t+∆t
dτ
t
dP
=
dτ
�
t+∆t
dτ F ext (τ ) .
t
The integral on the r.h.s. of this equation is called impulse. The statement of the second Newton law is then:
P (t + ∆t) − P (t) =
�
t+∆t
dτ F ext (τ )
(117)
t
namely that the net change in the total momentum of the system over a period of time equals to the impulse of the
external force in this time.
Let us now give a couple of examples of applications of this formulation of Newton law. These are examples
in which the mass of the system is varying in time, so (116) does not reduce to (114). We will first consider mass
accretion in general, and then focus on a particular example of rain drops falling vertically within a cloud. Finally
we will examine the equation of motion of a rocket.
Mathematics for Physics 3: Dynamics and Differential Equations
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Mass accretion
Consider a system of mass m(t) and velocity v(t), which accretes, between time t and time t + δt, an additional
mass δm having velocity u(t). We assume that the time interval δt is infinitesimally small. Let us write down the
momentum before and after the accetion event to be used on the l.h.s. of (117):
P (t) = m(t)v(t) + δm u(t)
(118)
P (t + δt) = m(t + δt)v(t + δt)
�
�
�
�
= m(t)v(t) + δt m(t)v̇(t) + ṁ(t)v(t) + O (δt)2
(119)
where in the second �equation
� (corresponding to the momentum after the accretion) we have expanded in δt and
neglected terms of O (δt)2 .
Using now Newton law in (117) we compare the difference between the momenta before and after the accretion
event to the impulse of the external force:
�
�
�
�� �
m(t)v(t) + δt m(t)v̇(t) + ṁ(t)v(t) − m(t)v(t) + δm u(t) = F (t)δt
�
��
� �
��
�
P (t+δt)
�
P (t)
�
2
where we neglected O (δt) corrections. Clearly the non-infinitesimal term m(t)v(t) drops in the difference on the
l.h.s.. Dividing by δt and then taking the limit δt → 0 (whereupon δm/δt → ṁ) we get:
�
�
m(t)v̇(t) + ṁ(t) v(t) − u(t) = F (t)
(120)
This is called the mass accretion equation.
The falling raindrop
As an example of mass accretion consider a raindrop falling vertically through a cloud, picking up water vapour as
it descends. For simplicity assume that the water vapour is stationary with respect to the ground (which we use as
our reference frame) so that u = 0. We further assume that the only external force is the one due to gravity, and
that it acts directly downwards:
F = −m(t)g ŷ
where ŷ is a unit vector pointing up. The problem is then one dimensional, with v =
Because we assumed u = 0, eq. (120) reduces to
m(t)v̇(t) + ṁ(t)v(t) = −m(t)g ŷ
=⇒
dy
ŷ.
dt
�
d�
m(t)v(t) = −m(t)g ŷ
dt
(121)
or if we use a scalar variable for the velocity component downwards: v = −wŷ, we have:
�
d�
m(t)w(t) = m(t)g
dt
(122)
This is the final result given what we assumed so far. Clearly, we have two unknown functions of the time the mass
m(t) and the velocity downwards w(t), which are coupled through the mass accretion relation (122). Two functions
cannot be fully determined by one constraint. One clearly needs to further model the rate of accretion in order to
solve the problem.
One simple model for the accretion rate is that the drops accumulate more vapour in proportionality with their
mass. If this is assumed one has,
ṁ(t) = αm(t)
where α is some constant. Under this assumption we can solve the problem. We obtain:
m(t)ẇ + αm(t)w(t) = m(t)g
(123)
ẇ + αw(t) = g
(124)
so the mass drops and we get the equation:
Interestingly, this is just the same equation we got (85) for linear drag. The solution is detailed there. Alternative
assumptions for the accretion rate will be discussed in the tutorial.
Mathematics for Physics 3: Dynamics and Differential Equations
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The rocket equation
Let us now turn to another application of the momentum-impulse form of the second Newton law in (117), namely
the case of a rocket. For simplicity we shall consider the case of motion in space where one can neglect external
forces (gravity). Eq. (117) then simply implies total momentum conservation:
P (t + ∆t) = P (t)
(125)
The rocket acceleration is based on throwing the exhaust gas backwards at a high speed. The motion is along the
thrust axis, which we can assume, for simplicity to be constant in time. The problem therefore reduces to one
dimensional, and we can drop the vector notation (we assume that the positive direction is forward, along the thrust
axis).
The basic mechanism is depicted in figure 6: a certain amount of gas of mass is thrown backwards, with velocity
u with respect to the rocket. We shall assume that this velocity is constant throughout the flight. Thus, according
to (125) the remaining body of the rocket moves forward at a higher speed v + dv.
Note that the total mass is conserved. So if at time t the mass of the rocket was m(t), at (infinitesimally) later
time t + δt the mass can be obtained as a Taylor expansion:
m(t + δt) � m(t) +
dm
δt + O((δt)2 ) = m(t) + dm
dt
dm
The mass is decreasing, i.e.
< 0 so dm < 0. Thus, as shown in figure 6, the amount of mass thrown backwards
dt
is −dm > 0.
At time t, before throwing the amount −dm the momentum in some inertial frame is
P (t) = m(t) v(t)
After throwing it
P (t + ∆t) = (m(t) + dm) (v(t) + dv) + (−dm)(v + dv − u)
where the second term accounts for the momentum of the exhaust gas of mass −dm, which moves with velocity −u
with respect to the rocket, or v + dv − u with respect to the inertial frame (where the rocket’s velocity is v + dv).
Comparing the momenta before and after:
(m(t) + dm) (v(t) + dv) + (−dm)(v + dv − u) = m(t) v(t)
and therefore, upon neglecting second order effects,
m(t)dv + udm = 0
Dividing by m(t) and integrating between the initial (i) and final (f ) times, we get:
� vf
� mf
dm
dv = − u
m
vi
mi
yielding:
∆v ≡ vf − vi = u ln
�
mi
mf
�
which is the so-called Rocket Equation. It states that the difference between the final and initial velocities is the
exhaust velocity times the logarithm of the ratio between the initial and final rocket mass.
Figure 6: Rocket acceleration