Download 27. Generalized Newton`s second law

Generalized Newton’s second law
Hiroki Okubo
P
The term F becomes the vector sum of all forces
acting on all particles of the isolated system from
sources external to the system, and f becomes the
vector sum of all forces on all particles produced
by the internal actions and reactions between particles. This last sum is identically zero since all
internal forces occur in pairs of equal and opposite
actions and reactions. By using Eq. (1), we have
00
ρi
m11
i
00
11
G
00
11
00
11
00
11
’
ρ
ri
i
ρ
r
O
rP P
X
Figure 1: Relative motion
F =m
d2 r
dt2
(4)
Equation (4) is the generalized Newton’s second
law of motion for a mass system and is called the
1 Generalized Newton’s sec- equation of motion of m. The equation states that
the resultant of the external forces on any system
ond law
of masses equals the total mass of the system times
We extend Newton’s second law of motion to cover the acceleration of the center of mass. This law
a general mass system which we model by consid- expresses the so-called principle of motion of the
ering n mass particles bounded by a closed surface mass center.
Equation (4) may be expressed in component
in space. Figure 1 shows a representative particle of mass mi of the system isolated with forces form using x-y-z coordinates. Thus,
F 1 , F 2 , F 3 ... acting on mi from sources external
X
d2 y X
d2 z
d2 x X
to the system boundary, and forces f 1 , f 2 , f 3 ...
Fy = m 2 ,
Fz = m 2
Fx = m 2 ,
dt
dt
dt
acting on mi from sources internal to the bound(5)
ary. The internal forces are forces of reaction with
other mass particles within the boundary. The particle of mass mi is located by its position vector r i 2
Work-energy
measured from the nonaccelerationg origin O. The
center of mass G of the isolated system of particles 2.1 Work-energy relation
is located by the position vector r which is given
For the entire system, the sum of the work-energy
by
P
X
equations written for all particles is
(U1−2 ) =
mr =
mi r i
(1) P
∆Ti , which may be represented by
Newton’s second law gives
T1 + U1−2 = T2
(6)
F 1 + F 2 + · · · + f 1 + f 2 + · · · = mi r̈ i (2)
P
where U1−2 =
(U1−2 ), the work done by all
forces,
and
∆T
is
the
change in the total kinetic
Similar equation may be written for each particle.
P
energy
T
=
T
of
the
system.
If these equations written for all particles of the
i
For
a
nonrigid
mechanical
system, a part of the
system are added together, the result is
work
done
by
the
external
forces
goes into changX
X
X
F+
f=
mi r̈ i
(3) ing the internal elastic potential energy Ve . If the
1
P
The time derivative P
of H O is Ḣ O = P
(r i × mi v̇ i ).
The summation is
(r i × mi v̇ i ) =
(r i × F i ),
which is the vector sum of the moments about O
of
P all forces acting on all particles of the system,
M O.
work done by the gravity forces is excluded from
the work term and is accounted for instead by the
changes in gravitational potential energy Vg , then
′
done on the system
we may equate the work U1−2
during an interval of motion to the change ∆E in
the mechanical energy of the system
′
T1 + Vg1 + Ve1 + U1−2
= T2 + Vg2 + Ve2
3.2.2
(7)
About the mass center G
The angular momentum of the mass system about
the mass center G is
X
HG =
ρi × mi ṙ i
(13)
We may write the velocity of the representative particle as
v i = v + ρ̇i
(8) By using Eq. (8) we obtain
X
where v is the velocity of the mass center G and ρ̇i
HG =
ρi × mi ρ̇i
(14)
is the velocity of mi with respect to a translating
reference frame moving with the mass center G. We The expression of Eq. (13) is called the absolute
write the kinetic energy of the system as
angular momentum, and the expression of Eq. (14)
is relative angular momentum. Differentiating Eq.
X1
X
1
1
mi v i · v i = mv 2 +
mi | ρ̇i |2 (9) (13) with respect to time gives
T =
2
2
2
X
P
because
mi ρi = 0. Equation (9) expresses the
Ḣ G =
ρi × mi r̈ i
(15)
fact that the total kinetic energy of a mass system
P
P
ρi × mi r̈ i = ρi × F i . Thus
equals the kinetic energy of mass-center translation The summation is
of the system as a whole plus the kinetic energy due
X
M G = Ḣ G
(16)
to motion of all particles relative to the mass center.
2.2
3
3.1
Kinetic energy expression
3.2.3
Impulse-momentum
About an arbitrary point P
The angular momentum about an arbitrary point
P as shown in Fig. 1 can be expressed as
X
X
HP =
ρ′i × mi ṙ i =
(ρ + ρi ) × mi ṙ i (17)
Linear momentum
The linear momentum of the system is defined as
the vector sum of the linear momenta of all of its
The rearranging gives
particles. By substituting Eq. (8), we obtain
X
H P = H G + ρ × mv
(18)
(10)
G=
mi (v + ρ̇i ) = mv
P
M P , we make use of the
The linear momentum of any system of constant In order to calculate
mass is the product of the mass and the velocity of principle of moment. Thus
X
its center of mass. The time derivative of G is the
M P = Ḣ G + ρ × ma
(19)
resultant external force acting on the system.
X
F = Ġ
(11) Equation (19) enables us to write the moment equation about any convenient moment center P .
3.2
3.2.1
Angular momentum
References
About a fixed point O
[1] J. L. Meriam and L. G. Kraige, (2001), Engineering Mechanics, Volume 2, Dynamics, 5th
edition, Wiley
The angular momentum of the mass system about
the fixed point O is defined as
X
HO =
(r i × mi v i )
(12)
2