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6.1 Classical Mechanics
In classical mechanics the trajectory of a particle moving under the influence of a force can be
completely determined from the initial position and velocity. The total energy of a many particle
system must be found from the positions and velocities of the particles as the sum of the kinetic
and potential energies. For a conservative system, the kinetic energy depends only on particle
velocities, while the potential energy depends only on their positions.
Example 6.1-1 (Rowley, “Statistical Mechanics for Thermophysical Property Calculation”)
Determine the trajectory of a 0.50 kg ball thrown from a height of 200 m into the air at an initial
velocity of 50 m/s and an angle of 45o with respect to the ground.
Solution


The components of Newton’s second law, F = m a , in the x and y directions are
Fx = m
d 2x
,
dt 2
Fy = m
d2y
dt 2
If frictional force is neglected, then there is no force in the x direction
m
dx
d 2x
=0
= C1  x = C1 t + C 2
2
dt
dt
The constants of integrations C1 and C2 can be determined from the initial conditions
t = 0, x = 0  C2 = 0
t = 0,
dx
= 50 cos(45o) = 35.355 m/s = C1  x = 35.355t
dt
The equation of motion in the y direction is
m
dy
d2y
=  mg 
=  gt + B1  y =  0.5gt2 + B1t + B2
2
dt
dt
The constants of integrations B1 and B2 can be determined from the initial conditions
t = 0, y = 200 m  B2 = 200 m
t = 0,
dy
= 50 sin(45o) = 35.355 m/s = B1
dt
y =  0.59.81t2 + 35.355 t + 200
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dy
=  0.59.81t + 35.355
dt
400
400
300
300
y(m)
y(m)
From these equations, the position and velocity of the ball can be calculated at any subsequent
time. Table 6.1-1 lists a Matlab program to determine the position of the ball at 5 equal time
intervals from the initial launching to the time the ball hits the ground. The trajectory of the ball
is mapped out in Figure 6.1-1.
200
100
100
0
100
200
x(m)
0
300
400
400
300
300
y(m)
y(m)
0
200
100
0
100
200
x(m)
300
0
100
200
x(m)
300
0
100
200
x(m)
300
200
0
100
200
x(m)
0
300
400
400
300
300
200
100
0
0
100
y(m)
y(m)
200
200
100
0
100
200
x(m)
0
300
Figure 6.1-1 Ball trajectory with indicated position at various times
__________ Table 6.1-1 Matlab program to plot the
%
% Trajactory of a ball
%
angle=pi/4;
vi=50;
yi=200;
vx=vi*cos(angle);vy=vi*sin(angle);
g=9.81;
tf=-(-vy-sqrt(vy*vy+2*g*yi))/g;
tb=[0:5]*tf/5;xb=vx*tb;yb=-.5*g*tb.*tb+vy*tb+yi;
subplot
tt=[0:60]*tf/60;
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ball trajectory ___________
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xx=vx*tt;yy=-.5*g*tt.*tt+vy*tt+yi;
xmax=vx*tf;ymax=2*yi;
for i=1:6
ii=12*(i-1)+1;
xp=xx(1:ii);yp=yy(1:ii);
subplot(3,2,i),plot(xp,yp,xb(i),yb(i),'o');axis([0 xmax 0 ymax]);
xlabel('x(m)');ylabel('y(m)')
end
Newton’s second law is convenient for calculating velocities and positions in Cartesian
coordinates. However the form of the equation may be different in other coordinate systems.
Newton’s equation of motion is not invariant under transformation from one coordinate system
to another.
y
m
r
M
x
Figure 6.1-2 Orbit of a planet about the sun
As an example we will determine the equation of motion for an object of mass m moving about a
fixed central object of mass M. The force between the two objects are given by Newton’s
gravitational law


MmGr
F =
r3

where the vector r is directed from M to m, and G is the gravitational constant. From Newton’s
second law of motion


F = ma



MmGr
d 2r
m 2 = F=
r3
dt

In rectangular coordinates the equations for the x and y components of the position vector r
become
MmGx
MmGy
d 2x
d2y
m 2 =
m
=
3
2
r
r3
dt
dt
where r 2  x 2  y 2 . For polar coordinates x = rcos, y = rsin
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d
dx
dr
= cos
 rsin
dt
dt
dt
d 
dr 
d 
d 
d 2x
=
 cos   
 r sin 

2
dt 
dt  dt 
dt 
dt
d dr
d dr
d 2x
d 2r
d 2
 d 
=

sin

+
cos


sin


rcos


rsin



dt dt
dt dt
dt 2
dt 2
dt 2
 dt 
2
d dr
d 2x
d 2r
d 2
 d 
=
{

r
}cos


{2
+
r
}sin


dt dt
dt 2
dt 2
dt 2
 dt 
2
d 2x
The acceleration 2 can be substituted into the x component of the equation of motion
dt
m
MmGx
MmGr cos 
MmG cos 
d 2x
=
= 
=
3
3
2
r
r
r2
dt
and rearranged to give
d dr
MG
d 2r
d 2
 d 
m{ 2  r 
+ r 2 }sin = 0
 + 2 }cos  m{2
dt dt
r
dt
dt
 dt 
2
Similarly, the y component of the equation of motion can be rearranged to
d dr
MG
d 2r
d 2
 d 
m{ 2  r 
+ r 2 }cos = 0
 + 2 }sin + m{2
dt dt
r
dt
dt
 dt 
2
Multiply the x component by cos, the y component by sin , and add the two resultant equations
together. The final form of the equation of motion in polar coordinates is
MG
d 2r
 d 
m{ 2  r 
 + 2 }=0
r
dt
 dt 
2
MmG
d 2r
 d 
= mr 
 
2
r2
dt
 dt 
2
m
MmGx
d 2x
Comparing the above equation with the equation in rectangular coordinate m 2 = 
,
r3
dt
2
 d 
an additional term mr 
 appears in polar coordinates. This term is known physically as the
 dt 
centrifugal force in circular motion.
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There are three common formulations of classical mechanics: the Newtonian, Lagrangian, and
Halmiltonian formulations. A very useful property of the Lagrangian and Halmiltonian
formulations is that they have the same form in any coordinate system. We will only discuss
Halmiltonian mechanics that uses position q and momentum p as the natural variables. The
Halmiltonian, H, is the total energy of the system, the sum of kinetic and potential energies. For
a single particle in a conservative system, H can be written as
H(p,q) = K (p) + V(q) = Kinetic energy + Potential energy
Since momentum is defined as mass time velocity, p = m q ( q =
dq
= Ivelocity), the
dt
Halmiltonian can be written as
H(p,q) =
p2
+ V(q)
2m
The Halmiltonian in terms of the velocity component q j (j = 1, 2, 3) is then
H(p,q) =
pj
2
 2m
+ V(q)
2pj
mq j
H
=
=
= q j
m
2m
p j

For a conservative system F =  V, therefore the force in the j direction is
Fj = 
V
q j
dq j
d ( mq j )
d( pj )
H
V
=
=  Fj =  m
=
=
=  p j
dt
dt
dt
q j
q j
All future states of the system will be determined from the initial conditions of particles’
momentum and position. The form for the intermolecular potential must also be known to
determine the potential energy.
Example 6.1-2 (Rowley, “Statistical Mechanics for Thermophysical Property Calculation”)
The particle in the mass-spring system in Figure 6.1-3 oscillates about the position x = 0> The
1
potential energy for this system is V = fx2. Determine the equation of motion for this one2
dimensional harmonic oscillator using the Hamiltoninan formulation. The particle is initially at
the equilibrium position with a velocity vo .
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Solution
Friction less surface
Figure 6.1-3 A harmonic oscillator: mass-spring system
The Hamiltonian for the system with only x direction and momentum p is given as
H(p,x) = K (p) + V(x)
 H 
 V 

 = 
 = fx =  p x =  m x
 x  p
 x  p
f
d 2x
 f 
= x  x = Asin  
2
m
dt
m
1/ 2
 f 
t + Bcos  
m
1/ 2
t
At t = 0, x = 0  B = 0
dx
 f 
At t = 0,
= vo = A  
dt
m
1/ 2
 f 
cos  
m
1/ 2
 f 
t = A 
m
1/ 2
1/ 2
m
The constant of integration is then A = vo  
f 
1/ 2
m
x = vo  
f 
 f 
sin  
m
and the equation of motion for the particle is
1/ 2
t
 f 
The period of oscillation is determined from  
m
The frequency of oscillation is  =
1/ 2
1
( f / m )1 / 2
=
t
2
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t = 2  t =
2
 f 
 
m
1/ 2