6.1 Classical Mechanics
Chapter6-1
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, = m, in the x and y directions are
Fx = m,Fy = m
If frictional force is neglected, then there is no force in the x direction
m = 0 = C1x = C1t + C2
The constants of integrations C1and C2 can be determined from the initial conditions
t = 0, x = 0 C2 = 0
t = 0, = 50 cos(45o) = 35.355 m/s = C1x = 35.355t
The equation of motion in the y direction is
m= mg = gt + B1 y = 0.5gt2 + B1t + B2
The constants of integrations B1and B2 can be determined from the initial conditions
t = 0, y = 200 m B2 = 200 m
t = 0, = 50 sin(45o) = 35.355 m/s = B1
y = 0.59.81t2 + 35.355 t + 200
= 0.59.81t + 35.355
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.
Figure 6.1-1 Ball trajectory with indicated position at various times
______Table 6.1-1Matlab program to plot the ball trajectory ______
%
% 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;
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.
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
=
where the vector is directed from M to m, and G is the gravitational constant. From Newton’s second law of motion
= m
m = =
In rectangular coordinates the equations for the x and y components of the position vector become
m= m =
where . For polar coordinates x = rcos, y = rsin
= cosrsin
=
= sin + cos sinrcosrsin
= {r}cos {2+ r}sin
The acceleration can be substituted into the x component of the equation of motion
m= = =
and rearranged to give
m{r+ }cos m{2+ r}sin = 0
Similarly, the y component of the equation of motion can be rearranged to
m{r+ }sin + m{2+ r}cos = 0
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
m{r+ } = 0
m = mr
Comparing the above equation with the equation in rectangular coordinate m= , an additional term mrappears in polar coordinates. This term is known physically as the centrifugal force in circular motion.
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( = = Ivelocity), the Halmiltonian can be written as
H(p,q) = + V(q)
The Halmiltonian in terms of the velocity component (j = 1, 2, 3) is then
H(p,q) = + V(q)
= = =
For a conservative system = V, therefore the force in the j direction is
Fj =
= = Fj = m= = =
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 potential energy for this system is V = fx2. Determine the equation of motion for this one-dimensional harmonic oscillator using the Hamiltoninan formulation. The particle is initially at the equilibrium position with a velocity vo.
Solution
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)
= = fx = = m
= x x = Asint + Bcost
At t = 0, x = 0 B = 0
At t = 0, = vo = Acost = A
The constant of integration is then A = voand the equation of motion for the particle is
x = vosint
The period of oscillation is determined from t = 2t =
The frequency of oscillation is = =
1