sec3.1.mws

Classical Mechanics with Maple

Section 3.1: Newton's Laws of Motion

Dr. Harald Kammerer
maple@jademountain.de

Initialisation

> restart;

> libname:="C:/mylib/m6dynlib","C:/mylib/m6dynfig",libname:

> with(linalg):with(plots):with(plottools):with(dynamics);with(figures_chapter_3);

Warning, the protected names norm and trace have been redefined and unprotected

Warning, the name changecoords has been redefined

Warning, the name arrow has been redefined

[LAGRANGE, MI, gravitational_energy, impact, kineti...

[Fig_3_1, Fig_3_2, Fig_3_3, Fig_3_4, Fig_3_5, Fig_3...

3.1 Newton's Laws

Newton's first law of motion states that a particle's velocity will not change unless a force is applied to the particle. This means that a body that is at a rest doesn't move until a force is applied. A moving body continues its motion with constant velocity unless a force is applied. This first law is a qualitative statement.

Newton's second law gets into a quantitative relationship between force and motion. In the form of an equation, the second law reads

F = m a

with the acceleration vector a , the mass m and the resultant of all acting forces on the particle F .

According to D'Alembert we can write this as

F - m a = 0

The term - m a can be regarded as a fictive force, the inertial force . This inertial force acts in the opposite direction as the acceleration. Then this second law takes the same shape as the equilibrium conditions in the solution of static problems.

The kilogram [kg] is used as the unit of mass. The unit of length and so the displacement is the meter [ m ]. Velocity is calculated as displacement divided by time and so its units are meters per second [m/s]. At last acceleration is calculated as velocity divided by time and so its units are meters per second per second, or meters per second squared [ m/(s^2) ]. The force sufficient to accelerate one kilogram by one meter per second squared is called one Newton [ N ].

Verbally spoken, Newton's second law says: "the resultant of all forces on a particle is in equilibrium with the inertial force".

To solve practical exercises it is usual to write three coordinate equations instead of the one vector equation. So we write for example in cartesian coordinates

Fx = m*xdd

Fy = m*ydd

Fz = m*zdd

(As in section 2, the derivative of a variable with respect to time is ususly denoted by a dot above the variable. Here it is denoted by an appended letter d. So we write for the coordinates of the acceleration in x-direction: xdd . Same is valid for the components in y- and z-direction.)

Desscriptions in polar or natural coordinates are also possible.

Newton's third law addresses the nature of forces. This third law states that the force resulting from the interaction of two bodies acts with equal magnitude on both of them and in the opposite directions. For every action, there is an equal and opposite reaction.

Forces that act on the particle due to geometrical boundaries that restrict the particle's motion are called passive forces .

Forces that are completely described by a force law are called active forces . Such forces are for example the weight force, the spring force and the sliding (friction) force.

Example: The Mathematical Pendulum

In this example we will analyse the motion of a mass particle with the mass m that hangs by a non-stretchable thread with length L . The motion of the mass particle is forced on a circlular trajectory around the suspension point. Here it makes sense to describe the motion with polar coordinates. In Fig. 1 the situation is shown.

> display(Fig_3_1(),scaling=constrained,axes=none,title="Figure 1");

[Maple Plot]

Friction and the mass of the thread are neglected. The motion is compledly describable with the angle f (in the figures: phi). So the system has one degree of freedom (dof). It is also usual to call such systems: single degree of freedom systems (sdof).

At first we cut the mass particle free and and consider all (active and passive) forces. The only active force is the weight force G of the mass m . The only passive force is the thread force S . We know the direction of S , because the thread can only transmit forces in its own direction.

Now we consider the mass particle in a displaced position f . The trajectory of the motion is known as a circle with radius L . From the kinematics in section 2 we know for the acceleration in polar coordinates

> a:=polar_acc(L,phi(t),t);

a := [-L*diff(phi(t),t)^2, L*diff(phi(t),`$`(t,2))]...

The radial component of the acceleration points from the mass particle to the suspension point. The inertial force Z belonging to it acts in the opposite direction, away from the suspension point. We write

> Z:=-m*a[1];

Z := m*L*diff(phi(t),t)^2

The tangential component of the acceleration points in the direction e f (ephi in the figures). Thus the inertial force belonging to it acts in the opposite direction. We write

> T:=-m*a[2];

T := -m*L*diff(phi(t),`$`(t,2))

In Fig. 2 we see all acting forces on the mass particle.

Notice: In the figure the forces Z and T are drawn in the positive coordinate direction. In reality the force T shows in the opposite direction as drawn in the figure, but this is respected by the negative sign in the definition of T .

> display(Fig_3_2(),scaling=constrained,axes=none,title="Figure 2");

[Maple Plot]

The equilibrium in tangential and radial direction yields

> EQt := -G*sin(phi(t))+T = 0

EQt := -G*sin(phi(t))-m*L*diff(phi(t),`$`(t,2)) = 0...

> EQr := G*cos(phi(t))-S+Z = 0

EQr := G*cos(phi(t))-S+m*L*diff(phi(t),t)^2 = 0

The force law for the weight force is

> FLG := G = m*g

G always points downwards. This is taken into account in Fig 2, so there is no negative sign in the force law.

Substituting the force law into both equilibrium conditions yields

> EQt:=subs(FLG,EQt);

EQt := -m*g*sin(phi(t))-m*L*diff(phi(t),`$`(t,2)) =...

> EQr:=subs(FLG,EQr);

EQr := m*g*cos(phi(t))-S+m*L*diff(phi(t),t)^2 = 0

For m <> 0 and L <> 0 we get from EQt the equation of motion for the system

> EOM:=expand(-EQt/(m*L));

EOM := 1/L*g*sin(phi(t))+diff(phi(t),`$`(t,2)) = 0

This is an ordinary differential equation of degree 2 with respect to f ( t ).

From EQr we get the force S in the thread as

> S:=solve(EQr,S);

S := m*g*cos(phi(t))+m*L*diff(phi(t),t)^2

After solving the equation of motion we can calculate the force S in dependence of the time t .

The derived equation of motion is nonlinear and cannot be easily solved. But for small angles f we can linarize the equation.

For f << 1 we make the substitutions

> subst1 := sin(phi(t)) = phi(t)

> subst2 := cos(phi(t)) = 1

One more substitution is usual. We write

g/L = omega^2

or in equivalent but more practicable manner in MAPLE

> subst3 := g = omega^2*L

So we get the linearized equation of motion

> LEOM:=subs({subst1,subst2,subst3},EOM);

LEOM := omega^2*phi(t)+diff(phi(t),`$`(t,2)) = 0

Now we get for the solution of LEOM

> sol:=dsolve(LEOM,phi(t));

sol := phi(t) = _C1*sin(omega*t)+_C2*cos(omega*t)

The integration constants _C1 and _C2 depend on the initial conditions.

At time t=0 the angle f (t) should be f0

> IC1:=phi(0)=phi0;

IC1 := phi(0) = phi0

and the angular velocity diff( f (t),t) should be f_ d _0

> IC2:=D(phi)(0)=phi_d_0;

IC2 := D(phi)(0) = phi_d_0

So we get for the complete solution

> Csol:=dsolve({LEOM,IC1,IC2},phi(t));

Csol := phi(t) = 1/omega*phi_d_0*sin(omega*t)+phi0*...

The expression w is called natural angle frequency . It is an important measure to describe periodic motions. The unit of w is 1/s (1 per second).

Sometimes one uses the natural frequency

f = omega/2/Pi

to describe the motion. f is the number of cycles of a motion within one second. The unit of f is 1 Hz (Hertz).

The reciprocal value of f is called the period.

T = 1/f

or

T = 2*Pi/omega

In our example the period is

> T:=expand(subs(1/omega=sqrt(L/g),2*Pi/omega));

T := 2*Pi*sqrt(L/g)

Here we make the third substitution subst3 back.

We see that in this system the periodic time is independent of the mass and the initial conditions.

Finally we consider the pendulum's motion with some concrete values. We know for the acceleration of gravity

> g := 9.81

The following values for the mass and the length of the thread can be changed. But don't forget that the solution is only valid for angles f << 1.

> L := 10

> m := 10

So we get for the motion

> assign(subs(1/omega=sqrt(L/g),Csol)):

> phi(t)

1.009637555*phi_d_0*sin(omega*t)+phi0*cos(omega*t)

The natural angle frequency of the motion is

> omega := sqrt(g/L)

omega := .9904544412

and the natural frequency is

> f := evalf(omega/2/Pi)

f := .1576357202

The period of the motion is

> evalf(T);

6.343739852

In Fig. 3 we see the time history of the motion over 10 periods for the following initial conditions

> phi0 := .1

> phi_d_0 := 0

> P1:=plot(phi(t),t=0..10*T, color=blue):

> display(P1,title="Figure 3");

[Maple Plot]

Fig. 4 shows the time history of the force S .

> P2:=plot(S,t=0..10*T, 0..m*g*1.2, color=black):

> display(P2,title="Figure 4");

[Maple Plot]

We see that the force in the thread is always positive. Further we see that the frequency of the thread force S is two times the frequency of the angle f . We don't discuss this phenomen further at this point because it is not subject of this course.

Fig. 5 shows a little animation of the motion of the pendulum. You can try what happens when you change the initial conditions or the mass or the length of the pendulum. To start the animation click on the picture and the "play" in the menu bar.

> display(Fig_3_3(L,m,eval(phi(t)),t,10*T),insequence=true,scaling=constrained,axes=none,title="Figure 5");

[Maple Plot]

General Strategy to Get The Equation of Motion

The above example shows the strategy to get the equation of motion for a dynamic system. Here we itemize the single steps:

1. Define an appropriate coordinate system and choose the relevant degrees of freedom.

2. Cut the mass particle free in a general position

3. Fix all passive forces with their known geometrical appointments.

4. Formulate the force law for all active forces.

5. Add the inertial forces against the direction of the positive acceleration components.

6. Formulate the equilibrium conditions in form of coordinate equations.

7. Mathematical analyses of the equation of motion by integration and suited to the initial conditions.

8. Discussion of the result.