sec4.1.html

sec4.1.mws

Initialisation
4.1 Systems of Mass Particles
- 4.1.1 The Center of Mass
  - Example: Motion of A Car With Trailer
  - The Case of Rigid Bodies
- 4.1.2 The Angular Momentum of Systems of Mass Particles

Classical Mechanics with Maple

Section 4.1: Systems of Mass Particles

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_4);

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

4.1 Systems of Mass Particles

Later in this section we consider the motions of rigid bodies. To transition from the mass particles in section 2 and section 3 we consider now systems of mass particles. We can imagine every real rigid body as a system of mass particles which are held together by inner forces. In the same manner we can consider the system of the earth and the moon as a system of two mass particles that are hold together by gravitational forces. In relation to the distance between them their own dimensions are approximately negligible. The following calculations are valid for all kinds of systems of mass particles, especially when the mass particles are close together in the case of rigid bodies. So the following considerations are very useful for later calculations for rigid bodies.

4.1.1 The Center of Mass

We consider a group of some mass particles, which are held together by rigid or elastic boundaries, as shown in Fig. 1.

> display(Fig_4_1()[1],scaling=constrained,axes=none,title="Figure 1");

When we cut the boundaries we have to add the corresponding forces between the mass particles, as shown in Fig. 2

> display(Fig_4_1()[2],scaling=constrained,axes=none,title="Figure 2");

Now we consider two neighboring mass particles and . According to Newton's Third Law , (see Fig. 3)

> display(Fig_4_1()[3],scaling=constrained,axes=none,title="Figure 3");

Additionally we define .

We consider now the freely cut mass particle . The resultant of all acting active forces on it is . The position vector of the mass particle is given by and its acceleration is given by with . From Newton's Second Law we get

When we consider all mass particles we get n such equations. We sumarize all of them and get

We know from the statics about the center of gravity of a system of particles

with the total mass of the system and the position vector of the center of gravity . The acceleration of the center of gravity is defined by . This yields for the left side of the equation above

On the right side of the equation we insert

where is the resultant of all active forces that are acting from the outside on all the n mass particles.

At last we have the double sum on the right side of the equation. This double sum is zero, because all inner forces are annul by pairs.

So we get the first axiom of the center of gravity as

This means that we can consider the motion of the center of gravity of a system of mass particles as the motion of a mass particle with the mass equal to the sum of all the single mass particles.

Example: Motion of A Car With Trailer

Now we consider the motion of a car with a trailer. Of course we consider a simplified model. Both the car and the trailer are considered as mass particles. The car has mass m1 , the trailer mass m2 . Both are moving without friction on the road. The interconnection between car and trailer is elastic and described by an elastic spring with the stiffness k . In Fig. 4 the situation is shown.

> display(Fig_4_2()[1],scaling=constrained,axes=none,title="Figure 4");

The distance between car and trailer is defined to be L when the spring is free of stress The Position of mass m1 is described by x1 , the position of mass m2 by x2 . The position of the center of gravity is described by xs. The accelerations are given by

and

First we use the axiom above to describe the motion of the center of gravity of the total system. Because there are no active forces in the horizontal direction we get for the equation of motion of the center of gravity

>

Dependent on the initial conditions is the center of gravity of the system moving with constant velocity. The inner force, here the spring force, has no influence on the motion of the center of gravity. We only need to consider it when we want to describe the motion of the separate mass particles. Cutting both masses free yields Fig. 5.

> display(Fig_4_2()[2],scaling=constrained,axes=none,title="Figure 5");

Here the spring force is denoted by F . So we get the two equations of motion

>

In the situation that the spring is unstressed the distance between both mass particles is defined as L . The relative displacement between m1 and m2 is given by

>

and the force law for the spring force is

>

We multiply both equations of motion with m2 respectively with m1 and calculate the difference of both. This yields

> eomr:=eom1*m2-eom2*m1;

After some rearrangement and substituting the relation for the spring force we get

> eomr1:=subs(subs2,sort(sort(expand(eomr/m1/m2),a1),a1));

From subs1 we get for the relative acceleration between car and trailer udd

>

> eomr2:=collect(expand(algsubs(subs3,eomr1)),u);

This equation is a differential equation of degree 2 with respect to the relative motion u between both mass particles. At this point, we don't want to go into detail about the solution of this kind of differential equation. But notice that this is a very important kind of differential equation which describes the nature of vibrations. Whithout special indication we had considered the linearized equation of motion of the mathematical pendulum and its solution in section 3. Because we use MAPLE for this course we can complete these example without supposing that everyone has the necessary mathematical backround to understand this solution in detail.

To solve the equation above we substitute

> subs4 := udd = diff(u(t),`$`(t,2))

and

>

Finally we use the natural angular frequency

>

Then we get from eomr2

> eomr3:=subs({subs4,subs5,subs6},eomr2);

eomr3 := omega^2*u(t)+diff(u(t),`$`(t,2)) = 0

Additionally we presume that we know the initial contitions of the motion. Let the relative displacement between car and trailer at time t=0 be u0 and the relative velocity at this time ud0 . Then we write

>

The next step is done by the appropriate MAPLE-command and we don't look at this process in detail. Please see the MAPLE Help for details about the use of the operator D and the function dsolve.

We get the solution for the relative motion by considerating the initial conditions ic1 and ic2 :

> sol:=dsolve({eomr3,ic1,ic2},u(t));

We see that the motion between car and trailer is a vibration. In reality we have this phenomenon for example in trains. The use of dampers reduces the vibrations between the wagons.

The Case of Rigid Bodies

When a system has constant distances between all mass particles, we denote the system as a rigid body . This means that the axiom of the center of gravity is also valid for rigid bodies. In such rigid bodies the center of gravity is fixed. Now we see that we made in the previous considerations no error, because we considered always the motion of the center of gravity.

In general of course the resultant Fres of all active forces doesn't point through the center of gravity ( cog ). But we can replace every resultant which doesn't point through the center of gravity by a parallel force trough the center of gravity and an appropriate moment Mres , as shown in Fig. 6

> display(Fig_4_3(),scaling=constrained,axes=none,title="Figure 6");

The next section deals with the effect of this moment on rigid bodies

4.1.2 The Angular Momentum of Systems of Mass Particles

> unassign('F'):

In section 3.3 we considered the linear momentum. Now we consider the case that there acts not only a resultant force on the center of gravity of a mass particle or a system of those but also a resultant moment (rotation), as shown in Fig. 6 above.

At first we have a look at the effect of a moment and the motion of a single mass particle . In Fig. 7 we see the mass particle m with its position vector r with respect to the inertial system with origin O . We consider here the special case of planar motion. We can do this by choosing the coordinate system accordingly. This makes the representation concise. But the following calculations are also valid for special situations.

> display(Fig_4_4(),scaling=constrained,axes=none,title="Figure 7");

[Maple Plot]

The position is described by the vector

>

The force F acts on the mass particle

>

Then the moment

>

is the moment on the mass particle acting force with respect to O .

Now we take Newtons second law

>

with the second derivative of the position vector r with respect to time rdd = diff(r,`$`(t,2)) .

>

Accordingly we write for the velocity vector rd = diff(r,t)

>

Taking the cross product of Newtons second law with the vector r yields

>

Notice that the cross product of two vectors yields a vector orthogonal to both of them.

For the left side

we can write

The term is 0 , because the cross product of two parallel vectors vanishes.

We define

as the angular momentum . So the axiom of angular momentum is given by

Next we consider a system of n mass particles . The angular momentum of the total system is

Lo := sum(crossprod(r[i],m[i]*rd[i]),i = 1 .. n)

Differentiating with respect to time yields

diff(Lo,t) = sum(crossprod(rd[i],m[i]*rd[i]),i = 1 ...

and because we know that we get

diff(Lo,t) = sum(crossprod(r[i],m[i]*rdd[i]),i = 1 ...

with the mass , acceleration and position of the mass particle .

The resultant force which acts on the particle should be , and all the inner passive forces are sum(F[j,i],j = 1 .. n) This means

m[i]*rdd[i] = F[i]+sum(F[j,i],j = 1 .. n)

Thus we get

diff(Lo,t) = sum(crossprod(r[i],F[i])+crossprod(r[i...

The sum of moments of all inner forces are zero, because the inner forces cancel each other.

We get the axiom of angular momentum

diff(Lo,t) = sum(crossprod(r[i],F[i]),i = 1 .. n)

Here is the resultant moment of all forces acting on the total system from outside with respect to the fixpoint O .

So we have deduced the axiom of angular momentum for systems of mass particles. The transition to rigid bodies happens as follows. We consider the rigid body as a system of mass particles with the position vector and velocity rd[i] = diff(r[i],t) . From the definition above for the system of mass particles we get by the transition from the sum to the integral

So we get for the axiom of the angular momentum for rigid bodies

This formulation was first time done by LEONHARD EULER. We need to continue this subject later. But first we need to make some additional considerations.