Orbit of Two Objects

Figure 1

Figure 1 shows the orbit of 2 objects.  We can find the force between the two objects using the equation:

We give μ (or reduced mass) a value of 5, and the angular momentum, l a value of 1.  We are then given the radius r, which is: 

We are given r in terms of theta.  This allows us to take the second derivative as shown in the equation above.  We plug our new value into the equation and solve. 

In this case, we will get a long and complicated equation in terms of theta, however we are looking for the force, f(r) so we need to convert all our thetas into r's. Once have turned all the theta's back into r's and simplify everything down, we will get some answer for F(r).  In this case, we get:

This equation shows the force acting between these two objects depend on the radius between them.

Finding Hamiltonian's equations of motion

We can find the Hamiltonian equations of motion for an an-harmonic oscillator.  We are given the potential energy to be:

where k and b are both constants.

First, we need to find the Lagrangian.  The equation for this is:

 where U is the potential energy and T is the kinetic energy.  We know the potential, but we need to find the kinetic energy.  For this and most cases, we can represent this by:

Next, we need to plug this into the Hamiltonian equation, which is:

where p is the momentum, q dot is the velocity and L is the Lagrangian.  We already know the Lagrangian, so we plug it into the equation.  We have to write all velocities in terms of momentum. If we use the equation:

,

then we can solve for x dot.  we get this value to be:

Now, we plug that into the Hamiltonian equation.  After simplifying, we get our Hamiltonian to equal:

Finally, we need to find the Hamiltonian equations of motion.  To find these, we use the following equations:

  

Plugging in our Hamiltonian, we will get the following equations as our equations of motion:

Computing Inertia Tensor

figure 1

Today, we will be computing the Inertia Tensor of figure 1.  The black circle represent a mass. In this problem, they all have the same mass. First, we need to set a coordinate axis.  We can set the point were all the poles connect as the origin, with the x-axis point east, the y-axis pointing north and the z-axis coming out of the screen.  In order to compute this tensor, we need the following equation:

Next,  we need to define the length of the rods.  In this case, they all have the same length. We will call this variable a. Now we can calculate this tensor.  The i and j  values from the equation above represent the position in our matrix, meaning which section we are trying to calculate.  For example, in a 3x3 matrix, the top left number will be I11. The one directly to its right will be I12, and so on.  After plugging the Iij into each portion of our matrix, we get the following matrix:

We can check our work by looking at each axis.  If we rotate around just the x-axis, we will see there are two masses that effect our rotation.  This is the same for the y-axis.  If we rotate around the z-axis, we will see all four masses effect the rotation, which is what we found.

Ball on a Merry-Go-Round

Let's say you and your friend were sitting on opposite sides of a merry go round, and you wanted to throw a ball over to him.  Sounds simple right? The answer is no. It will be very hard to throw a ball from one side of a merry go round when it is spinning.

We can model the motion of that ball.  If your friend is sitting on the opposite side of the merry go round, you will try to throw it straight over to him.  If you do try this, however the ball will not make it over to him.  If the merry go round is spinning clockwise, the ball will end up having a curving, and if you are spinning fast enough, you can catch your own ball.

Here is an example for you.  Let's say we have a merry go round with radius of 1.5 meters, and you do 1 rotation every 3-5 seconds.  If you throw the ball directly in front of you with a velocity of 3m/s, you will see the ball have the following path.



As you can see by the animation above, the ball will not reach your friend.  It will actually curve and if you are spinning fast enough, you will catch your own pass.

Coulomb's Law

Using Coulomb's Law allows us to calculate the motion of any charged particles.  This law is represented by the following equation.

Viewing this, we see F is the force, k is a constant, q1 and q2 are the charges on the particles, and r is the distance between them.   Let's assign both q1 and q2 the charge of +0.00005.  If we hold one charge at the origin in an x-y plane, and the other we put a distance of 0.01m away on the x-axis, then we can calculate the force between them.  We can infer since they are both a positive charge, they will repel each other.  I have added in a friction force so the particle decelerates. The following two graphs show the change in position and the change in velocity of the particle.

If we look at the velocity vs time graph, we see the particle accelerates for for a fraction of a second, then slows down.  This is because eventually, the friction force will overcome the force between the two particles and the second particle will come to a stop. 

The following animation represents the motion of the second particle.  Remember the first particle is at the origin, and we just assume it is next to the visible ball at the beginning of the animation.

Finding Center of mass

In order to find the center of mass of certain objects, we use:

where M is the mass, r is a vector and dm is the what we are integrating over, in this case the mass.  Because M is the mass, we have to integrate dm in order to find out what our mass is. This is shown in the equation:

Next we need to pick a shape.  A simple example would be to pick a rectangle.  This is shown in the following figure:

The Center of mass of a rectangle can be easily determined using logic.  We know that it is the exact middle because of symmetry.  

However, we can determine this using the equations above. We can put our coordinate system anywhere around the rectangle.  In this case, we will put it in the back left corner at the bottom of the rectangle.  

First, we need to find out what dm is.  In this case, dm is represented by:

In our case.  The density is rho and the volume is l * w * h.  After plugging this back into the integral, we figure out that the C.O.M is L/2, w/2, and h/2.  This is what we predicted above, but now you know the reasoning!

We can also do this with more complicated figures.. A cone for example.  

As you can see, this shape is much harder to determine what the center of mass is.  If we use the integral above, we can figure what it is quite easily.  If we plugged in the values of a cone in the above integral, we would find out that the C.O.M = h/4, where h is the height of the cone.

Normal mode

The following videos show a spring-ball system of balls with equal mass. There is a spring force that has a magnitude of k in between each of the balls.  The point of this system is to show all the different ways each ball can move to where the balls do not move up or down the x-axis.  Each of the following videos shows all the possible solutions to which the balls can move.

MatrixForm[BoxForm`e$]]]\);