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.

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Motion of a ball falling

This video represents the motion of a ball falling to the ground.  The motion is represented by -mg where m is the mass of the ball and g represents gravity (9.8m/(s^2)).  As the ball is dropped from reset, it slowly starts to fall at first, then picks up speed faster and faster until it hits the ground.  

This can also be represented by potential and kinetic energy.  Conservation of energy says that energy cannot be created or destroyed.  This means that As potential energy decreases, kinetic energy increases.  At the top before the ball is dropped,  this is the point of maximum potential energy.  The ball has no velocity or acceleration, so the kinetic energy at this point is 0.  As the ball is dropped, the velocity and acceleration of the ball increases causing it to have kinetic energy so the potential energy decreases.  

Projectile Motion of a baseball with wind resistance

This animation shows the projectile motion of a baseball with wind resistance.  The equation for wind resistance is represented by Fwind= -0.5CρAv^2.  C is the coefficient of performance, ρ represents the density of air, A represents the frontal area of the baseball and v represents the velocity of the wind.  Other forces acting on the baseball is shown by the equation Fgravity= -mg.  

The baseball has an initial velocity = {10m/s, 0, 10m/s}.  This means that the baseball has an initial velocity in both the positive x and z direction, but not in the y direction.  The forces opposing the baseball are wind and gravity which can be represented by the vector d = {Fwind, 0, Fgravity}.  These forces are both in the negative x and z direction, but not in the y direction.  This is not practical since wind does not only blow in one direction.  We can however set the opposition y force to 0 to better show how the baseball can be effected by the force of wind.  The wind opposes the velocity of the baseball, causing it to have a negative acceleration and changing direction of the baseball as you can see in the animation above.  

Simple Harmonic Motion

A good example of simple harmonic oscillation is in a spring system.  You can see an example of this in the video at the bottom of the blog.  These systems are represented by the formula: a+2βv ̇+w0^2x=Acos(wt). 2βv represents the dampening force, Acos(wt) represents the driving force and w0^2x represents the spring force.  The w0^2 shows the natural frequency (the frequency the oscillator will continuously oscillate at without any external force) of the oscillator while x represents the displacement.  If 2βv = 0 and Acos(wt) = 0, the motion of the spring is represented by:

This graph has no change in displacement because no outside force is acting on the system.  If 2βv does not equal 0, but Acos(wt) = 0.  The oscillator will start to get slower and slower until it comes to a complete stop.  This is because the dampening resists the motion of the spring.  This is represented by:

As you can see, the spring oscillates for a couple of seconds then comes to a complete stop.  This is because there is no driving force to help keep the system in motion.  If both 2βv and Acos(wt) do not equal 0, then the displacement of the spring will be off in the beginning but eventually even out and reach simple harmonic motion.  This is shown by the following graph:

(video obtained by physics animation channel on youtube)

Trajectory of a Basketball

This animation represents what a basketball looks like when you throw it from a high position relative to the ground.  If you just drop it, gravity will pull it straight down to the ground.  However, if you spin the basketball  towards you, the basketball will start to move forward as well as downward.  This is because F = w x v.  W is the angular velocity vector of the basketball while v is the velocity vector.  If you spin the basketball towards you and cross product this with the velocity downward, the basketball will have a resultant force away from you.  This simulation represents the basketball's path as it falls with both angular velocity and the gravitational pull to the earth.

Trajectory of a ball

This is a model of the trajectory of a ball on earth moving with an initial velocity vector of:
 { 10m/s, 10m/s, 10m/s}.