Wednesday November 17th 2010

No class today!

Mason sent out some emails to people but i guess he wasn't feeling well enough to go through his contacts so here is the message:

Family is sick today. (My daughter brought home a fever and gave it to all of us.) Please chat with your project partners and lay out a plan for how you will complete the project in the remaining time. We will start harmonic motion on Friday. Please do the Pre-Lab on
"Measuring Spring Constants"

Please pass this on to other students.

Cheers,
mmason

So there you have it guys.

Remember, Due Friday:

Pre-Lab: Measuring Spring Constants

Lab Write Up: Problem # 3 Torques, Energy, and Friction (Monday's lab)

We should prepare for a "Fiesta" aka Quiz

We should be spending some time on our Projects and making sure we spread the work evenly over the next couple of weeks.

Friday November 12, 2010

Torque

Definition: Torque is the tendency of a force to cause or change rotational motion of a body. Torque is calculated by multiplying Force and distance, so the SI units of torque are newton-meters, or N*m

Torque is represented by the greek symbol τ (tau)

τ=r×Fsinθ

When the net torque =0, it means there is no acceleration or the object is moving with a constant speed.

When an object undergoes static equilibrium it must satisfy 6 conditions:
∑ F_ x=0
∑ F_ y=0
∑ F_ z=0
∑ τ_ x=0
∑ τ_ y=0
∑ τ_ z=0

Problem 1:
Find the ∑ τ about 20 cm of a meterstick whose mass is 160 g
τ=r×Fsinθ
τ=-(.3m)*(.16kg)*(9.8m/s^2)=-.48mN
0.3m=distance from the 20cm mark to the center of the meterstick (.50m-.2m=.3m) because it is an object with a uniform mass-the center of mass is at .5m

.16kg total mass of the ruler

9.8m/s^2=acceleration due to gravity

NEGATIVE SIGN=using the right hand rule, the thumb is pointing in or clockwise; therefore, it is negative.

How much mass do we have to add in order to get equilibrium?
0.2m×(0.160kg+x(extra mass))=0.8m×(0.16kg)
x(Extra mass)=.48kg

PROBLEM 2
A man weighs 50 N, is standing on a board with uniform mass that weighs 10N and has a lenght of 5m. The board is supported with two cords (one at each end of the board)
that make a 90degree angle with respect to the board.
The man is standing two meters away from one of the cords.

What is the tension of the two cords?
T1=tension of a cord
T2=tension of the other cord
WP= person's weight=50N
WB=board's weight=10N

∑ F_ x=0
∑ F_ y= T1+T2-WP-WB=0
T1=60-T2

∑ F_ z=0
∑ τ= T1*(5m)-WP*(3m)-WB*(2.5)=0
T1=35N

T2=T1-60
T2=35

These tensions were found by using one of the ends of the board as the pivot point. The location of the pivot point does not change the answer, it can only reduce or increment the number of unknowns and the number of possible equations.

HOMEWORK

Mastering Physics
Prelab(Forces, Torques, and Energy) page 96
Project #3 was assigned:Choose a problem from the physics book and make a video of it, step-by step procedure on how everything was done, 10 min Power Point Presentation. Groups of 2 people only, pick your partner.

Advanced Gravitation Notes

http://profmason.com/?p=1392

Moments of Inertia (Oct. 25, 2010)

Moment of inertia, also called mass moment of inertia,or rotational inertia, is a measure of an object's resistance to changes to its rotation. It is the inertia of a rotating body with respect to its rotation. Moment of Inertia in SI units is kg·m².

The moment of inertia of an object about a given axis describes how difficult it is to change its angular motion about that axis. The moment of inertia depends upon how an object's mass is distributed relative to its pivot point.

For a point mass the moment of inertia is just the mass times the square of perpendicular distance to the rotation axis, I = mr². That point mass relationship becomes the basis for all other moments of inertia since any object can be built up from a collection of point masses.

For an object with an axis of symmetry the moment of inertia is some fraction of that which it would have if all the masswere at the radius r. For a continuous mass distributions it requires an infinite sum of all the mass moments which make up the whole thing, is the integration of all the mass.
In class we discussed that the dm can be expressed in many other forms for example:
dm= pdv
dm=(lambda)dL
dm=(m/L) dL
dm=(m/L) dr
dm=(lambda)dr

(lambda= linear density or the mass per unit length)

The moment inertia about the end of the rod can be calculated directly or obtained from the center of mass expression by using the Parallel Axis Theorem.
The moment of inertia about any axis parallel to that axis through the center of mass is called the Parallel Axis Theorem given by:
*Note if you know the moment of inertia of the center of mass you can find the moment of inertia about any pivot point using the parallel axis theorem, given that the axis of rotation is parallel.

Six steps to find Inertia....
1. Draw a diagram & choose the point of rotation.
2. Write the definition of moment of inertia.
3. Use density to find dm as a function of dr.
4. Plug in dm as a function of dr into the integral.
5. Choose appropriate bounds.
6. Integrate.

We also compared in class the similarities of liner and rotational in respect to inertia ...
ere is a brief summary diagram:





Here is an example of the derivation of a thin disk with x,y, and z axis.


*resources : class notes
http://hyperphysics.phy-astr.gsu.edu/HBASE/mi2.html

Class Friday October 22

Class on this day discussed mostly about rotational motion of a system.

Lecture:

Lecture in class we discussed about the pro and cons between low profile wheels and to thicker tires. Of course when we have lower profile tires the rims would be greater and therefore heavier and thus makes a car slower. Compared to a car with smaller rims and more tire the car goes faster because of the lighter weight and also because the radius of the wheel is smaller there fore requiring less force to spin. The moment of inertia of a single particle is also found to be just the product of its mass times the radius of that particle. The similarities with this compared to center mass is that in moment center mass it is MR however in inertia it is MR^2. we also had a demonstrations of different circular objects racing down a incline slope. The winner of the race was the solid circular because according to the equations of calculating the speed of the object at the bottom of the ramp we find that the values of c is smallest meaning it has the greatest speeds. The list of the order should follow as any solid sphere, any solid cylinder, any thin-walled hollow sphere, then any thin -walled hollow cylinder. We also discussed about problems of how to derive the moments of inertia of different objects such as sphere, cylinders, rods and it is important to note because the quiz on Friday will be based on those derivations.

Project:

We discussed what we should have by the end of the week and those include the need of having our planets in place and also have the launch. The need of having met up your groups are important because by this Friday which is tomorrow he is expecting to have our systems launch out from earth to at least the moon and to be within its orbit. If possible from there find into the mars orbit as well.

LAB:

During lab we had to make a video of one complete rotation by using a set of rotational apparatus that spins a horizontal beam on a stand. we draw then two dots on the beam and by making it visible we record data by using logger pro to capture the points of both circle that are plotted out then we analyze the differences in the graphs that were projected by the two dots.

The fear predates us!

Welcome to mars!

Here is a picture of the rocket as it is orbiting our friend the red planet.

This picture shows the orbits of the three main bodies, the earth is in blue, mars is in red and the rocket has a yellow trail.

As you can see in the following image, the perigee of mars orbit is 3.414e6, and the equatorial radius of mars is 3.396e6. This shows that at its closest point the rocket is orbiting a mere 18 km above the martian surface!!!
For this project there was a significant investment in mathematical modeling which had to be augmented as simulations were run. In an ideal Hohmann transfer the velocity for a shift to an elliptical orbit can be calculated, but in our simulation there was a problem. The earth keeps following the rocket!!! because of this, the rocket is continuously decelerated through it's elliptical trajectory while inside the earth's sphere of influence. In order to compensate for this fact we modified the rocket's trajectory (read: strapped more rockets on with duct tape), and were able to model a more realistic Hohmann transfer. Finally, we had issues with Mars being in the same place we predicted it to be, so we defined a variable to represent Mar's theta for the initial conditions.
-----------
If you look carefully at the second image, you will notice that mars is approximately 70-80 degrees out of phase with the earth, and this is absolutely essential. A Hohmann transfer has a very specific mission time associated with it, and once this is calculated it is a simple process to calculate how much out of phase the earth and mars must be in order to complete the transfer. Over all, this process was both challenging mentally and computationally, and I wish the rest of you good luck. I hope you are all looking forward to compensating for the effects of Jupiter and modeling rocket behavior!

TLI win

I managed to get it in a stable orbit around the moon, but I did it through trial and error. In the future, I'll have to be sure to use the formulas related to the Hohmann transfer orbit, otherwise the project would never get done!
Another level of complication was added because I also had the earth, moon & apollo going around the sun.
-Andy

Trans Lunar Injection = great success!
The most difficult aspect of this project was finding a stable orbit around the moon once the Apollo was within its sphere of influence.

If the TLI is accomplished using two separate rocket stages, the Apollo's velocity can be augmented once it is close enough to the moon. At that point you can 'fire rockets' and attempt to match the ideal velocity of the rocket to orbit the moon. The ideal velocity is a combination of the moon's velocity at the moment the orbit is to begin and the components of a stable orbit at that radius.

-Brian

Monday, October 11, 2010

What did we do in class today?

• We were put in a new group of 3 or 4 people for project #2: To Mars!
• The handout for project #2 was given
• Practiced Vpython programming (lab manual page 62-68)

Solving for Orbital Velocity of the Earth

This problem can be solved using two different approaches:

1.

2.

The way scientists determine the mass of the sun or other distant big planets is actually by observing the orbital velocity of the planet, and plugging in all the measurable quantities to the above formula.

In reality, there are plenty numbers of forces acting on the spaceship. However, we only calculate the gravitational forces on the spaceship due to the Earth, Mars, and the Sun for the purpose of this project, although there is another big planet that might also affect the spaceship like Jupiter.

You may design the landing system (Sky Crane) for an additional credit on your project.

This is the actual Sky Crane design of MSL:

You might want to visit these websites as a reference:

• http://en.wikipedia.org/wiki/Mars_Science_Laboratory
• http://en.wikipedia.org/wiki/Spirit_rover

Homework (due Wednesday, October 13, 2010):

• Write up Prelab (Conservation of Momentum)
• Mastering Physics (Assignment 17 Gravitation and Orbits)
• Start working on project #2

September 27

Power
While reading articles, it is important to read critically. There was a magazine article that talked about how the average refrigerator uses less power than a radio clock. The refrigerator used 2kw and it runs for 8 hours a day, the clock radio uses 1 watt for 24 hours. After making the proper calculations, this was proved wrong. The refrigerator uses more power than the clock radio.

P=W/t

W=Pt



Pr=2000w*8*3600

Pr=5.76x10^7J



Pc=1w*24*3600

Pc=86400J






SPRING
A spring with a mass was given an initial velocity and the spring started oscilating. The graphs of the spring oscilating was different for kinetic energy, gravitational energy, and force.
The graph for the force was linear (f=kx), the graphs for kinetic energy and gravitational energy were sinusodial (k=1/2mv^2). The sinusodial graphs had a constant pattern because the velocity was not changing significantly.

Lab
Kinetic Energy and Friction lab was done in class.

Announcements
Write up Kinetic Energy and Friction Lab or Kinetic Energy and Work (Due Friday)
Wednesday review for Midterm
Lab notebooks will be collected on Friday
Cart due on Monday Ocotber 4th