The Physics of Playing the Saxophone

Last Friday was the Stage Band winter concert, where I played the baritone Saxophone. There was a lot of physics involved in this concert.

First of all a lot of sound waves were being produced. A sound wave consists of condensed and rarefracted areas of high and low air pressure respectively. This wave is made up of particles that travel in simple harmonic motion. The note that the audience hears during a concert depends on the frequency of the note, which is one over the period of the sound wave.

While playing the saxophone, I am able to create music because of the vibrations going through my instrument. A saxophone has a mouth piece with a wooden reed that vibrates when air is blown through it. This air is then pushed through the rest of the instrument and different notes are made because for different notes, different keys are held closed, therefore changing the amount of air that is allowed to go through the instrument. For higher notes (higher frequency) air is allowed to escape towards beginning of the instrument and for lower notes (less frequency) air is not allowed to escape the instrument until it reaches the end. The less time the air spends in the instrument the higher the frequency and pitch, as it there is less time for the air to slow down.

Pan Pacific Robotics Competition

This Thursday, Friday, and Saturday was the Pan-Pacific Robotics Tournament. 86 teams from Hawaii, California, and China came and competed in a giant convention of physics! A majority of the designs involved some sort of scoop and basket that dumped balls over the wall in the center of the field. In order to collect the balls, the scoop must apply an upward force that is equal but opposite to the weight of the ball(s) that it is trying to pick up. This causes the balls to be lifted at a constant velocity. However, our scoop rotated around an axis so as the balls were moving they were changing in their angular position so they were accelerating because velocity is speed and direction. Also, as the balls were being raised their potential energy was increasing as PE is mgh. When the balls were dumped into the basket and were driven to the wall, they were being displaced, but since force was being applied perpendicularly to the ground (direction of motion) no work was being done. Work is Force times displacement.

Building Robots

This week in robotics we completely dismantled the bots we took to the Oahu Regional and began building two new ones for this week's Pan Pacific Championship. As I was trying to loosen a screw that was accidentally placed in the wrong place, I realized that my new knowledge of torque came in handy. In robotics we tighten (and loosen) screws with L shaped Allen Wrenches and I first I attempted to remove the screw by holding the short end of the wrench an twisting with the long end in the screw. Sadly, I realized that the screw was in too tight and I could not get it out. Then, naturally I flipped the wrench over and tried twisting the screw while holding the longer end. This worked well and immediately after it occurred to me that I had just used the principles of torque to solve my problem.

Torque is the product of force and a lever arm (perpendicular length or sometimes the radius), and by turning the wrench over, I got a grip on the wrench that was farther away than it had been, and therefore even though I was applying the same amount of force, I was getting more torque and was able to loosen the screw.

About ten minutes after this incident, we took the robot over to our practice field to see if it would be tall enough to knock a football down from the post that is on the field. While the robot was almost tall enough, it did not have quite enough force to knock down the ball. The ball sits on the post with a static friction force that is equal to the coefficient of static friction times the normal force exerted by the post on the ball. In order to knock down this ball, our robot would have to exert enough force to overcome the static friction and make the ball move.

Driving in Circles

For the past few months I have had my driver's permit and have been learning to drive. Learning about circular motion these past few weeks has helped me to explain some of the phenomena I feel while driving. Every time I drive I have to go through a roundabout on Ford Island in order to get home or leave the island. Here I have experienced circular motion. While turning through the curve I always seem to be pulled to the opposite side as the way I am turning. This is because when I turn my body continues in the straight line that it was originally moving while the car turns. The reason that I am able to turn with the car is because I am wearing a seat-belt which applies a force to pull me in the direction that the car is turning.

Also, as the car is turning, it has a tangential velocity that is equal to the radius of the circle times the angular velocity. Even though I may stay the same speed while turning in the circle, my acceleration is changing because acceleration is change in angular velocity which is speed and direction. Because I am continually changing direction, my acceleration is constantly changing.

Also, there is a net force directed radially inward of mv^2/r which is the centrepetal acceleration in this case.

Moving the Mats in the Lower Gym

This Wednesday was the East Oahu Robotics Tournament. It was a lot of fun and the day was filled with physics. The competition was in the Iolani Lower Gym, so afterward we stayed to help clean up. Our task was to roll up the floor coverings and put them into a box located by the door. It turns out that those mats are pretty heavy and take at least two people to lift one.

The mat has a weight (mass times gravity) directed downward and when it is on the ground, a normal force pointing upwards that is equal (but opposite) to the weight. In order for a person (or two

) to lift this mat they have to apply a force that is greater than the weight, thus resulting in an upward acceleration of the mat. This is shown in Newton's second law, F=ma (force =mass times acceleration.

However, once the mat is in the air a force equal and opposite to the weight must be applied to keep the mat stationary (F=ma, a=0, when upward and downward forces are equal). Also, though it may seem as though work is being done as the mat is moved to its box, the force is applied perpendicularly to the motion of the mat so no work is being done.

A Spinning CD

Have you ever opened a CD player while the CD was still spinning? I know I have, and with physics I can now explain in more detail the properties of the spinning CD. First, as you open the player, the CD begins to slow down, or accelerate in the negative direction, but continues to spin until it is stopped by the friction between it and the bottom of the CD player.

If you look at the words or pictures on the face of the CD, and track, lets say one letter as it spins, you can determine its tangential acceleration. Tangential acceleration is the radius of the circle on which the letter is spinning times the angular acceleration (the change in angular velocity over a certain time). Because the CD is slowing down the Tangential Acceleration vector is in the opposite direction as the Tangential Velocity vector and is tangent to the circle of motion.

Also, if there are words closer to the center of the CD than the word you are following, they are spinning with the same angular velocity (delta theta/time) as the letters father away, but because they have different radii, their tangential velocity (vt=(r)omega) and therefore tangential acceleration (at=r(alpha)) are different.

Small Boat Big Load

This morning I looked out my window and saw this small boat pulling this rather large dock. Let's analyze how this was done. The boat and its load were traveling to the right, in the positive direction so naturally the boat applied a force in the positive direction on the dock. There was only horizontal acceleration (not vertical) so the weight of the boat was equal to the normal force of the water. The acceleration of the boat is determined by Newton's Second Law that states that net force equals mass times acceleration. In this case the net force on the dock was the tension applied by the boat on the rope connecting it and the dock. This force was strong enough to overcome the (minimal) force of the resistance of the dock in the water.Also the current was flowing in the same direction as the boat was traveling so the current added to the velocity of the boat and dock together.

Here brings up another interesting point. I know realize that while I was watching the boat I was standing in a certain reference frame and that my perception of the velocity of the dock and boat was different than that of say someone either in the boat or in another boat floating in the current of the water.

Opening a Curtain

Wow! Physics certainly is in everything we do. I say this because opening a curtain is one of those things that we do everyday without thinking about how or why we are able to do it.

In fact, opening a mini-blind has a lot to do with physics. First, the curtain is set on a pulley that allows a person to open the blind from the ground and with less force than

would be required without the pulley. This pulley shows the physics concepts of tension and Newton's second law, net force equals mass times acceleration. In order to open the curtain a

person must apply force to their side of the pulley that is greater than the downward force

of the mass times gravity (weight) of the mini-blind. The force creates tension on the rope and because the rope is the same in the handle and the rest of the curtain the tension transfers to the curtain and as soon as the force becomes more than the weight, the curtain will begin to accelerate upwards.

Because curtains have a large surface area, the rope that serves as a handle on one side of the pulley must be evenly distributed throughout the curtain. This explains the web of string

you see in a curtain as the evenly distributed string

allows the upward acceleration due to tension to be supplied

throughout

the curtain, thus making both sides accelerate upwards at the same time.

Chariot Collisons

This chariot colliding into a wall and then into a pillow is an excellent example of impulse and momentum, two concepts that we are now learning in Physics. When I was younger, I used to love playing with legos and toy cars, and now with knowledge of physics I can better explain what happens when they collide with a still object.

When the chariot collides with the wall, the force of its impact is greater than the force of its impact

when it collides with the pillow. This is because force equals the impulse (change in momentum) divided by

the change in time. When the chariot hit the wall the time it takes to change momentum is less than when it hits the pillow because the pillow slows the time that the chariot takes to impact, and therefore it changes momentum at a slower rate. Because the impulse is the same in both scenarios, when the chariot collides with the wall, its force of impact is greater than when it collides with the pillow because the impulse is divided by a smaller time, therefore making the force greater.

Pushing A Vacuum

Pushing a vacuum, though we may not think so, requires a lot of physics. First, the vacuum itself provides an excellent opportunity to draw a free body diagram where all of the forces on the vacuum are

labeled (see diagram).

When pushing a vacuum, you must first apply enough force (which must be compnonentized to find the horizontal component, since you apply force to a vacuum at an angle) to overcome the force of static friction (Fsmax=us(coefficient of static friction) x Fn). This will cause

the vacuum to accelerate forwards as the force of kinetic friction comes into play (kinetic friction opposes the direction of motion of the vacuum). The acceleration of the vacuum can be found using Fnet=ma, F-fk=mass x acceleration, acceleration=(F-fk)/mass.

Also because I am moving the vacuum forward, I am doing work on the vacuum. Work= force x displacement. Because I am applying force at an angle I must componentize the force vector to find the horizontal component of force, so Work=Force(cos(theta)) x Displacement. However, if I were applying force on the vacuum, but it was not moving, even though it may feel as though I am working hard at trying to move the vacuum, I would not be doing any work because their is no displacement of the vacuum. In order for work to occur, their must be both displacement and force.

Raising a Flag

This week as I was driving home I realized that there are about ten United States flags within a mile of my house, and all of them are raised and lowered everyday. Why is it possible to raise these flags while standing on the ground? Pulleys of course, I've always answered myself, but in physics these past few weeks I have come to understand how and why a pulley works. In general, a pulley sits at the top of an object, in this case a flag pole, and has a rope on top of it, crossing to either side (in the case of a flag the ends of the rope are connected to make a circle so that the flag will slide straight up and down the pole, but

this has a minimal effect on the applicable physics concepts). In this case a flag is attached to one end of the rope and a person holds onto the other. In order to raise the flag the person must apply force to

his side of the rope. The application of this force creates tension in the rope. In order

to make the flag accelerate upwards, the tension in the rope must be greater than the weight (mass times gravity) of the flag (assuming that the rope has negligible mass). This can be explained by the equation net force=mass x acceleration. When the tension has overcome the weight of the flag, the flag will accelerate upwards until the person stops putting force on the rope, at which time their will no longer be tension on the rope and the flag will stop moving. Fnet=ma, 0=ma, a=0.

The Physics of Robotics

Recently I have realized that Robotics and Physics are strongly intertwined. I am on the Iolani School Robotics team, and in order to construct robots that can successfully complete the task required by the comepetition game, physics must be taken into account. In this year's game their is a square field that is divided in half by an eleven inch wall that has a four inch gap between the bottom of it and the floor of the playing field. On the field their are 31/2 inch high sphirical balls, 5 inch high footballs, and about 10 inch high round balls. You win this game by having the least number of balls on your side as possible at the end of the match. This means that our robot has to get as many balls off of our side and onto our opponent's side as possible during the two minute match.

We are currently building three robots to enter in next month's competition. The robot that I am working on is basically a mobile wall of conveyor belts designed to capture balls against the wall and push them up with the force of the rotation of the belts. We also have to panels that slide in and out, with rubber bands, on the front of our robot, used for capturing the balls in front of the conveyor.

Currently in physics we are studying force and motion. These concepts greatly affected our robots this week as we realized that the amount of force transfered to the ball from the conveyor was very important in achieving an end result of the ball traveling up and over the wall.

Our robot can easily roll the 31/2 inch balls under the gap in the wall and onto the other side, but when it comes to the footballs that are slightly larger than the gap, things get a little more tricky. If the force of the robot pushing against the wall is to great, the football will be wedged under the gap instead of going up and over the wall. The solution to this problem came from what we learned about force in physics this week. All we have to do in order to make sure the ball goes over the wall instead of getting wedged under it is to make sure that the rotating upward force of the conveyor belts is greater than the force of the robot pushing against the wall. Also, at the same time this forward force of the robot cannot be so strong as it pushes the ball under the wall.

The picture above is a picture of our unfinished robot. The wall of conveyor belts will go on the front (left in the picture) and the sliding panels will go on the sides where the sliders are located.

Emily's Physics blog

Wow! I can't believe we have already made it through four weeks of physics class! This class is definitely one of the hardest classes I have ever taken, but it is also exciting and different from chemistry and biology! I am excited to learn about physics because it is all about the way that the world works, friction, motion, gravity, etc.

So far, my favorite part of physics is the lab work. I enjoy the labs because they help me to visualize and understand concepts that I read about in the homework and they put the equations into practical use.

The homework and problem sets are pretty difficult and it takes a lot of hard work to get through them, but once I finally finish, I know that I have learned something new, and that is exciting!

Physics is really difficult, but so far it has been ok in terms of grades. I now that it is going to get harder and I am really worried about what is going to happen in the future, especially with the tests. I am worried about when the tests become a single period and also about the AP exam. But that I guess I will have to worry about when it comes. For now I think the hardest part is the homework, which is always a struggle to finish, and I'm glad I have the opportunity to ask other people for help on the problems I don't understand.

Ah, well, physics is going to be challenging, but I think it will be fun! My picture shows that physics is like a tight rope, I am nervous about it but so far it is ok.