SO, because everyone else decided to take a break, my brain also decided to go off on holiday for a while. In that case, I've decided to make this blog post 100% fun (yet still on topic), and what's more fun and physics related than comics? Hope these tickle your funny bone, have a great break everyone!!!
...And finally, a comic that Mr. Fullerton would enjoy:
Have a great break! Make sure to relax (and pretend that midterms aren't coming up)!
Credit to Mr. Powlin (who read this last year about the same time) and Snopes.com, where I found this humorous commentary once again. For those of you who did not hear this last Christmas or those who want to get into the spirit of the physics-filled holiday season, I thought I'd post this up for a few giggles. Happy Holidays, all! :snowman:
No known species of reindeer can fly. BUT there are 300,000species of living organisms yet to be classified, and while most of these areinsects and germs, this does not COMPLETELY rule out flying reindeer which onlySanta has ever seen.
There are two billion children (persons under 18) in the world.BUT since Santa doesn't appear to handle the Muslim, Hindu, Jewish and Buddhistchildren, that reduces the workload to 15% of the total — 378 million according to Population ReferenceBureau. At an average (census) rate of 3.5children per household,that's 91.8 million homes. One presumes there's at leastone good child in each.
Santa has 31 hours of Christmas to work with, thanks to thedifferent time zones and the rotation of the earth, assuming he travels east towest (which seems logical). This works out to 822.6visits per second.
This is to say that for each Christian household with goodchildren, Santa has 1/1000th of a second to park, hop out of the sleigh, jumpdown the chimney, fill the stockings, distribute the remaining presents underthe tree, eat whatever snacks have been left, get back up the chimney, get backinto the sleigh and move on to the next house. Assuming that each of these 91.8 million stops are evenly distributed aroundthe earth (which, of course, we know to be false but for the purposes of ourcalculations we will accept), we are now talking about .78 miles per household, a total trip of 75½ million miles, not counting stops to do whatmost of us must do at least once every 31hours, plus feeding andetc.
This means that Santa's sleigh is moving at 650 miles per second,3,000 times the speed of sound. For purposes of comparison, the fastestman-made vehicle on earth, the Ulysses space probe, moves at a poky27.4miles per second — a conventional reindeer can run, tops, 15 miles per hour.
If every one of the 91.8 million homes with good children were toput out a single chocolate chip cookie and an 8ounce glass of 2% milk, the total calories (needless to sayother vitamins and minerals) would be approximately 225 calories (100 for the cookie, give or take, and125 for the milk, give or take). Multiplying the number of calories per houseby the number of homes (225 x 91.8 x 1000000), we get the total number ofcalories Santa consumes that night, which is 20,655,000,000 calories. To breakit down further, 1 pound is equal to 3500 calories. Dividing our total number of caloriesby the number of calories in a pound (20655000000/3500) and we get the numberof pounds Santa gains, 5901428.6, which is 2950.7tons.
The payload on the sleigh adds another interesting element.Assuming that each child gets nothing more than a medium-sized lego set (twopounds), the sleigh is carrying 321,300 tons, not counting Santa, who isinvariably described as overweight. On land, conventional reindeer can pull nomore than 300 pounds.Even granting that "flying reindeer" (see above) could pull TEN TIMESthe normal amount, we cannot do the job with eight, or even nine. We need214,200 reindeer. This increases the payload (not even counting the weight ofthe sleigh) to 353,430 tons. Again, for comparison, this is four times theweight of the Queen Elizabeth.353,000 tons traveling at 650miles per second createsenormous air resistance — this willheat the reindeer up in the same fashion as spacecraft re-entering the earth's atmosphere. The lead pairof reindeer will absorb 14.3QUINTILLION joules ofenergy. Per second. Each.
In short, they will burst into flame almost instantaneously,exposing the reindeer behind them, and create deafening sonic booms in theirwake. The entire reindeer team will be vaporized within 4.26 thousandths of a second. Santa, meanwhile, will besubjected to centrifugal forces 17,500.06 times greater than gravity. A250-pound Santa (which seems ludicrously slim) would be pinned to the back ofhis sleigh by 4,315,015 pounds of force.
In conclusion: If Santa ever DID deliver presents on ChristmasEve, he's dead now.
*yawn* It's a beautiful Tuesday morning and you've awoken from camping in the jagged pass. You stow your tent into the key items pocket and continue on your trek to Lavaridge Town. You're on your merry way, thinking fondly of a dip in the hot springs, when the grass in front of you begins to rustle!
Oh my, a Spoink appeared! Adrenaline pulses through your veins as you shout, "Go, McNugget!" (Mc.Nugget is none other than your lvl 98 torchic).
You quickly break out your pokedex, which informs you that a spoink's tail can stretch up to .3 m from when it rests at equilibrium. You also remember hearing from a passing hiker that it's angular velocity between attacks is 6 rad/sec. Due to torchic's smaller size, it can only use scratch when spoink is closest to the ground. Asuming that torchic can strike as soon as you order him to attack, how soon after spoink starts oscillating from equilibrium position (moving up first, then down) should you tell torchic to use scratch?
...And for those of you who wonder why you didn't delete scratch for a cooler move, I think scratch is the bomb.
Respond now, a cookie is on the line! (This one is larger than the one awarded from the first challenge )
PRESSURE'S ON: First person to answer this correctly gets a cookie. :eagerness:
You're at the playground with a girl you babysit, little Tori McTorque. Being 9 years old and devious, Tori took you wallet and threatened to spend your babysitting money on ice cream and root beer. Kids these days! You chased her over to the see saw, where she and her friend (Lil' Newton) sat happily on one side. You have to think of a way out of this! You don't want all of your hard-earned cash to go to waste, do you? Because physics is always the answer, you decide to make a bet with the little devils.
You propose to Tori that, if you can make the see saw balance with three people on it on your first try, she must give back the wallet. If you fail, Tori will get unlimited ice cream privileges! D:
Knowing off the top of your head that the average weight of 9 year old children is roughly 28 kg and that Tori and Newton are sitting 1 m and .7 m away from the center of the see saw, how far should you place a 61 kg teenager with a mullet from the pivot point?
...can all be found at a fencing tournament! It's about time that fencing found it's way onto this forum. Fencing is an Olympic sport consisting of three weapons, epee, sabre and foil. In foil and epee, the opponent must hit their opponent's target area with their tip in order to score a touch. In sabre, the fencer may hit with the tip and/or the side of the blade to score a touch.
I stumbled upon these fencing related physics applications by Ann McBain Ezzell, an MIT alumini. GIve the questions a shot, but if nothing else, read through them as they are quite humorous.
A few comments on the question's content:
1. Fencers scream/yell during bouts. Odd, but true. A fencer may do this to celebrate a touch, frighten their opponent or convince the referee that they scored a touch. (Some sound like howls [Div 1 men's foil], others like pterodactyls [youth 12 women's epee]. My favorite yell is "YAZEE!," used frequently by a fencer at the University of Rochester).
2. Fencers sometimes thow their equipment when they are angry. If they do, the referee will likely black card the fencer and they are removed from the tournament (I've seen it happen, it's both frightening and comical).
3. Most of the terms used below are actual names of fencers, equiptment, etc. For example, Peter Westbrook is the founder of the Peter Westbrook Foundation in New York City, an organization allowing people to fence who normally would not be able to afford it.
Here is Ann's mock exam. I hope you have as much fun with this as I did!
FENCING PHYSICS FINAL
27 April 1989 - updated 11 December 1994
[Disclaimer: All similarities between real fencers and characters in this exam are purely intentional and completely without malice.]
Instructions: Answer all questions. Be sure to show your work (including, where appropriate, free body diagrams). Don't screw up the math. Except as noted, you may neglect air resistance and friction.
1. A 2m tall Italian epee fencer loses his last repechage bout by being pushed off the end of the strip (standard 14m length). He knocks his mask straight into the air and simultaneously kicks his reel, which had been positioned at the end line, towards the other end of the strip. The mask just touches the 6m high gym ceiling before starting its downward descent. The fencer sees the reel barely clear the head of the 1.75m tall referee, who is standing in front of the scoring table recording the result. Just as he is knocked unconscious by his plummeting mask, he sees the reel land at the feet of the chairman of the Directoire Technique, who had been watching the bout from the far end of the strip.
a) How long does it take the reel to reach the ground?
Calculate the initial magnitude and direction of the reel.
c) How long will it take after the fencer regains consciousness until he is expelled from the competition?
2. Claus Block is bouncing up and down two meters from his opponent's end of the strip. His reel has slipped to 1.5 meters in front of his end line, and the reel cord is attached to his waist 1m above the ground. The mass of the exposed portion of the reel cord is 500g. A standing wave of three loops is being produced in the reel cord.
a) If Claus hits the ground 10 times per second (it's the finals), what is the tension in the reel cord?
Assume that the tension in the reel remains as calculated in part (a). Where would Claus have to stand and bounce, relative to his initial position, to produce a standing wave with only two loops?
3. A brand-new Uhlmann epee point is constructed such that the total travel is exactly 1.5mm, and it just passes the 0.5mm shim test. When a test weight of 750g is gently dropped onto the tip, the scoring machine light comes on. After the machine resets, the light remains off. However, any further depression of the tip causes the light to come on.
a) Calculate the spring constant (k) for the point spring (you may neglect the mass of the tip).
A Russian point is dimensionally identical to the Uhlmann point, but friction in the point produces an extra 1N of resistive force. Since its owner cannot readily fix his weapons, the point spring must be strong enough to lift 2kg (as above), to ensure that his weapons will never fail on the strip.
Calculate the spring constant (k') required for this point spring.
The two weapons are fixed horizontally, tip to tip, then the retaining screws are removed to allow free movement of the tips. The two tips are displaced 0.5mm from their equilibrium position and then released.
c) Calculate the frequency of the resultant SHM. (Assume that the mass of 1 tip is 1g and that both tips move together.)
4. Yuri Rabinovich and his long-lost identical twin brother Pavel (each with mass 65 kg) are fencing sabre. With weapon arms half-extended, they launch simultaneous fleche attacks and lock bell guards in mid air. Just before impact, each is traveling at a speed of 5m/s. When their bodies pass, the centers of mass are 1m apart. The bell guards remain locked and their arms extend to full length (adding 1m to the distance between the centers of mass).
a) What is the angular momentum of the resultant tangle immediately following the collision?
When the arms are extended, what is their rotational frequency in revolutions per second?
5. In the midst of a team free-for-all, Frank MacKenzie (mass 90 kg) picks up Lara Tomasso (mass 65 kg) and attempts to hold her at arm's length (this would put her center of mass 1m from his center of mass). Frank has enough upper body strength to support a mass of 25kg in this manner.
a) Frank, being an engineer, starts to spin. After accelerating for 5 seconds at a constant rate, his arms are forming an angle of 5 degrees with the horizontal. Find his angular acceleration.
At this same acceleration, how long will it take until his arms are 2.5 degrees from the horizontal?
c) How long before his arms are perfectly horizontal?
d) How long will it be before Lara throws up?
6. a) The maximum length of a foil blade from tip to bell guard is 90cm. Taking the pivot point to be at the bell guard, calculate the torque produced by a force of 20N applied perpendicular to the blade at the following distances from the tip of the foil:
1) 85 cm
2) 50 cm
3) 10 cm
If you are able to produce a torque of 10Nm around your own bell guard, calculate the resultant torque around your opponent's bell guard if your blades are pushing at right angles to each other and the intersection point is 10 cm from your bell guard and 45 cm from your opponent's bell guard.
7. Assume that a foil blade (not including the tang) is a uniform rod of length 90cm, diameter 5mm and mass 150g. Your opponent beats your blade sharply 40cm from the tip, breaking the blade. She then immediately does a circle disengage and hits the free end of the broken piece with a 20N force for .01 second. Calculate the rotational frequency of the broken piece of blade as it spins off end over end. (The rotational inertia, I, for a uniform rod of length L is 1/12mL^2, with the axis of rotation at the center of the length of the rod.)
8. A golf ball of mass 46g hangs from an ideal string 1m in length. A diligent epee fencer practicing point control strikes the ball with sufficient force to cause the string to form an angle of 15 degrees with the vertical.
a) What is the velocity of the golf ball immediately following impact?
How long after impact will it take the ball to reach the point where it is closest to the fencer?
9. Peter Westbrook (mass 70kg), having temporarily forgotten the end-of-strip rules in the heat of the finals, retreats rapidly off the end of a raised piste 0.30m high. Fortunately for Peter, the regulation run-off incline of 2m has been included.
Unfortunately, he trips and ends up rolling ignominiously the entire length of the incline. Assume that Peter's body approximates a cylinder of 50cm diameter as he rolls without slipping down the incline. Further assume that he is not moving horizontally when he hits the top of the ramp.
a) If Peter is making 2 revolutions per second when he reaches the bottom of the incline, what was his angular momentum when he hit the top of the incline?
What torque is required to stop Peter's rolling at the bottom of the ramp in 1 second?
10. Isabelle Hamori shrieks in the heat of combat at 13,000 Hz. The gym is set up with pairs of two meter wide strips three meters apart, with six meters between each pair.
a) If Isabelle is fencing in the middle of strip 11 at the far end of the gym from the Bout Committee table, which is 10 meters from strip 1, how much longer will it take the Chairman of the Bout Committee to wince than Isabelle's referee, who is standing halfway between strips 10 and 11? (This is at the 1988 Chicago Nationals, where the ambient temperature is approximately 40 degrees C. Take the speed of sound in air at 20 degrees C to be 340 m/s and remember that the speed of sound is related to the square root of the temperature in degrees Kelvin.)
Isabelle's opponent is MJ O'Neill, also known for her dulcet tones on the strip. MJ screeches while fleching at Isabelle, who attempts to retreat, at full voice. The referee, who is maintaining his original position relative to Isabelle, notices that the combined shrieking is producing 2 beats per second. If MJ screeches at 12,980 Hz, what is her minimum velocity relative to Isabelle?
Leave it to an MIT student to make a kick-butt exam. All credit goes to Ms Ezzell!
So. I was reading my Biology textbook the other day and encountered something called "water potential." A simple summary of this term is water's potential energy , or it's capacity to perform work when free water moves from high water potential to low water potential. What? Physics in biology you say? Of course! :eagerness: Physics is everywhere.
Let's define water potential in depth. Water potential is given by the equation water potential (symbol = Greek letter psi) = potential due to solute concentration + potential due to pressure, or:
The potential due to solute concentration, or solute potential, is directly proportional to the number of dissolved solute molecules. Solute binds water therefore reducing the number of free water molecules and decreasing it's capacity to do work. Because of this, solute potential is negative.
potential due to pressure, or pressure potential, is the physical pressure on a solution. Tension due to pressure is a negative pressure potential, whereas an applied pressure creates a positive water potential.
The biology part of water potential is that it is essential to a cell's well being (plant cells in particular). The water potential determines the direction of movement of water in/out of a cell. For plant cells, it determines the shape and stiffness of the cell. A plant cell is flaccid initially. It becomes turgid when it intakes free water in that the pressure from the water pushes on the cell wall, making the cell swell. The cell becomes plasmolyzed when free water leaves the cell, causing the cell to shrivel and the cell membrane to pull away from the cell wall. This state is dangerous for a plant (commonly known as wilting) and the plant may die. These conditions are created by unequal water potentials of the cell vs the cell's surrounding environment. If a cell has a lower water potential than the surrounding solution, it will intake free water and become turgid. If the cell has a higher water potential than it's surrounding environment, it will expel free water and become plasmolyzed.
If you'd like to know more about water potential, didn't understand a thing I just said or would like background noise doing homework, the following link may be of use to you:
In an episode of Tom and Jerry from 1948, Tom once again has his face smashed in from a falling object. This time, the offender was a half-ounce canary wielding circular cage parts. The bird unfastened the cage bottom and let it drop onto the unsuspecting feline below, making Tom's face into a pancake. How much force does this pan actually make? Could it really damage a cat's face?
First, we must find the velocity of the pan when it hits Tom's face. We know that the pan falls from rest, its acceleration is 9.81 m/s2, and the time it takes to fall from the cage to Tom is roughly 3 seconds (1:09 -1:12 in the youtube video). Using the equation Vf = Vi + at, we find that the velocity of the pan just before it comes into contact with Tom's face is 29.43 m/s.
Let's estimate that the pan weighs roughly .1 kg (100 g). Also, let's estimate that the time it takes the plate to go from its initial velocity just before coming into contact with Tom's face and the time when it's final velocity reaches 0 m/s is roughly 1 tenth of a second (.1 s). We know that momentum is conserved in this situation and that (Force)(change in Time) = (mass)(change in velocity). Using this, we know that the change in velocity is -29.43 m/s, so the force of the pan on Tom is roughly 30 newtons. This is equal to roughly 6.7 lb of force.
It takes anywhere from 7 to 9 lb of force to break a human nose, so even though it's not likely that the bird cage would've smashed the kitty's face in, he might very well lose his sense of smell.
Here's a blast from the past composed of 40 % physics and 60% pain. Disclaimer: I do not promote domestic animal abuse, nor do I reccomend testing 7 lbs of force on your friend's nose.
So. I was thinking of what to carve on my pumpkin earlier and thought, "What's something that'll scare the pants off of anyone, even high schoolers?" Bingo, air resistance. Many of us were shaking in our boots when Mr. Fullerton derived a few drag-related equations, but looking back they're not too bad right?
Here's a little review. That long page really boils down to a few key equations:
Air resistance = Fdrag = bv = cv2 , where b and c are constants
V = VT ( 1 - e(-b/m) )
V = (mg/b)( 1 - e(-b/m) )
a = g e(-b/m)t
Some equations may contain the variable tau, which looks like a backwards J or a T with a tail. I will use T to represent tau. It is a time constant that can be substituted into the equations above. Tau = T = m/b.
I hope that helped a little. Sleep with the lights on tonight, folks. 'Tis the season!
How would I determine the drag coefficient of an organic shape, such as a blob of pudding or a chicken or a Looney Tunes character?
I wanted to do a blog post on the terminal velocity of Wile E. Coyote falling off of a cliff. I went back into my notes and found the following equations:
Air resistance = Fdrag = bv = cv2
V = VT ( 1 - e(-b/m) )
Notice the constants, b and c. I turned to google, thinking that the constants would be relatively easy to find.
It turns out, the equation for Vterminal is a little more complex than I thought.
Finding V terminal involves the mass of the object, acceleration due to gravity, the density of the medium that the object is traveling through, the area effected, and, of course, a drag coefficient. In my quest to find the drag coefficient, I found that the coefficient is related to the shape of the affected surface area. The lower the drag coefficient, the more easily the object can move through the air. The following table helps illustrate this:
That's fine and well if you're trying to find the terminal velocity of a UPS box falling from a cargo plane in air of know density, although there are a few complications in the Wile E. Coyote situation. My number one probelm is as follows: unless the furry critter assumes a fetal position and magically transforms his body into a perfect sphere, his coefficient is difficult to determine.
So far, no other particle has been able to move at the speed of light. However, human beings are capable of seeing light move. Ramesh Raskar and his team at MIT have developed a camera capable of capturing light at 1 trillion frames per second. This method, called fempto photography, can take slow motion videos of light in motion. Watch the video for a better explanation but for those of you in a rush below is a summary of MIT's amazing research.
As shown in the video, Raskar uses a laser to send a packet of photons through an object. Using fempto photography, the MIT team created videos of light traveling through a coca cola bottle and washing over a tomato.
The group presents promising applications of their technology, such as finding survivors in unsafe conditions or hiding beings as well as exploring inner organs by seeing around corners with light.
Perhaps the most interesting aspect of this video is featured in 9:20 - 10:04, in which time appears to be moving in reverse according to the camera's images. How is this possible? Watch to find out! Weird things happen when humans try to go faster than the speed of light
Infrasound is sound with a frequency lower than 20 Hz. Human hearing registers sounds from roughly 20 to Hz 20,000, though under certain circumstances the body will hear/feel sound at a lower frequency than 20Hz. Though the human ear does not normally register infrasound, these inaudible waves effect our everyday lives.
Infrasound is produced naturally by severe weather and other forms of nature. Storms, thunder, volcanic activity, earthquakes, avalanches, tsunamis, waves, wind, and even the aurora borealis emit these sound waves. (1) Because many animals can detect sounds of lower frequencies than humans, they are able to sense oncoming natural phenomenon. This explains why many animals move to higher ground or act oddly before a storm.
Animals also use infrasound to communicate. Whales, hippopotamuses, giraffes, elephants and alligators are among the group that do so. These animals are able to send messages over thousands of miles in this manner. Homing pidgeons and other birds are also able to sense these sounds. They use infrasound in combination with magnetic fields to create a mental map of their surroundings.
Infrasound is utilized by humans to locate earthquakes, specific types of rock, petroleum, and even artillery/nuclear weaponry. Low frequencies are also created by man-made objects. These noises are theorized to create disturbances for creatures that can register infrasound. The sounds may cause the animals to become disoriented, causing whales to beach themselves or pidgeons to lose their bearings. (2) Infrasound also causes irritation to human beings. In studies such as that of Professor Richard Wiseman, Richard Lord (acoustic scientist), and Vic Tandy, human exposure to infrasound may increase paranoia, physical discomfort and even nausia. (3)(4)
In Tandy's situation, halucinations and supernatural "sightings" were later found to be linked to low frequency readings in the area. Infrasound is also the factor behind mysterious humming, called "The Hum," in certain areas that are audible to some people yet nonexistent to others. The hums are often emitted by man-made or natural sources. For example in Kokomo, Indiana, the humming was later pinpointed in a local factory. A fan producing a 10 Hz noise was the source of the scare.
***Authors note: This post took a little bit more research than usual ^^; It was interesting, though! Enjoy
How to become an autodidact (defn: self-directed learner) :einstein)
Monday, we were given a few packets of work, some written directions and a "finish this before the test next week." Weird. A class with no teacher? A few groups popped up to grab a computer, others buried their noses in the textbook, and some started chatting leisurely with friends. It's not that we don't have a teacher, it's just that for the next few days, we're our own teachers.
For a few of us (including myself), this whole learning-on-our-own thing is a little bit intimidating. I've scoured the inner depths of google to put together an E2K countdown of ways to survive without teacher supervision. Try reading them over to pick up a helpful hint or two:
3.) Having trouble? All you need is a little help from your friends. Try forming small groups for support, review, and advice. Not only will confusion be dismissed, but it turns out that people retain information longer after they've taught a peer. It's a win-win situation. However, make sure to refrain from copying or relying on others to teach you. The point of self-directed learning is to learn independently, not to leech off of others.
2.) Use your resources. Between the textbook, aplusphysics.com, and the unit summary packets from class, you're bound to find an answer to your problem. READ THE TEXTBOOK. Key chapters are listed on the board, but if the answer still isn't there try skimming the glossary for key terms. Google and other internet searches are okay, but sites like Wikipedia often over complicate or give false information. Try the aplusphysics site. The video sections, tutorials, and course notes are great tools to reinforce concepts if the textbook was unclear, or even for review before the test. Don't forget about asking others (and making friends-- you can friend people on aplus, you know ).
1.) The number one suggestion for learning on one's own is manage your time. Procrastination is our number one enemy when it comes to self-guided learning. In order to avoid doing the web assign, worksheet, problem packet and 8 blog posts the night before they're due, take a few steps of precaution. Try making your own due dates. These are most functional if they're BEFORE the one Mr. Fullerton assigned. Making a calendar is a little over the top, but writing down goal dates to finish certain sections of the work is helpful. At least have mental due dates-- no one should be staying up 'til 1 the night before the test.
A few of you may ask, "Why is God's name did Mr. Fullerton do this to us? Does he hate us? Was it something we said?" This way of learning will benefit us greatly. For one thing, it's preparing us for college next year. Not all classes will have 20-30 kids. A large intro/lecture class could contain anywhere from 50 to 500 students. In otherwords, the teacher can't tend to hundreds of kids at once, making an independent learning style an essential survival tool. Even if the classes are small, there's no guarantee that the teacher is good at,well, teaching. In the event your teacher is a story teller, a newbie, or just plain unhelpful, knowing how to consult a textbook or another resource means the difference between passing and failing.
In other words, take a deep breath, plan briefly, and get to work. Have confidence in yourself as a student and don't be afraid to take autodidacticism for a test spin!
Have you ever wondered how trampolines work? Anything fun or worthwhile has physics behind it, so let’s take a peek at the gymnast’s best friend:
I hope you all enjoy my art skills. Read it and weep. :victorious:
The magic behind a trampoline can be explained in terms of energy. Let’s say that a child is bouncing up and down on the trampoline. When the child is at a maximum height, his/her potential energy due to gravity is at a maximum. Because PE= mgh, with acceleration due to gravity and mass constant, his/her PE is the greatest because height is at a maximum. However, their kinetic energy is at a minimum of 0 because the child has a velocity of zero and KE= (1/2) m v^2. When the child is in contact with the trampoline and is as low as he/she will travel, his/her PE due to gravity is now at a minimum of zero because the height is zero. However, at this point the child’s kinetic energy is greatest because the velocity at this point is at a maximum. In addition, the potential energy due to the trampoline’s springs is at a maximum. Uspring (potential energy of the spring) is greatest at this point because the displacement x of the spring is greatest at this point and Uspring = (1/2) k x^2. In other words, the spring is at its maximum stretch possible for the child and wants to return to its state of rest, so it sends the child back into the air.
If that AP B review didn't click, try watching the specimen V. vulpes exploring this bouncy apparatus. (CAUTION: Video has sound. If you're in the school library, please adjust volume level accordingly before proceeding).
Many of you are familiar with the children’s movie happy feet, about a whimsical penguin chick that just can’t stop dancing. Why don’t these birds fly instead of dance, you ask? Let’s use physics to figure out why Mumble is aerially challenged:
There are four main forces involved in avian air travel: lift, weight, drag, and thrust.
As shown by the diagram of a blue jay in flight (credit to http://www.lcse.umn.edu), lift opposes weight and thrust opposes drag. A bird is able to fly when lift is greater that weight and thrust is greater than drag. Read below for more on these forces:
1. Weight: Mass x acceleration due to gravity
2. Lift: This force can be explained using Bernoulli’s principle—as a fluid’s velocity increases, the pressure decreases and vise versa. Bird wings are in an airfoil shape with a bump on the top and a smooth bottom (like an air plane wing). The air is forced to move faster over the top of the wing than on the bottom because it has a longer distance to travel over the bump. Like faster moving fluids, faster moving air causes the pressure on top of the wing to be lower than on the bottom of the wing, allowing the bird to lift upward.
3. Drag: This force is caused by air resistance. The more aerodynamic the flier, the less drag that will act upon the flier.
4. Thrust: This is the force created to push the bird forward. Birds create thrust by the backward push of their wing, like humans do when we push backward with our arms to swim in a pool. Plane propellers and jet engines create thrust for a plane.
In short, the reason why Mumble cannot fly is because penguins store fat to keep themselves warm, increasing their weight. Their wings also are not the correct shape or size to produce enough lift to get into the air. Weight > Lift, therefore Mumble dances.
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