Physics 303 Motion of Falling Objects

Transcription

1 Physics 303 Motion of Falling Objects Before we start today s lesson, we need to clear up some items from our last program. First of all, did you find out if Justin Time was speeding or not? It turns out that a velocity of 32 meters per second is 71 miles per hour. On most roads, Justin would have a speeding ticket to go along with his wrecked car. After the last program ended, your teacher gave you some motion problems for practice. Your teacher has a copy of all the problems worked out, and you ve probably already gone over some of them. But we ll go over the two that often give students trouble. Let s start with the problem about the acceleration of light. Did it trick you? The two key pieces of information here are straight line and constant speed. What kind of motion does this describe? (Brief Pause) If you said uniform motion, you re right. So the answer is that the acceleration equals zero. When motion is uniform, acceleration is zero. The list of data from this problem includes initial velocity of 22 meters per second, a distance of 110 meters, and an acceleration of negative 1.4 m/s 2. If you didn t use the negative sign you got the wrong answer here. We want to solve for final velocity, so we ll use equation three. It has all our variables and does not involve time, which we don t know and don t want to know. All we have to do is take the square root of both sides and then plug and chug. The answer is 13 m/s. Here s where you need to check your answer to see if it is reasonable. Is it reasonable to decelerate a car and get this answer? Yes. But if you made a mistake and got a velocity greater than 22 m/s, you d catch the mistake and could fix it. All the motion we ve talked about so far has involved cars on roads, carts on hills, planes in the sky, and so forth. But linear motion also includes objects that are falling. And that s the topic of today s lesson, falling objects. (Read objectives on screen.) OK. We ve all seen objects accelerate as they fall. But studying this motion over the years has been confusing. The Greek philosopher, Aristotle, thought that heavier objects fall faster than lighter ones. That s because he could see a heavy object, like a hammer, falling faster than a light one, like a feather. See? But Aristotle was not a scientist and stopped there. In the early sixteen hundreds, Galileo Galilei studied the motion of falling objects. Legend has it that he dropped cannon balls of different sizes off the Leaning Tower of Pisa, but there is no evidence to confirm this. But we know that he did perform experiments, making use of a creative technique of imagining what would happen under ideal conditions. In the case of falling objects, the ideal condition would be the absence of air. 1

2 (astronaut on screen) In my left hand I have a feather and in my right hand I have a hammer. I think one of the reasons we got here today was because of a gentleman named Galileo a long time ago who made a significant discovery about falling objects in gravity fields. And we thought that where would be a better place to confirm his findings than on the moon. So we thought we d try it here for you. The feather happens to be appropriately a falcon feather. I ll drop the two of them and hopefully, they ll hit the ground at the same time. How about that? Mr. Galileo was correct in his findings. As a result of his experiments, Galileo proposed the idea that all objects fall with the same constant acceleration in the absence of air. Why? Galileo didn t know, but you ll learn why in the next unit. In this unit, remember, we re just describing motion, not talking about its causes. So will all our problems involving falling objects have to be in a vacuum? No. In many ordinary circumstances, air resistance is so small that we can consider it negligible. As long as the objects are rather compact and fall a relatively short distance, we can say that they all accelerate the same. Look at these two falling rocks. It s obvious that they are not in a vacuum, but we can say that they both fall with the same constant acceleration. (green chalkboard on screen) We say that objects falling with negligible or no air resistance are in free-fall. The acceleration of objects in free-fall is a constant. (skydivers on screen) The sports term free-fall is not the same as free-fall in physics. To sky divers, free-fall means falling without an open parachute. In this situation, air resistance is not negligible, but essential. It keeps the skydivers from accelerating as fast as objects do when air resistance in negligible. Don t get the sports term and the physics term confused. Now that we have our terms straightened out, the question is, What is this constant acceleration due to gravity? We ll need the actual number when we solve problems involving falling objects. So what do physicists do when we need determine a quantity? We do an experiment, of course. We don t want to use a tape timer for this experiment because the tape pulling through the timer would resist the motion of the ball. We want our ball to be in free fall. This time we ll drop a golf ball so that it falls in front of a meter stick, which will be taped to the wall. We ll do the experiment in a totally dark room and use a stroboscope to flash light at very short time intervals. All we have to do is take a time-exposed picture and we ll see the ball s position on the meter stick at each flash of the strobe. 2

3 (dark screen) Watch as the ball falls and the light flashes. Now watch the same thing in slow motion. (balls on dark screen) Now here s an example of a time exposure photo using a strobe-light. Remember that each picture of the ball represents a flash of the strobe-light, which happens in a definite time interval. You can see that during each successive time period, the ball is falling farther and farther, so the motion of the ball is accelerated. (photo on screen) It s difficult to see the exact positions on the meter stick, so we ve made a diagram that shows the first six positions of the ball. This will give us plenty of data to analyze. In a moment, your teacher will pause the tape and give you a copy of this diagram. You ll need to copy a table for recording your data and to find a ruler or some other straight edge. Then we ll come back and help you get started. (Pause Tape Now graphic) (table on screen) The first picture of the ball is labeled zero since this is the position at the first flash of the light. At this point, no time has elapsed. The strobe-light flashes every seconds, so the time for picture two is s, for picture three is s, and so forth. You can finish this column later. Next, we want to take a straight edge and draw a line across the bottom of each ball. Ball zero lines up with zero centimeters and number one is at the 0.80 cm mark. Keep going and record these positions in the total displacement column. Since we ll be graphing this data, two significant digits are plenty. Your teacher will pause the tape now and let you measure. Then go ahead and make a graph of total time on the x-axis and total displacement on the y-axis. Your teacher will remind you of the tasks you need to complete before we return. Here s what your graph of displacement versus time should look like. Take a few seconds now to tell your teacher how you think displacement and time might be related. I ll wait. How many of you saw that curve and surfed your way into the answer, inversely proportional? Taking a closer look would show you that as time increased, so did displacement. So we can t be looking at an inverse relationship here. If you suggested an exponential direct proportion, you happen to be right. Pat yourself on the back. (graph on screen) When you see a graph like this one, you should expect an exponential proportion. Since physicists are always looking for a straight line for an easy mathematical treatment, we usually try squaring the variable on the x-axis. 3

4 In this case, when we plot time squared versus displacement, we get that straight line. This tells us that displacement is directly proportional to time squared. Now look at your acceleration equations for motion starting at rest. The second equation is a mathematical expression of this proportion. We ve changed a proportion into an equation by inserting the constant, one-half a. (table on screen) Now that we ve analyzed our displacement and time data, let s find the instantaneous velocity of the ball when each picture was taken. Notice that I ve replaced the pictures of the balls with simple dots. This is just to make the measurement easier to see. To find the instantaneous velocity of the ball at this exact point, we ll need to look at the distance it moved in the shortest time period before it and after it. So for dot number one, we want to measure the distance from this position to this one, for one time period before it and one after it. The displacement is 3.1cm minus 0cm. Now since we ve already measured all these positions, let s go to your data table and find the instantaneous displacement just by subtracting. For ball number one, we use position two minus position 0. That s one flash of the strobe light after and one before ball one. I get 3.1) minus 0 cm for an instantaneous displacement of 3.1 cm. But what we really want is instantaneous velocity. All we have to do is take our 3.1 cm and divide by our short time interval of two flashes of the strobe light. That s 0.08 seconds. Notice that we ll divide by this same short time interval every time. One way to make this process easy is by covering up the position of the ball you are working with, for example number two. Then just look at the numbers after it and before it and subtract to find instantaneous displacement. Then divide by 0.08 seconds to find instantaneous velocity. Now you finish collecting data and then make a graph of time on the x-axis versus instantaneous velocity on the y-axis. Finally, draw a rise-run triangle and find the slope of the line. There is a correct answer for the slope, so do a good job. We ll look at the graph and talk about the slope when we come back. (Pause Tape Now graphic) (table on screen) Here s a sample of the type of data you should have gotten from the experiment. Because of estimated digits, yours may be a little different. And here s a sample graph of instantaneous velocity versus time. You probably suspected a straight-line graph since we asked you to find the slope. Let s find the slope of this line. I picked these two points because they re easy to read. My rise is from 60 to 140 centimeters per second, or 80 centimeters per second. And my run is from 0.06 seconds to 0.14 seconds or 0.08 s. 4

5 That makes my slope 80 centimeters per second divided by 0.08 seconds or 1,000 centimeters per second per second. Converting this value to MKS units, we get a slope of 10 meters per second squared. This is an experimental value. What is the actual value and what does it represent? Watch and see. (boy dropping ball on screen) Galileo performed experiments to measure the rate at which objects fall to earth. Thanks to Galileo, we now know that the earth s gravitational force causes an object to accelerate at a rate of 9.80 m/s 2. In other words, the object s velocity increases by 9.8 m/s each second. This acceleration due to the earth s gravity is represented by the symbol g. So that s it. You already know that the slope of a velocity versus time graph represents acceleration. When air resistance is negligible, all objects fall with the same constant acceleration. (green chalkboard on screen) We call this constant acceleration due to gravity. In most places on the earth, the value of this acceleration due to gravity, given the symbol g, is 9.80 m/s 2. So our experimental value of 10 meters per second squared was mighty close. In fact, I d say that, within bounds of experimental error, it was the same. When objects are in free-fall, we can use g in the place of a in any acceleration equation. Let s try some example problems involving g. Rhoda Tiller looks up to see a flowerpot fall off a ledge 9.2 meters above her head. How long does she have to react and move before being hit on the head? When you read this problem, it looks like you won t have enough information to work it, since you re only given one number, a displacement of 9.2 m. But two things are understood. The flowerpot starts at rest, so initial velocity is zero, and it falls freely, so acceleration is 9.80 m/s 2. We want to solve for time, so we ll use equation number two, starting at rest, and rearrange to get time equals the square root of two times d divided by a. You can use a for acceleration or change it to g for acceleration due to gravity. Either one will work, as long as you remember the value. The answer is 1.4 seconds. 9.2 meters is about three stories up. Rhoda had better react quickly. A rock is dropped from a cliff. What is the velocity of the rock after it has fallen for 3.7s? In this problem, we know time, initial velocity, and acceleration due to gravity, and we want to know final velocity. For this problem, we ll use acceleration equation number one, starting at rest. We don t need to 5

6 rearrange. All we need to do is plug and chug. The answer is 36 m/s. Let s make the last example problem a little more difficult, for those of you who need a challenge. What if the rock was thrown downward at a velocity of 11 meters per seconds? How will this affect the acceleration and final velocity after 3.7 s? Your teacher will pause the tape and give you the information again. After you ve come up with your answers, come back and we ll go check it out. (Pause Tape Now graphic) The only way this problem differs from the last is that the ball has an initial velocity of 11 meters per second in the same downward direction. Now how will this affect the acceleration of the ball? (Sound Effect: warning siren) Did we trick you? The answer is that once the ball leaves the person s hand, it is in free fall. So the acceleration does not change. It s still 9.80 m/s 2. To find final velocity, we ll use equation number one, with initial velocity included. Plugging and chugging gives us the answer, 47 m/s. Before you practice some more free fall problems, let s apply what you ve learned to this little game. Try this with a friend. Take a crisp one-dollar bill and hold it vertically, with the middle of the bill between your friend s fingers, like this. Your friend s fingers should be about five centimeters apart. Tell your friend to close his fingers when he sees you drop the bill and see if he can catch it. People have been known to place wagers on this trick, but of course we won t do that. Watch what happens to these students. (students on screen) You would think that closing your fingers to catch the dollar bill would be an easy task, but the vast majority of people simply cannot do it. Their eye-hand reaction time is too slow. So let s do an easy experiment to test your eye-hand reaction time and see if you could win the game. Here s the procedure you will follow. Hold your fingers five centimeters apart while your partner places a meter stick between them at the 50 centimeter mark. When you see your partner drop the stick, simply close your fingers to catch it. Measure the distance the stick has dropped, remembering that it started at the 50 centimeter mark. Repeat this and convert centimeters to meters. Then record both displacements in your data table. If the two trials agree within bounds of experimental error, you have enough data. If they are very 6

7 different, try again and throw out the one that was in error. After you get good data for one partner, swap places and repeat the experiment for the other partner. Now don t cheat. You want to determine your actual reaction time. You ll use an acceleration equation for free fall to determine the time it took the meter stick to fall as far as it did. Once you determine your reaction times, calculate your class average. Then we ll come back and find out if your times are fast enough to win the game. (Pause Tape Now graphic) The average eye-hand reaction time for people is 0.15 s. That s the time it takes us to react to something we see and move our fingers in response. Some of us are slower than others, and it doesn t seem to have anything to do with intelligence. Now let s see how far an object will free-fall from rest in 0.15s. We ll use equation number two. Displacement equals one-half g times time squared. Let s plug in 9.80 m/s 2 for g and 0.15 s for time, and chug out our answer. The answer is that during the average person s reaction time an object will fall 0.11 m or 11 cm. Now let s measure half the length of a dollar bill, since for the game, we start in the middle. The length is eight centimeters. That would make the bill fall three centimeters below your fingers before you could react and close them. Try the game with the people in your class with the fastest reaction time. Even they will miss the bill often. Now it s time for you to practice some simple free-fall problems. Your teacher will give them to you and will go over all the answers with you. Local Teachers, turn off the tape and give students problem set number one from facilitator's guide. (screen fades to black) (fact or fiction on screen) It s true. That s why shooting a gun up into the air is no way to celebrate anything any time. There have been many recorded instances of people being killed by this friendly fire. You ve heard that what goes up must come down. But let s talk about what happens when you toss a ball like this into the air. What happens to it on its way up, at the top of its flight, and on the way down? Tell your teacher. I ll wait. Did you say that the ball slows down on its way up? You re right. (green chalkboard on screen) 7

8 When the ball leaves my hand, it has whatever initial velocity I have given it. As it moves upward, it slows down 9.8 m/s each second. At the top of its flight, the ball actually stops. Then it changes direction and starts back down. On its way down, it picks up speed at a rate of 9.8 m/s each second as it falls. By the time it gets back to its starting point, it has accelerated to the same final velocity it had when it left my hand. A bullet fired straight up would move such a great distance that wind resistance would affect it s final velocity a little, but if the bullet hit you in the head, it would be about as bad as if had just been shot from the gun. Teacher A ball is tossed into the air and rises to a height of 12m. How long is the ball in the air? We could make this problem difficult by considering the trip as a whole. We d have to pick one direction as positive and then use negative signs for any displacement, velocity, or acceleration in the other direction. Why? Student Because all of those are vectors, and the trip involves two different directions. Let s take the easy way. Teacher OK. Pay close attention and we ll make it easy. We ll just break the ball s trip into two equal halves. What do the two halves have in common? Student The ball travels up and down the same distance, so d would be equal numbers. Teacher Right. What else is the same? Student Well, the time up and down is the same and the ball will be going the same speed when it gets here as it was when it left his hand. And 9.80 m/s 2 is used for both of them. Teacher OK. What about the velocity of the ball as it leaves the boy s hand and just as it hits? Student It would also be the same. Teacher Good. Now tell me this. We re asked to solve for total time. If we re going to use only half the flight of the ball, what will we have to do at the end? 8

9 Student We ll have to double it. Teacher Right. Let s use the falling part of the ball s motion, We re ready to write down our data. What do we know? Student We know that initial velocity is zero, the displacement is 12 meters, and g is 9.80 m/s 2. And we want to know time. Teacher Good. Which acceleration equation will we use? Student Number two. t will equal the square root of two times d divided by g. Would it be all right to use a instead of g? Teacher Sure, as long as you know what it represents. Now just plug and chug and tell me the answer. Student It s 1.6 seconds. But I need to double it to get the total time. The answer is 3.2 seconds. Don t forget to re-read the question before you make your answer final. Now it s time for you to try some free fall problems. Your teacher will give you the problems and we ll come back and go over one or two. Local Teachers, turn off the tape and give students problem set number two from facilitator's guide. (Pause Tape Now graphic) Let s go over this problem. To make it easy, we ll use the second half of the ball s trip, so we ll need to cut the time in half, too. That gives us this data: t is 1.8 s, initial velocity is zero, and g is 9.8 m/s 2. If we solve for the distance the ball falls, we ll know how high it went. Using equation number two, we plug and chug and get a displacement of 16 meters. We also want to calculate the final velocity of the ball when it reaches its starting point again. Now some of you might over-think this one and say zero, assuming that the ball is caught and stops. But we want its final velocity while it is still in free fall, just before it is caught. We ll try not to use the answer to part a of our problem, in case it s wrong. So we ll use the same 9

10 data as in part a and use equation number one: final velocity equals g times t. The answer is 18 meters per second. Your teacher will show you the solutions to the other problems and also give you some more acceleration problems to practice. Some involve free fall and others don t. Just remember to make a list of what you are given in the problem, what you know, and what you don t know. Pick an acceleration equation that includes these variables, and rearrange it to solve for your unknown. Then just plug and chug. Your teacher has a good review sheet to help you study for the unit test. Don t forget that the only way to study is to work every problem from scratch. Don t look at the answer to a problem until you ve tried it on your own because if you aren t capable of solving these problems then your grades will surely be in free-fall. Geronimo! 10