Lab_03 - Laboratory 3 Motion with Nonuniform Acceleration...

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Unformatted text preview: Laboratory 3 Motion with Nonuniform Acceleration Introduction In Laboratories 1 and 2, we measured the displacement of an object that was subject to a constant acceleration, which led to a quadratic dependence of the object’s position on time. In this laboratory, we will analyze motion where the acceleration is not uniform; this will demonstrate the power of Newton’s Second Law to predict behavior for more complicated physical situations. The example we will study is the acceleration of an object subject to a constant gravitational force combined with the retarding influence of air drag. We will also explore some of the features of motion with air drag, which will prepare you for the study of fluid mechanics in the future. Figure 1: Galileo Galilei (1564-1642) was an Italian physicist, mathematician, and astronomer who established that gravitational acceleration is constant at Earth’s surface. He also supported heliocentrism, which the Inquisition did not appreciate. As a result, he spent the last ten years of his life under house arrest. One famous experiment involving gravity and air resistance is usually attributed to Galileo Galilei (see Figure 1), although recent scholarship has shown that it was probably anecdotal. The popular story describes how Galileo, in an effort to get the intellectual establishment of the time to accept that objects of different masses - in the absence of air resistance - fall through the same distance in the same amount of time, dropped objects of various masses off the leaning tower of Pisa and 20 timed their falls. The story is almost certainly false. But it is true that, at the time, most intellectuals believed in the teachings of Aristotle, who claimed that heavy objects fall faster than lighter objects in direct proportion to their weight. Although Aristotle was correct under conditions of air resistance as you will see in a moment, Galileo was correct on principle: gravitational acceleration does not depend on the mass of the falling object. Theoretical Background: Gravity and Mass Mass is a measure of how hard it is to get something to move. Subject to the same force, a more massive object will accelerate more slowly than a less massive object. Imagine that the force of gravity were constant for all objects. If Fg = C, we would expect that a heavier object in freefall would accelerate more slowly downward than a lighter object. From Newton’s second law, F = ma, we would obtain the result that ag = C , m which, experiments have shown, is wrong. Why? Because the force of gravity is not constant. In fact, the force of gravity depends linearly on mass, so heavier objects are subject to a greater force than lighter objects. In reality, Fg = m C, so that Cm ag = = C. m The acceleration is constant, and the constant, C, is the familiar g (on Earth) - the constant acceleration due to gravity. This was Galileo’s result. Many students get tripped up because most of the forces we encounter in lecture - people pushing on boxes, horses pulling on carts, and the like - are constant. The horse can only pull so hard, so a heavier cart will accelerate more slowly than a lighter cart. But gravity is special in this regard: the heavier an object is, the harder the Earth’s gravitational field pulls on it, so the acceleration is always the same. Theoretical Background: Air Resistance One reason why people got tripped up on this mass dependence problem for so long is that it is very rare for an object at Earth’s surface to actually be in free-fall. It is true that, in vacuum, a feather would fall to the ground as fast as a bowling ball dropped from the same height, but this is obviously not true in the real world. The culprit, of course, is air resistance. We know from a variety of experiences that moving air can produce considerable force. However, there is no way to characterize air resistance with the same simplicity and precision as we can with 21 TurbulentFlow LaminarFlow Figure 2: A diagram illustrating the differences between laminar flow and turbulent flow around a Mylar cone. the gravitational force, because it depends both on an object’s geometry - how “aerodynamic” or “streamlined” it is - and its velocity. For this reason, virtually no problem you encounter in lecture will take air resistance into account. However, it is a very important factor that people who deal with “real” situations (such as engineers and, in particular, aerospace engineers) need to consider in their calculations, and we will see why in this laboratory. Problem 3.1 Describe how the force of air resistance is unlike the gravitational force. Specifically, what quantities does it depend on that the gravitational force does not? What quantity does the gravitational force depend on that the force of air resistance does not? There are two major types of air flow that contribute to drag on a falling object. The first is laminar flow. Laminar flow occurs when layers of air move past an object at the same speed. The layers do not interact, and no air molecules are exchanged among them. The other type of flow is turbulent flow, where the flow lines are disorganized and distributed in eddies that swirl around and exchange air molecules. In turbulent flow, air molecules traveling with different speeds bump into each other, and their speeds are constantly changing. Figure 2 contains a diagram that illustrates the differences between laminar and turbulent flow. As might be expected, the mathematics of laminar and turbulent flow are also distinct. A sphere in laminar flow experiences a force given by F = 6πη rv, (1) where r is the radius of the sphere, η is the viscosity of the air, and v is the object’s velocity relative to the air. Note that this equation only holds for a sphere - the calculations for other objects are too complex for this manual. The next equation, however, is true for any object. An object in turbulent 22 flow experiences a force [approximately] given by 1 F = ρ Cd Av2 , 2 (2) where ρ is the density of the air, A is the cross-sectional area of the object, and Cd is the object’s drag coefficient (which depends on its geometrical properties). Note that, in the laminar regime, air resistance is proportional to v, whereas in the turbulent regime it is proportional to v2 . In general, turbulent flow is mathematically difficult to describe, and the real expression for the force due to this flow is much more complicated. We will generally say that flow is “turbulent” if the exponent on v is greater than one; there may be v3 terms, v4 terms, etc. Equation 2 is just a good approximation. As an object falls and its velocity increases, the force from the air drag increasingly opposes gravity until there comes a point when the net force, and hence the acceleration, is zero. This occurs when Mg = F = cvn where c is the constant term in Equation 1 or 2 and vn is the velocity raised to some power n, which varies depending on whether the air flow is laminar (n = 1) or turbulent (n > 1). The velocity under these conditions - the terminal velocity, which we will call vT - is therefore a constant. We can obtain the terminal velocity in both the laminar and turbulent flow regimes by solving for v in the equation above. For laminar flow, Mg vT = , (3) 6πη r while for turbulent flow vT = ￿ Mg c ￿(1/n) . (4) This is an important result, because the terminal velocity is something we can easily measure. Equation 4 is what’s called a power law. Later, you will see it’s easier to pin down n in this model if we apply a few simple mathematical tricks to our experimental data. Problem 3.2 Is the force of air resistance greater for a faster-moving or slower-moving object, assuming that the two objects have the same mass and geometric shape? Problem 3.3 How can Equations 3 and 4 (and Figures 3 and 4) help explain the fact that, in situations where air resistance plays a role, a heavier object dropped from the leaning tower of Pisa would actually hit the ground before a lighter object of the same geometry? 23 Figure 3: Position vs. time data for objects with different masses. The position of the heaviest object follows a nearly-parabolic path, while the lighter objects approach a linear path. Figure 4: Velocity vs. time data for objects with different masses. The lightest objects clearly reach limiting velocities soon after being dropped, while heavier objects take much longer to attain their limiting velocities. Theoretical Background: Least-Squares Fitting Least-squares fitting is a technique, based on the statistical concepts we discussed in Laboratory 1, that allows experimenters to determine how well a particular model fits their data. In addition, the technique can be used to adjust the parameters of the model in such a way as to provide the closest possible fit to all of the data points simultaneously. 24 Figure 5 shows a picture of some noisy data with a least-squares fit line through it. LoggerPro finds best-fit lines automatically via its user interface, which you have used in previous laboratories. Least-squares fitting can be used, as mentioned, with any type of model. It depends on the concept of a residual, which is the distance between an individual data point and the corresponding data point predicted by the model. Although this distance can be hard to calculate for a multidimensional model, we will typically be dealing with models that can be plotted on a y vs. x graph. In this case, the formula for a residual can be expressed mathematically as residual for data point i = actual value of point i − fit value for point i, δi = yi − y f iti . (5) A magnified picture of a region of Figure 5 is shown in Figure 6. The lengths of the green lines are the values of the residuals for some of the data points. Points below the fit line will have negative residuals, while points above the fit line will have positive residuals. Figure 5: An example of noisy data with a least-squares fit line that best describes the global behavior. Figure 6: Diagram of how residuals are measured - not all residuals are shown. 25 The essence of the least-squares method is that it minimizes the sum of the squared residuals, ∑ i ￿ y f iti − yi ￿2 . (6) We square the residuals before we add them up because we do not want negative and positive residuals to cancel each other out. The least-squares method, as performed by LoggerPro, varies the parameters of the model y f it (x) until the sum in Equation 6 is minimized. For example, in Figure 5, the fit model is a line. The individual data points have the form (x, y), and the fit line has the formula y f it (x) = m x + b. LoggerPro varies the parameters m and b until the sum of the squared residuals is as small as possible, and it then plots the resultant line on your graph. If you were using a quadratic model instead, where y f it (x) = A x2 + B x + C, LoggerPro would have to vary three parameters instead of two to get the best fit. Problem 3.4 Draw a graph with some made-up data points on it. Now draw an approximate fit line/curve through those data points. Label three of the residuals. How does the least-squares method adjust the fit line to give you the best possible fit? Use your own words. Problem 3.5 Would summing the residuals and then squaring the sum give you the same result that you obtain by squaring the residuals first? Why or why not? Power Law Model Fitting The expression in Equation 4 is a power law, that is to say n could in principle take on any real number other than 0. We could try distinguishing between the laminar and turbulent flow regimes by plotting vT vs. m. If the flow is laminar, then it will have a linear trend (since n = 1 for laminar flow). On the other hand, If the flow is turbulent, then there is a continuum of values n can take, and hence a continuum of model equations to which the trend could fit. This means that we can’t find the best model without already knowing the precise value of n, but it is the value of n that we are trying to find by fitting the data to a model! To further complicate matters, this method is not very helpful when using a limited range of masses - reason being is that a linear trend will fit most data sets over a limited domain (think tangent line to a curve). What we do instead is use mathematical properties of the logarithm to force our data to follow a linear trend, where the slope of the linear trend gives us information about the value of n. Now instead of having an infinite number of possible models where the information about the flow is 26 encoded in the exact model equation used to fit the data, we have a single model equation for all flow types, and the information about the type of flow is encoded in one of the fitting parameters of the model itself. To get this ‘all-encompassing’ model, you must make first take the logarithm of Equation 4, then use of the following two log properties ln(a + b) = ln(a) + ln(b), ln(a)b = b ln(a) (7) (8) to further disect the right hand side of the resulting equation. Collect terms in the appropriate fashion, and you’re left with the expression ln vT = 1 ln M + b, n (9) where b ≡ (1/n) ln(g/c). Equation 9 is an equation for a line, where 1/n is the slope of the line and b is the y-intercept. To see a linear trend in your experimental data, the above form suggests we plot ln vT vs ln M , and the slope of the fitted line will relate to the flow parameter, n. This is exactly what you will do in this experiment to determine if your objects are experiencing laminar or turbulent flow. Problem 3.6 Derive Equation 9 by first taking the natural logarithm of Equation 4, and then use the log properties in Equations 7 and 8 to get it into the proper linear form. Experimental Background: Dropping Different Objects The experimental apparatus for this experiment consists of an ultrasonic ranging system and a vacuum release system identical to the one used in Laboratory 1. Please see that laboratory for a photograph of the apparatus. In this laboratory, you will drop four Mylar cones with different masses. As shown in Figure 7, there are three different interchangeable release tubes with different-sized conical cavities that will hold the various objects to be dropped. These can be exchanged by depressing the small metal release on the upper plastic fitting that holds the release tube in place. You will use the two smaller adapters to drop the cones. 27 Figure 7: The vacuum release system requires different release adapters for different object geometries. Figure 8: The proper positioning of a Mylar cone on the vacuum release system. Experiments Experiment 1: Motion of a Conical Object with Air Resistance Here we drop four cones with different masses. Each cone consists of a sheet of Mylar folded into a cone with a ball bearing at its apex. When dropped from a height of approximately 1.5 meters, the cones should reach their terminal velocities well before they hit the ground. We can use multiple measurements of mass vs. terminal velocity to determine whether the cones are experiencing laminar or turbulent air resistance. 28 1. Use the smaller-diameter vacuum release adapters to hold the cones in place above the ultrasonic sensor, exactly as you did for the rubber ball in Laboratory 1. (See Figure 8). Start with the heaviest cone. Its mass is written on it in pen. 2. Set up LoggerPro to take 4 seconds of data at 10 samples per second. You can do this using the template file nonuniform.cmbl from the “LoggerPro Files” folder. 3. Use the trigger to release the cone, and take position vs. time data on its way down. 4. The graph should flatten out near the end of the cone’s fall. The slope of the graph in this region is the terminal velocity. Calculate the cone’s terminal velocity by performing a linear fit. See Appendix A Section 1.2 for how to perform a linear fit in LoggerPro. Problem 3.7 In general, how can you determine a reasonable region (on the p vs t graph) to perform the fitting in order to get a somewhat accurate value of terminal velocity? 5. Copy the graph with the linear fit into your lab report. Also list the mass of the cone and its terminal velocity on the data table in your report. 6. Repeat this procedure three more times with the same cone. Then repeat everything again with the three other cones. In the end, you should have sixteen sets of mass vs. terminal velocity data (4 cones × 4 trials each). Note that you do not need to paste the graphs for the final three trials of each cone into your lab report. 7. Calculate ln(mass) and ln(v¯ ) for each data point. T 8. Calculate the mean, standard deviation, error of the mean, and 95% confidence interval for each of your four measurements of terminal velocity (one confidence interval for each cone). See Appendix A Section 2.2 if you get stuck. Experiment 2: Model Fitting and Analysis Deciding whether an object is experiencing laminar or turbulent air resistance is not a trivial task. We want to see which model of air flow (laminar or turbulent) best matches our observed data. How do we determine which model is best? That is what we will investigate in this section. 1. Below the Experiment 2 heading in your lab report, you will see a graph. This graph contains the data on ln(mass) and ln(terminal velocity) for the cones. 2. Fit a straight line to the ln(mass) vs. ln(terminal velocity) graph using Excel’s feature Trendline, which is similar to LoggerPro’s Linear Fit. See Appendix A Section 2.4 for instructions on how to use Trendline. Include the fit equation on the graph. What does the slope of this line correspond to? Make sure you understand this before moving on. 29 Figure 9: The position vs. time curve for an object experiencing air resistance. Observe that the position vs. time graph begins to approach a straight line for this object rather than continuing to follow a parabolic arc, as it did with the rubber balls we dropped in Laboratory 1. This is what one would expect for objects that have reached their limiting velocities. Problem 3.8 When you perform a linear fit on data in LoggerPro, a text box appears with some information about your fit. One of the quantities in the box is the RMSE, or root-mean-squared error, which is related to Equation 6. What does this quantity tell you about the quality of your fit? Use terms and concepts from the theoretical section on least-squares fitting. Problem 3.9 Does terminal velocity appear to increase or decrease with mass? Does this confirm what you learned in the theoretical section of this lab? How does this help explain the observation that a bowling ball hits the earth before a styrofoam ball of the same shape, dropped from the same height? 3. Record your value for n for the cones using the online data collection tool (see Appendix A Section 3). 4. When all of the class data has been collected, calculate a 95% confidence interval for n for the air resistance experienced by the cones (see Appendix A Section 2.1). Problem 3.10 Using your graph of ln(mass) vs. ln(terminal velocity), decide whether the cones experience laminar or turbulent air resistance. How can you tell? Does the confidence interval of the class data include the value 1.0? Supposing it did include 1, what would be your conclusion about whether the object was experiencing laminar or turbulent air flow? Be careful about your wording. In statistics, we can only reject hypotheses, never prove them to be true. 30 Problem 3.11 How would the graph of position vs. time for a single cone change under each of the following conditions? (a) The cone’s mass was increased but its shape was kept the same. (b) The cone’s mass was kept the same but its diameter was increased. (c) The cone fell through vacuum instead of through air. (d) The air density in the classroom was increased. 31 ...
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This note was uploaded on 09/13/2011 for the course ECONOMICS 101 taught by Professor Gerson during the Spring '11 term at University of Michigan.

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