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Transcript
Drop Tower Physics William A. (TOBY) Dittrich [email protected] Physics Dept. SY ST 312 Portland Community College P.O. Box 19000 Portland, Oregon 97280 The drop towers of yesteryear were used to make lead shot for muskets, as described in The Physics Teacher1 in April 2012. A modern drop tower is essentially an elevator designed so that the cable can “break” on demand creating an environment with micro gravity for a short period of time, currently up to nine seconds. Using these drop towers one can investigate various physical systems operating in this near zero g environment. The resulting “Drop Tower Physics” is a new and exciting way to challenge students with a physical example that requires solid knowledge of many basic physics principles and it forces the student to practice the scientific method. This is done by asking a simple question: “How would a simple toy behave when it is suddenly exposed to a zero-‐g environment?” The student must then postulate a particular behavior, test the hypothesis against physical principles, and if the hypothesis conforms to these chosen physical laws, the student can formulate a final conclusion. At that point having access to a drop tower is very convenient, in that the student can then experimentally test their conclusion. The purpose of this discussion is to explain the response of these physical systems (“toys”) to the zero g environment and to provide video demonstrations of this behavior to support in class discussions of Drop Tower Physics. Drop Tower Physics is not Amusement Park Physics 2 because the physical systems dropped into zero g experience the microgravity for a duration of time much longer than the period of weightlessness in a carnival ride. It is also not equivalent to the familiar NASA program of Toys in Space 3. On the International Space Station the experiments, like the spherical ball of liquid and so forth, exist for the duration in microgravity. With Drop Tower Physics, the transition from normal gravity to microgravity is the interesting and unique phenomenon demanding exploration. Student interest soars! A glimpse at Drop Tower Physics was seen in the American Journal of Physics4 when a falling slinky was evaluated and photographed. Drop towers have been constructed at various places around the world and their descriptive characteristics are displayed in Table 1. In Portland Oregon, the Dryden Drop Tower 5 is conveniently located at the Engineering School at Portland State University (Figure 1). The Dryden Tower is 31.1 meters high and provides a micro-‐g environment of a little over 2 seconds. Using the Dryden Tower the toys listed in Table 1 have been experimentally investigated and filmed for classroom, and the resulting videos links are posted. This enables teachers to augment their discussions of these physical examples in elementary physics curricula by posing the above question and then having the experimental results available for class viewing. 1 Figure 1: Dryden Drop Tower Table 1: Drop Towers Location Height Free (m) Fall Time (s) Portland,OR Bremen,Germany GRC Nasa Glenn MORE 31.1 146 2.1 9.3 2.2 Trajectory Down(D) Up & Down (UD) D D and UD D O DDT Characteristics: O Tower height: 31.1m (102ft) O Free fall distance: 22.2m (73ft) O Low-‐g time: 2.13 sec. O g-‐level: < 10-‐3go O Deceleration distance: ~ 3.5m O Drag Shield mass: 115kg O Experiment mass: < 50kg O Peak deceleration: 15go O Average deceleration: 8.5go O Automated Retrieval: 5 min. O O Design Features: O Cable-‐Guided Drag Shield Approach O Open Enclosure O Magnetic Field/Eddy Current Deceleration O Air Collet Release Mechanism O Single Floor, Single Person Operation O Aesthetic, Public Space Size of Box Braking LWH Magnetic (m) (M) Air Bag (AB) M AB pellets .96x.84x.40 AB website [email protected] ZARM GRC drop 2 Table 2: Drop Tower Physics Video Demonstrations Physical Experimental Video Example (Toy) Floating Cork https://www.youtube.com/watch?v=cDAVOD9Vqz8 (stationary) Stack of Coins http://youtube/cj6v3Y-‐DY7g Simple https://www.youtube.com/watch?feature=player_embedded&v=Xft0ZMLHpp4 Pendulum Stretched http://www.youtube.com/watch?v=RelTmkLiZfc Elastic Weighted Band Mass Spring In progress Oscillator Ball Rotating https://www.youtube.com/watch?v=OPgDfgKetNY inside a Bowl https://www.youtube.com/watch?v=QtrazIJaZcM https://www.youtube.com/watch?v=XWHOukjp8Xc Gyroscope https://www.youtube.com/watch?v=ibLKUJxqtN0 Conical http://youtu.be/Rw9qLvwrtIQ (fast) Pendulum http://youtu.be/2UZh49zIeFA (medium speed) http://youtu.be/bSte1AdzVBI (slow speed) Why build these towers in the first place? The answer can be found by considering the first of many examples to follow. The question here is: “How would a floating spherical cork behave when suddenly dropped?” Rather then provide the answer directly, please consider this question yourself and formulate your final conclusion. The answer is below, but don’t spoil the fun – consider the question, make a hypothesis and test that hypothesis with your knowledge of physics. If the hypothesis fails, try again! Isn’t that the scientific method without experimentation? This experiment can be easily performed by viewing the video posted in Table 2. This video shows the floating cork being dropped in the Dryden Drop Tower. The experimental outcome may surprise you, and evoke further physics contemplation and theorizing for you and your students. When and if you give up, read the answer below! -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ 1. ANSWER-‐ Floating Cork : The floating cork has no weight, and the buoyant force supporting the cork also disappears since the” weight of the displaced fluid” disappears as well. One would then think that the cork would remain in place, however the smaller fluid forces of surface tension and viscous drag become dominant and they govern the behavior of the cork! 3 Consider the cork to be less than half submerged. Since the bottom of the cork is wetted and the angle of the wetting on both sides of the lower portion of the sphere produces a downward force from surface tension, the cork will begin to move downward. It is easily explained why the surface tension force is downward. Consider the meniscus formed when water is in a glass. At the glass water interface the water is pulled upward by the glass and, in accordance with Newton’s 3rd Law, the water pulls down on the glass. Since the cork wets in the same manner as glass, the water pulls down on the cork. Once the cork becomes more than half submerged, the upper portion of the cork becomes wetted and this produces a smaller upward force. Once the cork becomes completely submerged, the upward and downward surface tension forces are equal and this produces a net force equal to zero. At this point the cork has gained downward momentum from the impulse developed by the wetting force. As the cork moves down, work is done by the viscous drag decreasing the kinetic energy until it is zero. At this point the cork is a certain distance under the surface of the liquid, where it remains for the duration of the free fall. -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ From this floating cork example, one learns that the dominant forces in micro-‐ gravity are caused by surface tension and viscosity. For this reason, very interesting phenomenon occur in fluid flows. This is of great interest to NASA, and hence funding for the development of the drop towers and their use performing zero gravity research. This research is featured in a workshop in Washington DC, at this address: http://www.cvent.com/events/nasa-‐materials-‐lab-‐workshop/event-‐ summary-‐f0ee4d29123a453b94c511ab660b81fa.aspx).Drop Tower Physics is an unusual spinoff from this research effort. Consider, for example, a container of water with a short pipe attached at the bottom. In introductory physics courses, the normal derivation utilizing the Bernoulli Equation is the velocity of efflux of the fluid is given by the familiar v=√ 2gh formula. When this tank is dropped, the hydrostatic pressure driving the flow becomes zero, but the flow continues. This flow is now sustained by the wetting force created by surface tension. The equation for this force is given as: ΔP = Patm – {2σ cos θ/ R} σ= Surface Tension force (N) θ = Angle of contact between liquid and solid R = Radius of tube (m) This change in pressure causes the flow shown in this video. Now if the pipe is tapered the viscous forces act in such a way to produce droplets of equal size and equal space distribution emerging from the pipe as shown in Droplet Ejection at PSU's Dryden Tower. Another arrangement where there is a small hole open to a gas like air can produce a uniform stream of bubbles passing down the pipe thus creating a very uniform and controllable two phase flow fluid flow. It is interesting to note that the bubbles (gas with density less than the fluid) behaves in a similar 4 way as the floating cork (where density is greater than the fluid). This two phase flow is shown in this video and another one . As a result, a tank of liquid in space can be drained into a second smaller tank without the use of a pump even if it is located “above” the other tank! The next illustrative example that demonstrates an operational peculiarity of the drop towers is the case of a stack of coins being dropped. What would happen to the stack? The video in Table 2 holds yet another very intriguing surprise! -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ 2. ANSWER Stack of Coins: The weight of the coins is zero and the normal force is correspondingly zero. Since the coins are stationary when dropped, one would conclude that they will remain stationary. However, they do not. They jump inexplicably into the air momentarily and then land back on the stack when the free fall ends. The most likely cause of this behavior is that when one places an object on a surface with weight acting, the surface is elastically deformed storing elastic potential energy. When dropped this energy is released causing the stack of coins to jump upward. If a measurement of acceleration versus time for the drop box is made, the graph of acceleration versus time is displayed in Figure 2. These early transient micro-‐accelerations are again most likely caused by released elastic potential energy in the drop box itself. The video of the jumping stack of coins is shown here . Figure 3 shows the accelerations versus time for the typical drop including braking from magnetic damping or collision with an air bag system. -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ Figure 2: Acceleration versus Time from Release 5 Figure 3: Acceleration versus Time including Braking Consider now the simple pendulum released when the pendulum is moving downward or upward, or for simplicity when the pendulum is at the lowest point in its oscillation. -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ 3. ANSWER Simple Pendulum: The restoring force for the pendulum, mg sin θ, is absent after drop. The pendulum does have kinetic energy and it cannot be converted to gravitational potential energy so that the kinetic energy remains constant. The result is that the pendulum motion transitions into uniform circular motion. This video is posted at here . -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ The stretched and weighted elastic band, standing still much like a power line, provides another example for contemplation of a reaction to dropping. After loading with uniform weights the band would stretch and take the approximate shape of a catenary. What happens to it when it is dropped? -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ 4. ANSWER Stretched Weighted Elastic Band: The elastic band, before dropping, has a balance between elastic and gravitational forces. It has elastic potential energy. When dropped the gravitational force becomes effectively zero, and the elastic potential energy turns to kinetic energy 6 making the system act like an oscillator. Where did the kinetic energy come from? It came from the elastic potential energy, of course. But where did the elastic potential energy come from? The person who lifted the weights up vertically onto the band! When the band stops falling, the amplitude of the oscillation again increases as the downward kinetic energy of the falling mass spring oscillator adds additional elastic energy to the oscillation. Then damping reduces the oscillations until the band becomes quiescent at the original shape, approximately a catenary. In preparation of the video for this example difficulty was encountered in making the horizontal elastic band perform adequately. Therefore, the posted video is a vertical elastic weighted band which does demonstrate the described behavior. -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ The behavior of the above elastic band suggests yet another physical system to drop, a mass spring oscillator. -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ 5. ANSWER Mass-‐Spring Oscillator: The mass spring oscillator continues to transfer elastic energy into kinetic energy and visa versa as if it were in an elastic potential well unaffected by the lack of weight. But the question lurking behind a quick answer -‐ nothing changes -‐ is found when one ponders where the equilibrium position of the simple harmonic oscillator is located. When weight is present, the equilibrium position is determined by the balance between the gravitational and spring forces (xo=mg/k). When loading the spring to this equilibrium position, elastic energy is imparted to the spring, this energy acts like a “ground state energy” so to speak. Additional elastic energy is given to the system to set it into oscillation. Thus when the oscillator is dropped, this initial stored energy associated with the difference between the un-‐stretched and stretched equilibrium length of the spring is released and becomes available for kinetic energy conversion. The result, for an ideal spring whose spring constant is unchanged under compression or extension, is an increased amplitude of oscillation about the un-‐stretched length of the spring. When the drop is finished, damping will remove the energy of the original oscillation, and the spring will settle back into the original stretched equilibrium position. This video shows also that the oscillator gains considerable energy during the braking acceleration for the drop box. A better video of the mass spring oscillator is being created now. -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ A common classroom example of an object moving in a potential energy well is the familiar ball rolling around inside a bowl. For simplicity let us consider a ball that is initially rotating horizontally undergoing circular motion. -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ 6. ANSWER Ball rotating in a bowl The rotating ball has kinetic energy and is spinning under the effect of a balance of weight and normal force as seen in Figure 4. The familiar resolved free body diagram indicates how the weight of the ball creates the normal force and the normal force provides the centripetal acceleration for the circular motion (similar to a banking 7 airplane). However, when dropped the weight is effectively zero, and therefore one might think that the normal force would also be zero. Figure 4: Free Body: Diagram Ball in Bowl One might think that with the absence of weight produces a rather bleak free body diagram of just a dot with no forces on it. However, the normal force remains due to the inertia of the moving ball! This provides an opportunity to discuss inertia in a new and fascinating way. 8 Figure 5: Free Body Diagram with d’Alembert’s Principle (Inertial Reaction Outward on wall) First, this is experimental proof that inertia exists in space even though there is essentially no weight in deep space ( a concept that is hard to grasp for conceptual physics students). This is also where d’Alembert’s Principle is useful. The rolling ball has inertia and the inertial reaction of the ball pushing on the wall is present, as represented by the outward dotted “force” added to the free body diagrams. This is shown in Figure 5. Now when the weight is removed during free fall, the normal force remains due to the inertial reaction of the ball. This normal force ( Figure 6) has a vertical component which is now unbalanced, and therefore the ball begins an upward spiral until it flies out of the bowl! Figure 6: Free Body Diagram after drop -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ The gyroscope is the favorite class demonstration regarding torque and angular momentum. The rotating bicycle wheel is the most common form and an excellent video of this demonstration is posted at here. What would happen to a gyroscope should it be released into precession and then dropped? -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ 7. ANSWER Gyroscope Once again, when considering the dropped precessing gyroscope, the easy answer holds certain questions that elucidate the educational pedagogy of teaching the principles associated with this subject. The easy answer is, of course, that the gyroscope’s weight acting at the center of mass of the gyroscope is effectively zero and that renders the torque acting to be zero also. Thus the time rate of change of the 9 angular momentum of the gyroscope must be zero, and the gyroscope stops precessing. While this is true, and it is shown experimentally in the posted video , how does this behavior precisely conform to the Law of Conservation of Angular Momentum? A close examination of the spinning bicycle wheel demonstration reveals that there are two components of angular momentum of the precessing wheel, the spin angular momentum (in the x-‐y plane) and the precessional angular momentum (in the z direction) as shown in Figure 7. Figure 7: Total angular Momentum Vector (Precessing Bicycle Wheel) Physics textbooks discuss the spin angular momentum and its time rate of change due to torque, but fail to mention that the total angular momentum vector is a result of both components (much like an electron with both orbital and spin angular momentum). Only one textbook has been found to show the true total angular momentum vector as shown in Figure 7 – Intermediate College Mechanics by Christie6. The easy answer explains why the spin angular momentum remains constant in space, but what happened to the precessional angular momentum vector (in the z direction) when the precession ceases? Conservation of Angular Momentum demands that it must be imparted to another body! This forces the student to “Define the System” whenever considering an application of the conservation laws. When the precessing wheel demonstration is performed one is usually standing on the floor, however consider the performance while standing on an ideal rotatable platform. If you hold the wheel above your head with both hands and release it with one hand it begins to precess, but in order to obey the Law of Conservation of Angular Momentum about the z axis you also begin rotating on the platform in the opposite 10 direction. The rates of rotation of both the wheel and the system of the person and platform are controlled by their respective moments of inertia about the z axis. By grabbing the un-‐held handle of the bicycle wheel again to stop the precession, a torque is exerted on the hand in the direction so as to cease all rotation (in the ideal case of no friction present in the wheel and platform). In the case of the precessing gyroscope in the drop box, when one releases the gyroscope into precession, a reaction torque is exerted onto the box which is attached to the earth. So that the precessional angular momentum of the gyroscope upward is balanced by the downward angular momentum vector imparted to the earth. The drop box is then released and is no longer connected to the earth. The system for Angular Momentum Conservation then becomes the drop box and the gyroscope. When the gyroscope stops precessing it also imparts a torque to the box. The drop box then rotates with the same angular momentum as the precessional gyroscope had about the z axis. One notes in the above video that the gyroscope leaps into the air with extraordinary vigor when entering zero g. This could be a result of the push the gyroscope applies to the box to conserve angular momentum (as well as a elastic surface energy release). To conserve angular momentum when the drop box stops falling , this angular momentum is once again delivered to the earth, so that the net change in angular momentum from beginning to end is zero. The important point to be taken from this example is that the total angular momentum of the precessing top does not lie along the axis of the top but that it has also a precessional component. Secondly, and more importantly, one must consider this to be a two body problem for complete understanding. One must also carefully define the system, which is a very valuable lesson to remind students. Before leaving the gyroscope, an interesting effect is seen in the MIT video. It is easily noticed that the bicycle wheel handle that is rotating drops as the demonstration continues. As Clark Maxwell might say “ What is the go of this?” perhaps one of your bright students can answer that question. -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ The gyroscope reminds one that these examples are two body problems. Now lets consider the falling pendulum as a two body problem. As the pendulum falls the center of mass of the pendulum bob and the drop box must fall vertically, since no external force acts on the system. If the drop box can be considered to be infinitely massive, the pendulum support falls vertically. However, if the masses are comparable, Figure 8 shows the path of the center of mass of the drop box. Again, the centrifugal force of the swinging bob pulls on the box and provides the acceleration in the x or y direction as the system accelerates downward in the negative z direction. The need to consider the examples as two body problems arises again when we consider the final Drop Tower Physics example. 11 Figure 8: Path of Drop Box COM (Two Body Problem) The conical pendulum is a very useful example of both oscillation and circular motion, and is often used in both the texts and classrooms as challenges for students. When dropping a conical pendulum swinging around in a circle inside a drop box, the problem of discerning what will happen is indeed a challenge. Please once more accept this challenge, even if it makes you think for a long period of time. -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ 8. ANSWER Conical Pendulum: The conical pendulum bob has linear momentum at the moment of drop. Linear momentum must be conserved and because it no longer has any effective weight it would attempt to move in a straight line. However, the drop box has mass and the bob would pull on it with a tensional force. If we consider the box mass to be small enough in comparison to the mass of the rotating bob, the resulting motion would be such that the box and mass would rotate about their center of mass while the center of mass would move off horizontally in a straight line. The two masses would fly off in the manner of a thrown bolo. A video of a thrown bolo is seen here. 12 Interestingly enough, if the drop box is considered to be infinitely massive – one might think that the conical pendulum would proceed to swing around in a circle unaltered by the drop. However, the rotating mass’s inertia produces a tension in the rope due to the inertial reaction of the bob’s circular motion. Just like the case of the ball in the bowl, the tension has a vertical component causing the bob to spiral upward while continuing to circle around maintaining angular momentum and kinetic energy. The vertical component of tension decreases in magnitude (T sin Φ) as the bob spirals upward. The vertical component of the tension force becomes zero when the mass is horizontal (Φ=0) but the mass continues to spiral upward since it has a vertical component of momentum. However, when the mass moves above the horizontal plane the vertical component of tension force is now downward causing the mass to accelerate downward. The bob’s vertical motion is reduces until it stops and reverses direction spiraling downward. This up and down spiraling motion would be like simple harmonic motion if the angle Φ was very small (sin Φ = Φ). When viewed from the side one would see the superposition of a vertical simple harmonic oscillation in the z direction and one simple harmonic oscillation from the circular motion in the x-‐y plane. Combining these two harmonic oscillations would produce the trace of a lissajous figure! -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ Finally, let us consider another example that is often discussed in elementary level physics classes studying mechanics. This example is that of a Yo-‐Yo. For simplicity consider the ideal Yo-‐Yo with no energy loss and sufficient friction between the string and rotating mass so that no slippage occurs ( no “sleeping”). These assumptions lead to the Yo-‐Yo moving down and up along the string indefinitely. What would happen to this ideal Yo-‐Yo if it was dropped as the rotating mass was descending at the time of drop? -‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐-‐ 9. ANSWER Ideal Yo-‐Yo: Before drop, the gravitational potential energy of the mass is converted to translational and rotational kinetic energy on the way down which in turn is returned to GPE on the way back up (a simple and elegant demonstration of conservation of energy). Now without weight, the total kinetic energy of the mass will not change as it descends and ascends without increasing linear or rotational speed. The Yo-‐Yo proceeds to strike the players hand conceivably imparting a painful Yo-‐Yo burn as it skids to a stop while pressing on the hand! Drop Tower Physics is a fascinating and exciting challenge for teachers and students alike. Likewise it is a challenge to experimentally test these hypotheses or predictions in a Drop Tower. The drop tower in Bremen, Germany with its 9 second micro gravity environment would be very helpful experimentally. In any case, this new area of Physics Education provides us all with many insights and fun along the way! Students, of course, will always ask “ Can I get in the drop tower and go for a 13 ride? The answer here is theoretically yes, you would survive the deceleration at the bottom if the braking period with nearly 10g’s of deceleration was suitably designed and equipped. However, practically the teacher’s answer will always be an emphatic NO! References: 1.The Physics of Shot Towers, The Physics Teacher, Vol 50 No 4 April 2012 page 218 2. http://www.learner.org/interactives/parkphysics/ 3. http://www.nasa.gov/audience/foreducators/microgravity/home/toys-‐in-‐ space.html 4. Modeling a Falling Slinky, AJP,Vol 80 No 12, December 2012, page 1051 5. http://www.ddt.pdx.edu 6. Intermediate College Mechanics, D.E. Christie, McGraw-‐Hill Book Co. 1952, p141 Surface tension video Tufts University http://www.youtube.com/watch?feature=endscreen&NR=1&v=DkEhPltiqmo http://www.youtube.com/watch?NR=1&v=yiixltf_HKw http://www.youtube.com/watch?v=5d6efCcwkWs PSU droplet video http://www.youtube.com/watch?v=JXKM6D9rPis 14
