Most people said that they understand force and acceleration. GOOD!

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1 Questions and Answers on Dynamics 3/17/98 Thanks to the students who submitted questions and comments! I have grouped them by topic and shortened some of the long questions/comments. Please feel free to submit more comments & questions in class or via . -Penny Axelrad 3/25/1998 Most people said that they understand force and acceleration. GOOD! The key thing here is drawing careful free body and kinetic diagrams and writing down the 2 or 3 required equations of motion carefully. Watch your definitions of +/- directions. Label the point around which you are summing moments AND make sure that is the one you use in the equations. Also, when using a geometrical constraint, like v=ωr, pay attention to the definitions of each of the variables and their signs. In general, more problem solving is needed. I need more practice doing problems. I understand the concepts but need more practice deciding which approach to use. Practice problems. There are some excellent sources of additional practice problems for dynamics. The two review chapters in your book (R1 &R2) have many good problems presented in a random order to help you practice deciding what approach to take. In fact, questions 2 & 3 on Quiz #4 came straight out of R2. You are welcome to see either of the instructors or any of the TA's to look at the solutions to any problems in our book that you have worked on your own. This is a great way to get extra practice in doing problems. I am reluctant to assign any extra homework because I think that different people need extra practice in different areas. Another great source of extra practice problems are Schaum's Outlines or other similar study aides. Look for one on "Mechanics" or "Dynamics" or "Systems and Controls" to find the topics you need. These books will provide lists of procedures, many more worked out problems that a conventional textbook and many, many supplementary problems with answers. Another choice is to find a friend's copy of other dynamics books like those by Beer and Johnson, or Bedford and Fowler. Beer and Johnson has been particularly popular with students in the past, so you might try it as a supplemental reading. It has very good guidelines for problem solving and very clearly worked examples. Again, you are welcome to come to office hours for help on any dynamics problems you find to solve. Finally, use your classmates as a source of practice problems. Try to make up a test and solution set for each other. Sometimes practicing making up the questions can help you sort out what the key topics and solution methods are. If you come up with a really good question, please pass it on to us and we might use it on the final or for future classes. ASEN 2003 QA.doc 03/25/98 1

2 What is the difference between a particle and a rigid body? When does kinetic energy have two parts.. Basically including angular components make things more complicated. PARTICLE A "particle" is what we call an object which can be treated for purposes of analysis, as if ALL of its mass was concentrated exactly at the center (r=0). A particle is defined to have zero moment of inertia about its center of mass. All other properties of a particle derive from this assumption. A particle can only translate; it cannot rotate. A particle has zero angular momentum about its center of mass. The angular momentum of a particle about any other point is thus given by r m v, where r is the vector from the point of interest to the particle (c.m.), m is the mass of the particle, and v is the absolute velocity of the particle (more on this later). The kinetic energy of a particle is given by 1/2 m v 2., where v is the velocity of the particle with respect to an inertial coordinate system. Because the particle cannot rotate and its moment of inertia about the center of mass is zero, there is no 1/2 I ω 2 term. Because all the mass of the particle is at the center, all forces applied to the particle act through the mass center. Thus, in drawing a free body diagram for a particle problem, all forces may be shown acting at the c.m. Furthermore, moments cannot be applied to a particle. RIGID BODY A rigid body has a non-zero moment of inertia about its center of mass. It can rotate as well as translate. The angular momentum of a rigid body about any point, P is equal to the sum of its angular momentum about its c.m. plus the angular momentum of the c.m. about point P, H = Iω + r mv Again, both the angular velocity and the linear velocity of the c.m. must be referred to an inertial reference frame. Because a rigid body can undergo translation and rotation, a free body diagram involving a rigid body must show the forces acting at the point they are applied. A force applied on one end of a rigid body has a different effect than on the other. Forces that act through the center of mass, of course do not produce a moment about the center of mass. The kinetic diagram must show both the ma and the Iα terms. The kinetic energy for a rigid body has two terms 1/2 m v 2 and 1/2 I ω 2. The "v" refers to the velocity of the center of mass. The second term is due to the rotational motion of the rigid body about its center of mass. The I should be the moment of inertia about the center of mass. ASEN 2003 QA.doc 03/25/98 2

3 Could you clarify the different equations for Force/Acceleration, Impulse/Momentum What to look for in order to decide how to solve problems using: work, energy, momentum Do not understand momentum method. I have a hard time deciding which equations to use. I do not understand how to use work and energy to find values. There are so many different method of solving problems: and so many different situations, it's hard to keep track of what rules apply where, when, and what. I have difficulties with work couples. I do not understand angular impulse and momentum. I feel impulse and momentum were not covered very much, although it didn't seem too difficult in the homework. GENERAL APPROACH Given a dynamics problem, how do you decide to approach it? Here is my approach (when I am doing things right ): 1) Draw a free body and kinetic diagram. Draw axes and write down any geometrical constraints. 2) Write down what I need to find. 3) Write down what I think I can determine easily. 4) Decide which method gets me most directly from what I know to what I need to find. Examples of key steps: 1) Free body diagram - Check for: weight, tension, friction, normal force, applied forces and moments. Kinetic diagram - ma x, ma y, Iα make sure to define +/- directions consistently. 2) Define the question carefully. Do I need t, d, v, a,? Do I need to check forces? Make sure I totally understand what is asked. Define mathematically what is meant by, does something slip? does something fall off a corner? does it leave a surface?. A good example of this was the pencil example in a recent problem set where you were asked to find the point that does not move. To formulate this mathematically we could say find the point for which the absolute velocity = 0. 3) Do I know forces, moments, velocities, angular velocities, etc. Pay particular attention to any vector quantities that I know and whether velocities are absolute or relative to a moving reference frame. 4) What approach do I try? If I know speeds at start and end, and I am interested in any kind of distance, work and energy is a good bet. If vector velocities or positions are of interest, force and acceleration are a good guess. If no external forces are acting on the system, conservation of momentum is my first choice. If time is involved, force and acceleration or impulse and momentum should be considered. I generally would only use impulse and momentum when there is an impact. Another hint in using force and acceleration is to select the reference point for summing moments so as to minimize the number of extra variables that must be solved for along the way. A zero velocity point, like a no-slip contact point is usually a good bet. By the way, for most problems considered in this class you will not find a huge difference in the difficulty of solving by different methods. So, pick the approach that you think will work the best. If it starts getting ugly, maybe reconsider. Along the way, make sure you always understand how each step is leading to the required result. ASEN 2003 QA.doc 03/25/98 3

4 FORCE AND ACCELERATION - Summary Force and acceleration may be used to solve almost all of the problems we consider in this course, although it is not always the most direct. The key to this method is to construct a clear and correct Free Body Diagram and Kinetic Diagram. The method then involves setting the forces and moments in the FBD equal to the kinetic forces (ma, Ia) in the kinetic diagram. An additional equation due to geometry or known friction force is then added to complete a system of 3 or 4 linear equations in 3 or 4 unknowns which include accelerations, angular accelerations and forces (like tension, normal force, etc). To simplify the solution process, try to set up the solution of this system to solve directly for the one you are interested in. (In particular, for roll w/no slip problems, you can sometimes solve for one variable right away by using the zero velocity point for summing moments.) Once the acceleration or angular acceleration is known, velocities, angular velocities, and distances may be solved by integration. This is especially easy if the accelerations are constant. WORK AND ENERGY - Summary Work and energy is the most direct method to solve problems for which forces are acting over known distances, or begin & end velocities are known and the distance is of interest. The basic equation states that the initial kinetic energy of a system plus the work done on the system equals the final kinetic energy of the system. If the system is at rest, the KE is zero so that simplifies the initial calculations a lot. Work done by gravity is always the weight times the change in height. Work done by an external force is force times the distance over which the force acts. This can be a little tricky and requires careful thought. Work done by friction is µn times the distance that the surface slips, which is not necessarily the same as the distance that the center of mass moves. A friction force, rolling resistance, or tension that just supports a roll w/no slip contact point does no work. The contact point has zero velocity, thus these forces do not act over a distance. Work done by a rope or cable in pulling something or lowering it via a pulley, is the tension times the distance the object is moved. Work done by a spring is the spring constant times the distance moved. In this case especially, be careful about the sign of the work done. If a spring is stretched or compressed away from its unstretched length, the work done by the spring is negative, because it opposes the motion. If the spring is returning to its natural length, it is assisting in the movement and the work done is positive. Work done by a moment is the moment times the angle over which the moment acts. This should make sense because the work done by a force and work done by a moment must have the same units (kg m 2 /s 2 ). In applying work and energy methods to angular motion be careful to consider moments about the center of mass or about a fixed point only. In the special case where the only forces acting on the body are gravity and springs, the system is conservative. In this case the sum of the kinetic and potential energy must stay the same. Note that you can define the reference point for the PE any way you like, so the absolute value of the total energy doesn't really matter. In any problem involving a person or a motor, energy is usually not conserved even though momentum is conserved. Energy or work can be imparted to the system through the action of a person's muscles or the expenditure of fuel in a motor or engine. MOMENTUM AND IMPULSE - Summary Momentum conservation (linear and angular) is most useful when considering a system in which no external forces act in the direction of interest. In this case the momentum of the whole system is conserved although individual elements of the system (e.g. a person or a platform) can exchange momentum. When considering momentum, make sure you decide whether to deal with an individual object or the system as a whole. Linear momentum is defined as mv G where m is the mass of the object and v G is the vector velocity of the center of mass of the object with respect to an inertial reference frame. The linear momentum for a system ASEN 2003 QA.doc 03/25/98 4

5 of objects is just the vector sum of their linear momenta. The phrase "velocity of the center of mass" is particularly important for a rigid body. Remember that because a rigid body can rotate as well as translate, different parts of the body have different velocities. Linear momentum refers to the center of mass. In the example of the man running on the platform in Quiz#4, the linear momentum of the platform is zero because its center is fixed. The linear momentum of the man changes as he runs around the edge of the platform (v is a vector quantity, so even though his speed does not change, the direction does change). What is causing his change in linear momentum? - the force of the support at the center of the platform. The platform spins freely, meaning that the support does not exert a moment on the platform. However, in order for him to be able to run on the platform and not have it fly off in the opposite direction, the support does provide an equal and opposite force to the platform. Linear momentum is not conserved in this case. Note that the units of linear momentum are kg-m/s or slug-ft/s. Angular momentum for planar motion is always in the positive or negative "z" direction. The angular momentum of a rigid body about a point P is defined by I G ω+r G/P mv G where I G is the moment of inertia of the body about its center of mass and r G/P is the position vector of the center of mass with respect to point P. For a system comprising more than one rigid body or particle, the angular momentum of the whole system is computed by summing the contributions of each element. If there are no external forces acting on a body, angular momentum is conserved. If there are external forces, but they all act through one point (like the center of mass), angular momentum about that point is conserved. In the Quiz#4 example, the only external force is supplied by the support point, which does not exert a moment on the platform. Since this force acts at the center of the platform, the angular momentum of the platform+man system about this center point is conserved. Since the platform and man start from rest, the initial angular momentum is zero. When the man runs counter clockwise he picks up positive angular momentum. Thus the platform must have a negative angular momentum which is equal in magnitude. Remember the angular momentum of the man about the center point must be computed using his absolute velocity. Momentum and impulse equations are useful for determining the outcome of an impact. Earlier in this course we spent some time studying particle impacts. Because all the mass of a particle is concentrated at the center of mass, the point of impact does not matter - only the magnitude and direction of the particle velocities. It is important that you understand how to solve these types of problems with particles. Rigid body impacts are more complex, because the point of impact and the velocity vectors affect the subsequent motion. To solve any kind of realistic problem with rigid body impacts it is not practical to use the methods presented in this class. Generalized equations based on methods of Lagrangian dynamics are used. In your textbook a number of examples are shown in the chapters on impulse and momentum which use momentum equations to solve problems involving constant forces acting over a period of time; for example, things rolling or sliding on inclined planes, or weights in pulleys. In this case we say that the initial linear or angular momentum plus the integral of the force or moment with respect to time equals the final linear or angular momentum. If the force or moment is constant, this looks like an easy way to solve a problem that asks what the final velocity is after a certain time, or to find the time necessary to achieve a particular velocity. In reality, using this method is the same as using forces and accelerations, solving for acceleration and integrating over time (i.e. multiplying if the acceleration is constant). So, be careful in using this method. As in the force and acceleration method, draw careful diagrams. Remember that momentum is a vector quantity, get the signs right, and make sure that all your forces are, in fact, constant. If you find this method confusing, there is absolutely no reason to use it for solving this type of problem. Just use force and acceleration to find the accelerations (or angular accelerations) and integrate to get velocity, angular rate, or distance. (Make sure you do understand how to use momentum methods with no external forces and when there are impacts.) ASEN 2003 QA.doc 03/25/98 5

6 I could use some more practice determining and using different reference frames. The coriolis term is another area that is fuzzy for me. I am not sure where, exactly, it comes from. Relative motion problems I don t understand questions that involve relative radius, velocities, accelerations Moving reference frame problems created real pain for me. RELATIVE MOTION AND ROTATING REFERENCE FRAMES - Summary In defining the motion of a particle or body, in most cases we must determine how the body is moving with respect to an inertial (or absolute) reference frame. An inertial frame is one that does not accelerate. This implies that it also does not rotate. On the other hand, it is often convenient to measure or represent the motion of one object with respect to a moving frame. For example, to determine the angular momentum of a person in a teacup amusement park ride, it would be difficult to write down directly their velocity vector with respect to inertial space. The kinematic equations given in Chapters 12 and 16.8 in your book provide a systematic means to express the position, velocity, and acceleration vectors of a particle or a point on a rigid body with respect to an inertial frame. This topic can be conceptually difficult and to really "get it" you must construct a mental image for yourself of how it works. Try going back to the ant on the spinning plate, or if that doesn't work for you try to develop your own model that you can return to whenever the topic comes up. Once you have a solid mental image of what is going on, go back and slog through the equations and examples in the chapter and make sure you can connect them to your reference image. Getting the details right when faced with a problem will require lots of practice. Here are some thoughts: 1) Any time you use a velocity or acceleration in an equation make sure you understand exactly what it refers to. Be very careful about your notation - this may be a pain, but will help you get it right. Understand what is meant by r B/A, v B, (v B/A ) xyz. 2) Define a reference point and coordinate system. The reference point and coordinate system should be selected so that you can easily determine the position and velocity of the point of interest. This helps to break down the problem into simpler pieces. For example, in the yo-yo despinner problem picking the reference point to be the point at which the chord is tangent to the spacecraft makes it very easy to define the position of the yo-yo mass. It is not immediately obvious what the motion of this point is, but now we have separated this difficulty from also considering the motion of the end mass. Once you have the reference point and coordinate frame selected you should be able to immediately write down the position vector of the point of interest with respect to the reference point and the velocity of this point with respect to the reference coordinate frame. If you can't do this, your choice of reference point and frame were probably not the best. 3) Describe the motion of the reference point and coordinate frame. Now figure out what the motion of your reference point and coordinate frame are. Probably the first step is to find the rotation rate of the frame. For most problems in this class, the angular rate will be a constant, but check it. If the reference point is not at the center of the rotating frame, its velocity and acceleration must be determined by considering the rotating motion. If it gets too complicated you can introduce a second rotating reference frame; although this usually won't occur until we go to 3-D motion in our Junior level course. 4) Now put it all together using the equations in Chapter 12 or To understand where each term in the full up acceleration expression comes from (especially the coriolis term), you really need to just work through the derivation carefully. If you have trouble with this please see me or one of the TA's for some help. ASEN 2003 QA.doc 03/25/98 6