Ch9 Page 1 Chapter 9: Momentum Tuesday, September 17, 2013 10:00 PM In this chapter and the next one, we'll explore alternative perspectives to Newton's second law. The concepts of momentum and energy provide pathways into deeper understandings of Newtonian mechanics. The basic understanding of Newton's second law that we have discussed so far is that the acceleration of an object is caused by the net force acting on it. However, there are situations where it is difficult to measure the force acting on an object, and so therefore it's difficult to apply Newton's second law. For example, the force may act for only a very short time, or over a very short distance, or both. This is the case in collisions, for example; kicking a ball, hitting a baseball with a bat, a slap shot in hockey, hitting a tennis ball with a racquet, a car collision, etc. In such situations, it's typical that the force varies dramatically over a short time; this is much more complicated than the situations we've dealt with so far, where we often assumed that the force acting is constant (which is a good approximation in some situations).
Ch9 Page 2 How then do we analyze such situations? Well, we can think about averaging the force to make things simple. But should we average the force over time or over distance? Each of the ideas has advantages and useful properties; averaging the force over time leads to the concepts of impulse and momentum, whereas averaging the force over distance leads to the concepts of work and energy, as we shall see. Then, as we shall see, Newton's second law of motion can be expressed in terms of these new concepts in the following ways: Impulse = change in momentum Work = change in kinetic energy We'll also see that there are conservation principles for momentum and energy. Conservation principles are very useful in physics; the world is a very complicated place, and so if you can identify some quantities that are constant throughout the complicated processes that you are analyzing, then it gives you something to hang on to. Conservation principles are useful in solving physics problems, and often allow one to solve problems more simply and more directly than using Newton's laws of motion by themselves. If you dig deeper into things, you'll find that conservation principles are a consequence of certain somewhat abstract symmetry principles. For example, the fact that momentum is conserved is a consequence of the fact that the fundamental laws of physics are invariant with respect to spatial translations. In other words, if you do an experiment here in St. Catharines, and then slid your apparatus over to Buffalo, or Toronto, or anywhere, then you will find that the experimental results are the same. This is a very deep connection between underlying symmetries of the universe and conservation principles; our deepest
Ch9 Page 3 understandings of the universe are currently expressed in these terms of symmetry and invariance. Other examples: Conservation of energy is a consequence of invariance with respect to time translations, and conservation of angular momentum is a consequence of invariance with respect to spatial rotations. This is the story in Newtonian mechanics, but similar, and similarly deep (some would say more fundamental) symmetry principles apply in quantum mechanics, too, and you'll encounter them eventually if you continue your studies in this direction. Back to the story of how to cope with forces when they vary wildly in magnitude over very short times or over very short distances. Ok, let's now explore the affect of impulse on motion. Recall from above that the definition of impulse is the average force acting on an object times the time interval over which the force acts: How does impulse affect motion? Consider the following calculation: This calculation suggests that the quantity mv may be important, and so it's worthwhile giving it a name; we call it momentum. The relation that we've just derived above, which describes the effect of impulse on motion, is called the impulse-momentum theorem.
Ch9 Page 4 examples: tennis, baseball, hockey, catching an egg or water-balloon or a hard-thrown ball Example: A baseball of mass 150 g is thrown towards home plate with a speed of 100 km/h. The batter hits the ball with an impulse of 10 N s so that it reverses its motion. Determine the speed at which the ball leaves the bat.
Example: A tennis ball of mass 90 g arrives at your racquet with a speed of 80 km/h and you hit it directly back at a speed of 60 km/h. (a) Determine the impulse that you exert on the ball. (b) Determine the magnitude of the average force that you exert on the ball if it is in contact with your racquet for 14 ms. Ch9 Page 5
Ch9 Page 6 Principle of conservation of momentum Consider the impulse-momentum theorem, If it happens that the impulse is zero, then the change in momentum will also be zero. In other words, if the impulse acting on a system is zero, then momentum is
Ch9 Page 7 conserved. Yet another way to say this is that if the net external force on a system is zero, then the total momentum of the system is conserved. The principle of conservation of momentum is a generalization of Newton's third law of motion. That is, the principle of conservation of momentum is considered to be more fundamental than Newton's third law of motion, and furthermore Newton's third law of motion can be derived from the principle of conservation of momentum. Furthermore, the principle of conservation of momentum is more general than Newton's third law of motion because the former applies to light and fields as well as particles, whereas the latter applies only to particles. Furthermore, because force is the derivative of momentum, one can get by without the force concept as long as one uses momentum. This is true to an even greater extent in more advanced approaches to physics (quantum mechanics, relativity, field theories) where force is not very useful, but momentum is extremely useful. The principle of conservation is useful in analyzing collisions, explosions, etc. Example: A car of mass 1000 kg travelling east at 40 km/h collides with a car of mass 1500 kg going west at 50 km/h in a completely inelastic collision (i.e. they stick together). Determine the velocity of the car/car combination immediately after the collision.
Ch9 Page 8 The velocity of the two cars together after the collision is 14 km/h to the West. Example: A truck of mass 2000 kg travelling east at 60 km/h collides with a car of mass 1000 kg going north at 80 km/h in a completely inelastic collision (i.e. they stick together). Determine the velocity of the car/truck combination immediately after the collision. Solution: This is a two-dimensional problem, so we'll adopt the usual math-class conventions:
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Ch9 Page 10 This is the final velocity of the two cars together. If you prefer to express the velocity in terms of its magnitude and direction, then: Thus, the final velocity of the car-truck combination is 48.1 km/h at an angle of 33.7 degrees North of East. Example: A rubber ball of mass 5 kg travelling to the right at 10 m/s collides with a ball of mass 10 kg going to the left at 4 m/s. After the collision, the first ball moves at a speed of 2 m/s to the left. Determine the velocity of the second ball immediately after the collision.
Ch9 Page 11 Thus, the velocity of the second ball after the collision is 2 m/s to the right. Angular momentum One perspective on Newton's second law is This led to identifying momentum as a concept of interest. Is there a rotational analogue of momentum?
Ch9 Page 12 Principle of conservation of angular momentum: If the net external torque acting on a system is zero, then the angular momentum of the system is conserved. Example: a spinning figure skater; note how the angular speed of the skater changes as she brings her arms in or out. When she brings her arms in, her moment of inertia decreases, so her angular speed increases as her angular momentum stays constant. When she wishes to slow down, she extends her arms again, which increases her angular momentum and decreases her angular speed, again maintaining her constant angular momentum.
Ch9 Page 13 Behaviour of a spinning bicycle wheel when external torques are applied. demonstrations Note that angular momentum is a vector, and so is torque; using the right-hand rule we can predict how a spinning wheel behaves when various torques are applied. Example: Precession of a spinning bicycle wheel: http://www.youtube.com/watch?v=8h98bgrzpom Example: "Push Steering" (also known as counter steering)
Ch9 Page 14 Check the following video for an explanation of how push steering works; ignore the first two minutes of physics explanation, and concentrate on the motions of the front wheel of the motorcycle. We do this unconsciously, as our bodies just do the right thing to keep our balance without us thinking consciously about it. But if you pay close attention next time you are on a bicycle (or motorcycle) then you'll become aware of the effect. The effect is more noticeable on a motorcycle because motorcycles are heavier than bicycles. Bicycles are lighter, so we can do most of the work by leaning our bodies, whereas you really have to use push steering on a motorcycle or be in danger of falling. http://www.youtube.com/embed/c848r9xwrjc?rel=0
Ch9 Page 15 Using conservation of angular momentum to solve a problem: Example: Problem 71 The figure shows a 100-g puck revolving in a 20-cm circle on a frictionless table. The string passes through a hole in the centre of the table and is tied to two 200-g masses. (a) What speed does the puck need to support the two masses? (b) The lower mass is a light bag filled with sand. Suppose a hole is poked in the bag and the sand slowly leaks out as the puck is revolving. Determine the speed of the puck and the radius of its path after all the sand is gone.
Ch9 Page 16 Solution: (a) Draw a free-body diagram for the two hanging masses considered as one object. Then use the free-body diagram to determine the tension in the string, assuming that the speed of the puck is sufficient to keep the hanging masses still. Finally determine the speed at which the puck must move so that the tension in the string provides all the centripetal force needed to keep the puck in a circle of radius 20 cm. (b) The hanging mass decreases by a factor of 2, so the tension in the string also decreases by a factor of 2. Thus, the tension in the string is T = 0.2g. Let V be the final speed of the puck and let R be the final radius of the puck's circular path. Then
Ch9 Page 17 Our task is to determine V and R, but we have only one equation for two unknowns. We need another independent equation. The other equation can be obtained by noting that the falling sand does not have any angular momentum with respect to the central axis of the puck's revolution. This means that if we consider the hanging mass, the sand, and the puck to be one single system, then its angular momentum is conserved. Neither the hanging weight nor the sand have any angular momentum, so this means that the angular momentum of the puck is conserved. Thus, Substituting this expression into equation (1) and solving for R, we obtain
Ch9 Page 18 The final speed of the puck is Thus, when the sand leaks out of the bag, the puck slows down and moves in a larger circle. Is this reasonable? For example, what would happen if the other hanging mass slipped off the string?