Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Energy Suppose the body on our track is acted upon by a force which depends on the coordinate x in a known way: Consider the quantity x V(x) = − I F(x´) dx´ F = F(x) , where F(x) is a specified function of x. (It doesn’t depend on the velocity or on the time.) The differential equation which determines x(t) is m d 2 x(t) = F(x(t)) . dt2 Maybe there’s a quantity that is always the same as time goes on. Such a quantity is called a constant of the motion, and gives us powerful information even in the absence of an explicit solution. The most important constant of the motion, and the one we will talk about the most, is called the total energy. Let’s see how it is defined and how it is used. . This is the negative of the area under the graph of F(x´), from an arbitrary initial position x´=x0 to x´=x : Force F(x') 0101010101 Now, the dependence of F on x could be very complicated, perhaps so complicated that the only way we can solve the relevant differential equation is by using some elaborate numerical procedure. Is there an easy way to get at least some information about the motion, without resorting to a computer? x0 AREA = −V(x) x0 x x' Note that x´ is just a dummy variable. The function V(x) does not depend on x´. Notice also that the force is just the negative of the slope of V(x): dV(x) F(x) = − dx (by the Fundamental Theorem of Calculus.) Example: the harmonic oscillator This all might seem pretty abstract, until you see a specific example. For the harmonic oscillator F(x) = – kx . Then V(x) is just quadratic in x: x V(x) = − I ( − kx´) dx´ x0 = 1 1 k x 2 − k x 20 2 2 = 1 k x 2 + arbitrary constant . 2 V(x) change, when x changes this way? d dV(x(t)) dx(t) V(x(t)) = dt dt dx = − F(x(t)) dx(t) . dt (The first step uses the chain rule.) This may be more compactly written as dV = − Fv , dt where it is understood that everything is evaluated at x(t). Next, let’s consider a completely different quantity, which we will symbolize by T (not to be confused with the period of periodic motion, which is sometimes also symbolized by T !): T= 1 mv2 2 . The above example highlights the fact that the potential energy is only defined to within an arbitrary additive constant. This reflects the arbitrariness of the choice of starting point for the integral. Another way of saying this is that we may choose to make the potential energy equal zero at any desired point. where the last step uses Newton’s law ma=F. Now, as the body moves, the value of x changes in time according to x(t). How does the value of We notice that V and T change in equal but opposite ways over time. If V loses a bit, then T How does it change in time? dT dv = mv = mva = Fv , dt dt gains the same amount, and vice versa. Another way to say this is that their sum, T + V , is a constant. This constant is called the total energy: E=T+V . When the external force is not a function of velocity or time, the total energy is conserved: dE =0 dt . The quantity T is called the kinetic energy - it is the energy associated with the motion of the body. The quantity V is called the potential energy - it is the amount of energy stored up in whatever is causing the force on the body. We note that the kinetic energy is not an intrinsic property of a body. It depends on the frame of reference. For example, we can always view a body from a frame in which the body is at rest. Its kinetic energy is then zero, which is not true in other frames. Similarly, the potential energy is not an intrinsic property, because we are free to add any constant value to it. Despite these shortcomings, the concept of energy is extremely useful because it is con- served in any inertial frame, as long as the force depends only on position. The energy provides us with one of our main tools (sometimes our only tool) to get information about motion. To find the dimension of energy, we look at an equation which defines an energy in terms of more fundamental quantities. A good choice is, T= 1 2 mv . 2 Then it’s easy to see that the dimension of energy is [E] = mass (length)2 (time)− 2 . In the system of units we are using, the unit of energy is called the Joule (sounds like “jewel”): 1 Joule = 1 kg m2 s-2 . Example: the harmonic oscillator A very instructive example is the simple harmonic oscillator. We worked out the potential energy above; we are free to take the constant to be zero, corresponding to zero potential energy when the spring is not stretched. Then the amount of energy stored in the spring when it is stretched by a distance x is V(x) = kx2/2. The farther the spring is stretched, the more energy is stored in the spring and the less is available for kinetic energy; the velocity therefore decreases. Eventually, all the energy is stored in the spring and the body comes to rest; this occurs at one of the turning points of the oscillator. The next plot shows the kinetic, potential and total energies versus distance, for the oscillator. The resources for this section contain a movie which displays the exchange of potential and kinetic energies during the motion. It is worth studying closely. Well, right away we know the total energy of the body, because at the beginning the kinetic energy was zero. The total energy is equal to the amount of potential energy at the beginning: E= 1 2 kx0 . 2 (Recall that we worked out the potential energy earlier, as an example.) The body reaches its maximum velocity when the potential energy is a minimum; that is, at x=0. At that point, all of its energy is kinetic. Hence, 1 1 m v 2max = k x 20 , 2 2 Energy total or kinetic v max = potential x Sub-example: what is the maximum velocity? Here is an example of the kind of problem that is easily solved using energy. Suppose we start off the oscillator by pulling it to some initial point x0 and releasing it with zero initial velocity. What will be its maximum velocity? k x . m 0 You can easily verify this answer from the detailed treatment of the harmonic oscillator presented earlier. Example: gravitational potential near earth Let’s analyze the gravitational force near the surface of the earth, from the point of view of energy. If we orient our coordinates with positive values pointing upwards, then the force due to gravity is F = – mg , This agrees with an earlier result. the minus sign indicating that the force is downwards. In summary, the relation E= The potential energy is x V(x) = − I ( − mg) dx´ x0 = mgx + constant . We see that the potential energy grows linearly with x. The higher the body goes, the less energy is available for kinetic energy, so the body slows down. Sub-example: how high does the body go? relates the velocity v and the position x, allowing you to solve for one in terms of the other if you wish. And you can do this without needing to know x(t). Application: periodic motion What property is shared by every periodic motion? The body stops and turns around at both ends. Here is a typical case: Suppose we start it at x=0, with velocity v0. How high does it go? We may choose V to be zero at any particular value of x we desire, by adjusting the constant. It is convenient to let V=0 at x=0; then the constant is zero. The initial potential energy of the body is therefore zero, and its total energy is E=mv02/2. At the top, its kinetic energy is zero, and all the energy is potential. Hence 1 2 mv 0 = mg xmax, or 2 xmax v 20 = . 2g 1 mv 2 + V(x) 2 Energy V(x) E total kinetic potential x min x x max Using energy methods, we can arrive at a closed expression for the period of such a motion. Sol- ving the above expression for v, we obtain dx = dt 2 á E − V(x) é . m Here we took the positive square root; we’ll multiply by two for the return journey below. Rearranging, we get dt = m 2 dx . E − V(x) Let’s integrate both sides of this expression, from one turning point to the other. The integral of the left-hand side will give half the period, and therefore the full period is T= xmax 2m I xmin dx E − V(x) . Dissipation In previous sections, we considered motion in the presence of a damping force –bv, where b is a positive constant. For example, we considered the damped harmonic oscillator, and saw that its oscillations become smaller and smaller as time goes on. Evidently, energy is lost by the moving body. This energy loss is called dissipation. What is the rate of energy loss? Let’s not make any assumptions about the total force acting on the body, except that it is of the form Ftot = F(x) − bv . As before, we define the total energy of the moving body as the sum of its kinetic plus potential energy: E= Exercise The potential energy for the simple harmonic oscillator is V(x) = 1 m ω20 x 2 . 2 Using the above formula, verify our earlier expression T=2π/ω 0 for the period. 1 mv 2 + V(x) , 2 where V is the potential energy due to F : F(x) = − dV(x) . dx Differentiating the total energy, we find dE dv dx dV = mv + = v(ma − F) . dt dt dt dx The differential equation governing the motion is ma = F(x) − bv , where b is the positive damping constant. Hence, the rate of change of total energy is dE = − bv 2 . dt This is negative, indicating that energy is lost when the velocity is nonzero. Where does the energy go? It is transferred to whatever is causing the damping force. For example, if the damping is due to friction with a track, then the lost energy is transferred to the material of the track. It heats up as a result. The total energy of body plus track is conserved, of course. Advanced example of use of energy method: gravity including variation with height Here is another interesting example of the use of energy methods. Let’s analyze the motion of a body falling in the gravitational field of the earth, taking into account the fact that the force varies with distance. (In an earlier section, we took the gravitational force to be constant, which is a good approximation near the earth’s surface.) Although we have not yet studied gravity, you may know the result for the force law already. If we measure x upwards from the center of the earth, the force exerted by the earth on a body at or above the earth’s surface is GMm F= − 2 . x The minus sign indicates that the force points downwards. G is Newton’s constant, M is the mass of the earth, and m is the mass of the body. The potential energy is V(x) = − GMm + constant , x and we choose the constant to be zero by fixing V=0 as x becomes infinite. The total energy of the body is therefore E= 1 2 GMm mv − . x 2 Now, suppose the body starts at x=x0 with zero initial velocity. Then its total energy is E = –GMm/x0. Solving for the square of the velocity, we find åäåå dx ã dt ä ëì 2 ë ìì = 2GM åååå 1 − 1 ìììì . í ã x x0 í This has an interesting solution, one that you probably couldn’t derive without a lot of work and inspiration. It involves a new variable η (read “eta”), which is related to the time by t= force is accounted for, because the force is stronger, the further down the body gets. x0 x 0 (η + sin η) . 2GM 2 In terms of this new variable, you can easily verify (try it!) that the solution is given by x= x0 (1 + cos η) . 2 x(t) is the curve traced out by a point on an imaginary circle of diameter x0 “rolling” through angle η along the t axis. Such a curve is called a cycloid. x 0101010101010101 1001010101 x0 η x0 η 2 These facts are illustrated in the following figure, which compares the paths taken with constant gravitational force and with the true force which varies with distance: x0 2 x0 η sin 2 Second, according to the above figure the body apparently reaches infinite velocity at the center of the earth! However, this is not true because this force law ceases to be valid once the surface of the earth is reached. (We will explain the latter point when we talk about the gravitational interaction in more detail later.) If you bored a hole into the earth so the body could continue falling even after it reached the earth’s surface, the force law that we used to derive the solution would no longer apply. x(t) 2GM t x0 The resources for this section contain a movie illustrating the trajectory. There are two things to note. First, the body falls faster when the variation of the gravitational constant force varying force surface of earth t