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Physics 430: Lecture 7 Kinetic and Potential Energy Dale E. Gary NJIT Physics Department.

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Presentation on theme: "Physics 430: Lecture 7 Kinetic and Potential Energy Dale E. Gary NJIT Physics Department."— Presentation transcript:

1 Physics 430: Lecture 7 Kinetic and Potential Energy Dale E. Gary NJIT Physics Department

2 September 22, 2009  We are now going to take up the conservation of energy, and its implications. You have all seen this before, but now we will use a powerful, more mathematical description.  You will see that the discussion is more complicated that the other conservation laws for linear and angular momentum. The main reason is that each type of momentum comes in only one flavor, whereas there are many forms of energy (kinetic, several kinds of potential, thermal, etc.).  Processes transform one type of energy into another, and it is only the total energy that is conserved, hence the additional complication.  We will be introducing new mathematical tools of vector calculus, such as the gradient and the curl, which you may be familiar with, or not. I will give you the needed background as they come up. Chapter 4—Energy

3 September 22, 2009  An obvious form of energy is energy of motion, or kinetic energy. We will use the symbol T, which is perhaps strange to you but is very much standard in Classical Mechanics. The kinetic energy of a particle of mass m traveling at speed v is defined to be:  Consider such a particle moving on some trajectory through space while its kinetic energy changes, on moving from position r 1 to r 1 + dr. We can take the time derivative of the kinetic energy, after writing, so that  But the first term on the right is the force. Thus, we can write the derivative of kinetic energy as  Finally, multiplying both sides by dt and noting that v dt = dr, we have 4.1 Kinetic Energy and Work r1r1 r1+drr1+dr drdr Work-KE Theorem

4 September 22, 2009  The equation just derived is only valid for an infinitesimal displacement, but we can extend this to macroscopic displacements by integrating, to get: which says that the change in kinetic energy of a particle is equal to the sum of force (in the direction of the displacement) times the incremental displacement.  However, note that this is the displacement along the path of the particle. Such an integral is called a line integral. In evaluating the integral, it is usually possible to convert it into an ordinary integral over a single variable, as in the following example (which we will look at in a moment).  With the notation of the line integral where the last is a definition, defining the work done by F in moving from point 1 to point 2. Note that F is the net force on the particle, but we can also add up the work done by each force separately and write: Line Integrals and Work

5 September 22, 2009  Evaluate the line integral for the work done by the 2-d force F = (y, 2x) going from the origin O to the point P = (1, 1) along each of the three paths: a) OQ then QP b) OP along x = y c) OP along a circle  Path a):  Path b): Example 4.1: Three Line Integrals P O Q

6 September 22, 2009  Path c): This is a tricky one. Path c can be expressed as so This is a parametric equation, using  as a parameter along the path. With this parameter, F = (sin , 2(1  cos  )). With this substitution:  The point here is that the line integral depends on the path, in general (but not for special kinds of forces, which we will introduce in a moment). Example 4.1: Three Line Integrals

7 September 22, 2009  We must now introduce the concept of potential energy corresponding to the forces on an object. As you know, not every force lends itself to a corresponding potential energy. Those that do are called conservative forces.  There are two conditions that a force must satisfy to be considered a conservative force.  The first condition for a force F to be conservative is that F depends only on the position r of the object on which it acts. It cannot depend on velocity, time, or any other parameter.  Although this is restrictive, there are plenty of forces that satisfy this condition, such as gravity, the spring force, the electric force. You can often see this directly, such as for the gravitational force: 4.2 Potential Energy and Conservative Forces Depends only on r.

8 September 22, 2009  The second condition for a force to be conservative concerns the work done by the force as the object on which it acts moves between two points r 1 and r 2 (or just points 1 and 2, for short)  Reusing our earlier figure, we saw in Example 4.1 that the force described there was NOT conservative, because it did different amounts of work for the three paths a, b, and c.  Forces involving friction, obviously are not conservative, because if you were sliding a box, say, on a surface with friction along the three paths shown, the friction would do work, where L is different for the three paths. Such forces are non-conservative. Non-Conservative Forces 2 1

9 September 22, 2009  The force of gravity, on the other hand, has the property that the work done is independent of the path. You know that if the height of point 1 and point 2 differ by an amount  h, then you will drop in height by h no matter what path you take. In fact independent of path.  The conditions for a force to be conservative, then, are: Conservative Forces Conditions for a Force to be Conservative A force F acting on a particle is conservative if and only if it satisfies two conditions: 1. F depends only on the particle’s position r (and not on the velocity v, or the time t, or any other variable); that is, F = F(r). 2. For any two points 1 and 2, the work W(1  2) done by F is the same for all paths between 1 and 2.

10 September 22, 2009  The reason that forces meeting these conditions are called conservative is that, if all of the forces on an object are conservative we can define a quantity called potential energy, denoted U(r), a function only of position, with the property that the total mechanical energy is constant, i.e. is conserved.  To define the potential energy, we must first choose a reference point r o, at which U is defined to be zero. (For gravity, we typically choose the reference point to be ground level.) Then U(r), the potential energy, at any arbitrary point r, is defined to be  In words, U(r) is minus the work done by F when the particle moves from the reference point r o to the point r. Potential Energy

11 September 22, 2009  Statement of the problem: A charge q is placed in a uniform electric field pointing in the x direction with strength E o, so that the force on q is. Show that this force is conservative and find the corresponding potential energy.  Solution: The work done by F in going between any two points 1 and 2 along any path (which is negative potential energy) is: This work done is independent of the path, because the electric force depends only on position, i.e. the force is conservative. To find the corresponding potential energy, we must first choose a reference point at which U is zero. A natural choice is to choose our origin (the point 1), in which case the potential energy is You may recall that  E o x is the electric potential V, so that qV is the potential energy. Example 4.2: Potential Energy of a Charge in a Uniform Electric Field

12 September 22, 2009  The potential energy can be defined even when more than one force is acting, so long as all of the forces are conservative. An important example is when both gravity F grav and a spring force F spr are acting (so long as the spring obeys Hooke’s Law, F(r) =  kr ).  The work-kinetic energy theorem says that if we move an object subject to these two forces along some path, the forces will do work independent of the path (depending only on the two end-points of the path) given by  Rearrangement shows that hence total mechanical energy is conserved. Extended to n such forces: Several Forces Principle of Conservation of Energy for One Particle If all of the n forces F i (i=1…n) acting on a particle are conservative, each with its corresponding potential energy U i (r), the total mechanical energy defined as is constant in time.

13 September 22, 2009  As we have seen, not all forces are conservative, meaning we cannot define a corresponding potential energy. As you might guess, in that case we cannot define a conserved mechanical energy.  Nevertheless, if there are some conservative forces acting, for which a potential energy can be defined, then we can divide the forces into a conservative part F cons, and a nonconservative part F nc, such that which allows us to write  What this says is that mechanical energy ( T + U ) is no longer conserved, but any changes in mechanical energy are precisely equal to the work done by the nonconservative forces.  In many problems, the only nonconservative force is friction, which acts in the direction opposite the motion so that the work is negative. Nonconservative Forces

14 September 22, 2009 Example 4.3: Block Sliding Down an Incline  We did this problem using forces in lecture 2. Let’s now apply these ideas of energy to arrive at the same result.  As before, we have to identify the forces, and set a coordinate system, but this time we write down the potential and kinetic energies in the problem.  The kinetic energy, as always, is T = ½ mv 2.  The gravitational potential energy is U = mgy, where we can set y = 0 (and hence U = 0 ) at the ground level.  The friction force does negative work W fric =  fd, but recall that f =  N where. Putting all of this together, becomes where d is the distance along the incline, and  y is the change in height.  If the block starts out with zero initial velocity at the top of the incline, and we ask what is the speed v at the bottom, then  y =  h =  d sin , so or  mgmg N f h

15 September 22, 2009 Comparison with Example 1.1  When we did this problem using forces, we obtained equation of motion from which, after integration, we got the expression  Comparing with our just derived expression they may seem quite different. What is happening is that using forces we can get the velocity versus time, whereas with energy we are only getting the speed at the end points. Energy considerations are very powerful if you just want to know the result at a particular point, in which case you can ignore the details of the motion in getting there. If you instead need to know the path taken, or the details along the path, you have to use the tools of Newton’s Laws.  However, we will find in a few weeks that these energy considerations do contain all of the information of Newton’s Laws, and we will build the tools necessary in Lagrangian mechanics to get the equation of motion starting from energy. This allows us to attack much more complicated problems. For this reason, it is important to get good at energy problems. Here is an example you probably have seen before.

16 September 22, 2009  Statement of the problem: (a) The force exerted by a one-dimensional spring, fixed at one end, is F =  kx, where x is the displacement of the other end from its equilibrium position. Assuming that this force is conservative (which it is) show that the corresponding potential energy is U = ½ kx 2, if we choose U = 0 at its equilibrium position.  Solution to (a): We start with the definition of potential energy: But we choose U = 0 at x = x 1, which amounts to choosing x 1 = 0, so that Problem 4.9

17 September 22, 2009  Statement of the problem: (b) Suppose this spring is hung vertically from the ceiling with a mass m suspended from the other end, and constrained to move in the vertical direction only. Find the extension x o of the new equilibrium position with the suspended mass. Show that the total potential energy (spring plus gravity) has the same form ½ ky 2 if we use the coordinate y equal to the displacement measured from the new equilibrium position at x = x o (and redefine our reference point so that U = 0 at y = 0 ).  Solution to (b): The new equilibrium position is reached when the force of the stretched spring kx o equals the force of gravity on the mass mg. Thus To define the potential energy at the new equilibrium position, we have to examine the work done in displacing the mass a distance y : Problem 4.9, cont’d xoxo y=0

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