EN4: Dynamics and Vibrations

Brown University

Thus far the course has focussed on using Newton's Law to solve problems. The essential idea involves the idealization of the system of interest as a particle, identification the forces that act on the particle, and integration the resulting equations to obtain the trajectory of the particle. In principle these are the only steps required to solve any problem involving a single particle. However, in practice it is realized that the rote application of Newton's equations gives rise to systems of equations that does not afford a simple and tractable solution. It is, therefore, in quest of simpler methods that the work and energy concepts are introduced into mechanics. It may be noted that these concepts are of great importance in mechanics and physics especially in dealing with systems with many particles (such as the gas in the lecture hall which is composed of about 10²⁵ atoms).

A first hand understanding of the difficulties involved in using Newton's law may be

obtained by inspecting the following problem. Here the motion of a roller coaster trolley is to be obtained. Specifically, if the roller coaster starts from the point A with negligible speed compute the speed of the roller coaster at the point B. The profile of the track is given as the function . The roller coaster is, of course, located on earth and therefore gravitational effects have to be considered.

Challenge: Assuming , compute the velocity of the roller coaster at . The mass of the roller coaster may be taken to be m, and the track to be frictionless. The roller coaster is released from rest at . No other concepts and principles other than those covered in this course so far may be used!

If the solution of the above problem is attempted, it will be realized very quickly that the math gets to be quite complicated. This problem illustrates the need to consider other methods that will eliminate these irksome mathematical difficulties.

This section contains the definitions of work and kinetic energy, and the relationship between them in the form of the work energy theorem. The great simplification of the solution process (at least for a certain class of problems) will become apparent in the examples that are considered

To understand the concept of work, it is necessary to develop some intuition about it. To this end, a force F acting on a particle and displacing it through a small distance dr as shown in the figure may be considered. The work done by the force must somehow be related to the "effort put in" by the agency that is responsible for the application of the force. This effort must be related to the magnitude of the applied force, i.e., if the applied force is larger, the work done must be larger (for a given dr). Similarly, the "effort" must be proportional to the distance moved, i.e., the work done by the given force F will be larger if it moves through a larger distance. In other words, the magnitude of "work done" should increase with the magnitude of the force and the displacement produced by that force. It seems reasonable to define the infinitesimal work done by a force F moving a body through an infinitesimal distance dr to be

.

This quantity satisfies the intuitive requirements that were imposed before. The total work done in going from the point A to the point B, may now be obtained by an integration of the infinitesimal work from A to B, i.e.,

To understand better the concept of work, a force F may be considered, that acts on a particle and causes it to displace by an amount dr. Note that the above definition of work may be rewritten as , where is the angle made between the force vector and the displacement. Thus a very important point may be realized: Only the component of the force along the direction of the actual displacement does work. Referring to the figure above, only the component of the force along the vector dr (shown as a brown arrow parallel to dr) does work in moving the particle through dr. This is a very important point and will prove to be immensely useful as will be seen below. Note also that if the angle made by the force is bigger than a right angle (positive or negative), the work done would be negative. In particular, if the force is perpendicular to the displacement, no work is done.

1. It must be emphasized that the work done is a scalar quantity that was defined. In other words, it contains no new information other than the force and the displacement. The only difference is that the information is contained in a different, and sometimes very useful, form.
2. It must be noted that the work done depends on the force, and the path through which the particle moved under the influence of the force. There may be many paths that connect point A and B, and the work done in going from point A to B will, in general depend on the particular path chosen.
3. The units of work: In SI metric units, the unit of work is Nm which is otherwise known as a Joule.

Thus far particle has been assigned one intrinsic property, i.e., its mass. The position of the particle as a function of time is described by its position-vector, and its velocity and acceleration may be obtained by differentiation of the position vector. In this section a new quantity that combines the intrinsic property of the particle - its mass - is combined with a kinematic property - its velocity - to obtain a new quantity called the Kinetic energy. The kinetic energy of a particle may be intuitively understood to be as the energy it possesses do to its motion.

The kinetic energy of a particle of mass m, moving with a velocity v with respect to a fixed set of axes is defined as

.

This is, like work, a scalar quantity. The factor of one-half is just a matter of convention, and is introduced anticipating the simplifications in the formulae derived later. A little reflection would reveal that kinetic energy has the same units as those of work! This is not an accident, as will be seen in the next section.

In the last two sections two new quantities were introduced. This section will obtain a connection between the two quantities. It will be shown that the work done in moving a particle between two points will be equal to the change in its kinetic energy. This is known as the Work-Energy Theorem, is of great importance in physics.

To prove the statement made above, the expression for the work done in moving the particle from point A to point B may be revisited.

Using Newton's Law, and following a chain of mathematical arguments stated below,

the statement of the work-energy theorem is proved!

In this section, we shall apply the work-energy theorem to solve some problems that would be hard to solve using Newton's Law. Work-energy methods are especially useful in solving problems where the force acting on the particle is known as a function of position.

Example 1: A ball of mass m is dropped from a height h above the floor. Determine the velocity of the ball just before it hits the floor.

Solution: Consider a coordinate system such that the horizontal direction is the i-axis and the vertical direction is the j-axis. The initial coordinates (point A) of the ball may be taken to be (0,h). Let v be the speed of the ball just before it hits the floor at the point B with coordinates (0,0).

Step 1: Compute the work done. To do this it is required to know the force acting on the ball. In this case there is only the force of gravity which is equal to -mgj. The infinitesimal change in the position vector can be obtained as dr = dy j. Performing the work integral gives

.

So the work done by gravity is mgh.

Step 2: Compute the change in kinetic energy. The initial velocity of the ball is zero and therefore T_A = 0. Since we have assumed the velocity at point B to be -vj, the kinetic energy at the point B is , which gives that the change in kinetic energy is .

Step 3: Apply work-energy theorem. Thus the work done by gravity must equal the change in the kinetic energy of the body, or,

which implies that the velocity of the ball just before it hits the floor is . Note that the answer is independent of the mass of the ball.

This problem may be solved alternatively using Newton's Law. If this is done, it will be seen that the same answer will be obtained.

The above example illustrates the steps that are involved in using the work-energy theorem. However, it does not illustrate the usefulness of this theorem, since the answer could have been obtained, by use of Newton's Law, without difficulty. The next example will look at a more non-trivial example of the use of the work-energy principle.

Example 2: Back to the roller coaster challenge. Assuming , compute the velocity of the roller coaster at . The mass of the roller coaster may be taken to be m, and the track to be frictionless. The roller coaster is released from rest at .

Step 1: Compute the work done. The point A has coordinates and the point B has coordinates . The are two forces acting on the roller coaster, the first being the force of gravity which is equal to -mgj, and the normal reaction N from the track (which is unknown). Thus the work done in going from point A to point B is

.

Note that in going doing the math, the fact that is used; in other words, the normal reaction does no work. Also and hence the result.

Step 2: Compute the change in the kinetic energy. It is clear from the argument presented in the previous example that the change is kinetic energy is equal to the kinetic energy at the point B and is equal to .

Step 3: Use work-energy theorem. If this calculation is performed, it will be found that the speed of the roller coaster is .

Note that the result looks very similar to the previous example. This is not an accident! There is a deeper reason why these two results look similar. This point shall be revisited later.

This example provides a good illustration of advantages in the use of the work-energy method. This solution of this problem using Newton's Law would have required much more mathematical effort.

A few more examples of the application of the work-energy method will be considered below.

Example 3: A car of mass m is travelling with velocity v when the driver jams the brakes. If the coefficient of friction is , compute the distance s traveled by the car before it stops?

Step 1: Compute the work done. What is the force that does work? It is friction. To compute the work done by friction, we go through the usual procedure of drawing the free body diagram for the car etc. In the end one finds that the friction force is .

Let us now assume that the car stops after moving a distance s. The total work done will therefore be equal to

Step 2: Compute the change in kinetic energy. It can be shown trivially that the change in the kinetic energy is .

Step 3: Apply the work energy theorem. Using the work energy theorem, one find that .

In the examples so far, the forces that acted on the bodies were constants. The following examples will illustrate a non-constant force.

Example 4: A spring of stiffness k and undistorted length d is attached to a block of mass m. It is then stretched such that the spring has length 2d. Compute the velocity of the block when the spring regains its unstretched length d.

Step 1: Compute the work done. In this case the only force of relevance acting on the block is the spring force which may be taken to be . Performing the integral for work gives

Step 2: Compute change in the kinetic energy. Again this is equal to .

Step 3: Use work energy theorem to compute the velocity. It is equal to .

These four examples provide illustrations of the use of the work-energy theorem. In general, it involves three steps:

• Calculate the work done. To do this, the starting and end points of interest along the path of motion have to be identified. The forces that act on the particle have to be identified. Finally, the integral has to be evaluated.
• Compute the change in kinetic energy.
• Use work-energy theorem and compute the unknown quantity. It may be a distance or an unknown velocity.

In many practical applications it is more interesting to consider the rate at which work is done. For example, in the design of a motor a very important parameter is the power of the motor. Expressions such as "40 Horse Power Engine" are common, and these indicate the rate at which these machines are able to perform work. From the definition of infinitesimal work, the definition of power can be immediately derived as

.

The SI unit of power is Joules/seconds. This has a special name called "Watt" after James Watt.