Systems with a Single Particle or Object

We first consider a system with a single particle or object. Returning to our development of Equation 8.2, recall that we first separated all the forces acting on a particle into conservative and non-conservative types, and wrote the work done by each type of force as a separate term in the work-energy theorem. We then replaced the work done by the conservative forces by the change in the potential energy of the particle, combining it with the change in the particle’s kinetic energy to get Equation 8.2. Now, we write this equation without the middle step and define the sum of the kinetic and potential energies, $K+U=E;$ to be the mechanical energy of the particle.

This statement expresses the concept of energy conservation for a classical particle as long as there is no non-conservative work. Recall that a classical particle is just a point mass, is nonrelativistic, and obeys Newton’s laws of motion. In Relativity, we will see that conservation of energy still applies to a non-classical particle, but for that to happen, we have to make a slight adjustment to the definition of energy.

It is sometimes convenient to separate the case where the work done by non-conservative forces is zero, either because no such forces are assumed present, or, like the normal force, they do zero work when the motion is parallel to the surface. Then

In this case, the conservation of mechanical energy can be expressed as follows: The mechanical energy of a particle does not change if all the non-conservative forces that may act on it do no work. Understanding the concept of energy conservation is the important thing, not the particular equation you use to express it.

Example 8.7

Simple Pendulum

A particle of mass m is hung from the ceiling by a massless string of length 1.0 m, as shown in Figure 8.7. The particle is released from rest, when the angle between the string and the downward vertical direction is $30\text{°.}$ What is its speed when it reaches the lowest point of its arc?

Figure 8.7 A particle hung from a string constitutes a simple pendulum. It is shown when released from rest, along with some distances used in analyzing the motion.

Strategy

Using our problem-solving strategy, the first step is to define that we are interested in the particle-Earth system. Second, only the gravitational force is acting on the particle, which is conservative (step 3). We neglect air resistance in the problem, and no work is done by the string tension, which is perpendicular to the arc of the motion. Therefore, the mechanical energy of the system is conserved, as represented by Equation 8.13, $0=\text{Δ}\left(K+U\right)$. Because the particle starts from rest, the increase in the kinetic energy is just the kinetic energy at the lowest point. This increase in kinetic energy equals the decrease in the gravitational potential energy, which we can calculate from the geometry. In step 4, we choose a reference point for zero gravitational potential energy to be at the lowest vertical point the particle achieves, which is mid-swing. Lastly, in step 5, we set the sum of energies at the highest point (initial) of the swing to the lowest point (final) of the swing to ultimately solve for the final speed.

Solution

We are neglecting non-conservative forces, so we write the energy conservation formula relating the particle at the highest point (initial) and the lowest point in the swing (final) as

$$K_{\text{i}}+U_{\text{i}}=K_{\text{f}}+U_{\text{f}}.$$

Since the particle is released from rest, the initial kinetic energy is zero. At the lowest point, we define the gravitational potential energy to be zero. Therefore our conservation of energy formula reduces to

$$\begin{array}{ccc}\hfill 0+mgh& =\hfill & \frac{1}{2}m{v}^2+0\hfill \\ \hfill v& =\hfill & \sqrt{2gh}.\hfill \end{array}$$

The vertical height of the particle is not given directly in the problem. This can be solved for by using trigonometry and two givens: the length of the pendulum and the angle through which the particle is vertically pulled up. Looking at the diagram, the vertical dashed line is the length of the pendulum string. The vertical height is labeled h. The other partial length of the vertical string can be calculated with trigonometry. That piece is solved for by

$$\text{cos}\theta =x\text{/}L,x=L\text{cos}\theta .$$

Therefore, by looking at the two parts of the string, we can solve for the height h,

$$\begin{array}{ccc}\hfill x+h& =\hfill & L\hfill \\ \hfill L\text{cos}\theta +h& =\hfill & L\hfill \\ \hfill h& =\hfill & L-L\text{cos}\theta =L(1-\text{cos}\theta ).\hfill \end{array}$$

We substitute this height into the previous expression solved for speed to calculate our result:

$$v=\sqrt{2gL\left(1-\text{cos}\theta \right)}=\sqrt{2\left(9.8{\text{m/s}}^2\right)\left(1\text{m}\right)\left(1-\text{cos}30\text{°}\right)}=1.62\text{m/s}.$$

Significance

We found the speed directly from the conservation of mechanical energy, without having to solve the differential equation for the motion of a pendulum (see Oscillations). We can approach this problem in terms of bar graphs of total energy. Initially, the particle has all potential energy, being at the highest point, and no kinetic energy. When the particle crosses the lowest point at the bottom of the swing, the energy moves from the potential energy column to the kinetic energy column. Therefore, we can imagine a progression of this transfer as the particle moves between its highest point, lowest point of the swing, and back to the highest point (Figure 8.8). As the particle travels from the lowest point in the swing to the highest point on the far right hand side of the diagram, the energy bars go in reverse order from (c) to (b) to (a).

Figure 8.8 Bar graphs representing the total energy (E), potential energy (U), and kinetic energy (K) of the particle in different positions. (a) The total energy of the system equals the potential energy and the kinetic energy is zero, which is found at the highest point the particle reaches. (b) The particle is midway between the highest and lowest point, so the kinetic energy plus potential energy bar graphs equal the total energy. (c) The particle is at the lowest point of the swing, so the kinetic energy bar graph is the highest and equal to the total energy of the system.

Example 8.8

Air Resistance on a Falling Object

A helicopter is hovering at an altitude of $1\text{km}$ when a panel from its underside breaks loose and plummets to the ground (Figure 8.9). The mass of the panel is $15\text{kg},$ and it hits the ground with a speed of $45\text{m}\text{/}\text{s}$. How much mechanical energy was dissipated by air resistance during the panel’s descent?

Figure 8.9 A helicopter loses a panel that falls until it reaches terminal velocity of 45 m/s. How much did air resistance contribute to the dissipation of energy in this problem?

Strategy

Step 1: Here only one body is being investigated.

Step 2: Gravitational force is acting on the panel, as well as air resistance, which is stated in the problem.

Step 3: Gravitational force is conservative; however, the non-conservative force of air resistance does negative work on the falling panel, so we can use the conservation of mechanical energy, in the form expressed by Equation 8.12, to find the energy dissipated. This energy is the magnitude of the work:

$$\text{Δ}{E}_{\text{diss}}=\left|W_{\text{nc,if}}\right|=\left|\text{Δ}{\left(K+U\right)}_{\text{if}}\right|.$$

Step 4: The initial kinetic energy, at $y_{\text{i}}=1\text{km},$ is zero. We set the gravitational potential energy to zero at ground level out of convenience.

Step 5: The non-conservative work is set equal to the energies to solve for the work dissipated by air resistance.

Solution

The mechanical energy dissipated by air resistance is the algebraic sum of the gain in the kinetic energy and loss in potential energy. Therefore the calculation of this energy is

$$\begin{array}{cc}\hfill \text{Δ}{E}_{\text{diss}}& =\left|K_{\text{f}}-K_{\text{i}}+U_{\text{f}}-U_{\text{i}}\right|\hfill \\ & =\left|\frac{1}{2}\left(15\text{kg}\right){\left(45\text{m/s}\right)}^2-0+0-\left(15\text{kg}\right)\left(9.8{\text{m/s}}^2\right)\left(1000\text{m}\right)\right|=130\text{kJ}.\hfill \end{array}$$

Significance

Most of the initial mechanical energy of the panel $\left(U_{\text{i}}\right)$, 147 kJ, was lost to air resistance. Notice that we were able to calculate the energy dissipated without knowing what the force of air resistance was, only that it was dissipative.

In these examples, we were able to use conservation of energy to calculate the speed of a particle just at particular points in its motion. But the method of analyzing particle motion, starting from energy conservation, is more powerful than that. More advanced treatments of the theory of mechanics allow you to calculate the full time dependence of a particle’s motion, for a given potential energy. In fact, it is often the case that a better model for particle motion is provided by the form of its kinetic and potential energies, rather than an equation for force acting on it. (This is especially true for the quantum mechanical description of particles like electrons or atoms.)

We can illustrate some of the simplest features of this energy-based approach by considering a particle in one-dimensional motion, with potential energy U(x) and no non-conservative interactions present. Equation 8.12 and the definition of velocity require

Separate the variables x and t and integrate, from an initial time $t=0$ to an arbitrary time, to get

If you can do the integral in Equation 8.14, then you can solve for x as a function of t.

We will look at another more physically appropriate example of the use of Equation 8.13 after we have explored some further implications that can be drawn from the functional form of a particle’s potential energy.

Systems with Several Particles or Objects

Systems generally consist of more than one particle or object. However, the conservation of mechanical energy, in one of the forms in Equation 8.12 or Equation 8.13, is a fundamental law of physics and applies to any system. You just have to include the kinetic and potential energies of all the particles, and the work done by all the non-conservative forces acting on them. Until you learn more about the dynamics of systems composed of many particles, in Linear Momentum and Collisions, Fixed-Axis Rotation, and Angular Momentum, it is better to postpone discussing the application of energy conservation to then.