第8章位置エネルギーとエネルギー保存

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?

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.

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

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).

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?

Step 1: Here only one body is being investigated.

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

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.