Autodefinición del rol que cumple el docente en el

than 5 m/sec and b₂60 N-sec/m when the velocity is above 5 m/sec. For sailboat B the proportionality constants are b₃ 100 N-sec/m before planing when the velocity is less than 6 m/sec and b₄50 N-sec/m when the velocity is above 6 m/sec. If the first leg of the race is 500 m long, which sailboat will be leading at the end of the first leg?

24. Rocket Flight. A model rocket having initial mass m₀ kg is launched vertically from the ground. The rocket expels gas at a constant rate of kg/sec and at a constant velocity of m/sec relative to the rocket.

Assume that the magnitude of the gravitational force is proportional to the mass with proportionality con-stant g.Because the mass is not constant, Newton’s second law leads to the equation

where is the velocity of the rocket,xis its height above the ground, and is the mass of the rocket at tsec after launch. If the initial velocity is zero, solve the above equation to determine the velocity of the rocket and its height above ground for 25. Escape Velocity. According to Newton’s law of gravitation, the attractive force between two objects varies inversely as the square of the dis-tances between them. That is,

where M₁and M₂are the masses of the objects,ris the distance between them (center to center),F_gis the attractive force, and Gis the constant of propor-tionality. Consider a projectile of constant mass m being fired vertically from Earth (see Figure 3.12).

Let trepresent time and ythe velocity of the pro-jectile.

F_gGM₁M₂

/

0t 6 m₀

/

m₀at ydx

/

Am₀atBdy

dt ab gAm₀atB , b

a

Section 3.5 Electrical Circuits 117

r R

M

m Earth

Figure 3.12 Projectile escaping from Earth

(a) Show that the motion of the projectile, under Earth’s gravitational force, is governed by the equation

where r is the distance between the projectile and the center of Earth,Ris the radius of Earth, Mis the mass of Earth, and

(b) Use the fact that to obtain

(c) If the projectile leaves Earth’s surface with velocity y₀, show that

(d) Use the result of part (c) to show that the veloc-ity of the projectile remains positive if and only

if The velocity is

called the escape velocityof Earth.

(e) If g9.81 m/sec²and R6370 km for Earth, what is Earth’s escape velocity?

(f ) If the acceleration due to gravity for the Moon is and the radius of the Moon is R_m 1738 km, what is the escape velocity of the Moon?

g_mg

/

y_e 22gR y²₀2gR 7 0.

y²2gR²

r y²₀2gR . ydy

dr

gR² r² .

dr

/

gGM

/

dy dt

gR² r² ,

118 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

L R

Voltage source

Inductance

C

E E

R

Voltage source

(a) (b)

Capacitance

Resistance Resistance

Figure 3.13 (a) RL circuit and (b) RC circuit

†Historical Footnote:Gustav Robert Kirchhoff (1824–1887) was a German physicist noted for his research in spectrum analysis, optics, and electricity.

The physical principles governing electrical circuits were formulated by G. R. Kirchhoff^† in 1859. They are the following:

1. Kirchhoff’s current law The algebraic sum of the currents flowing into any junction point must be zero.

2. Kirchhoff’s voltage law The algebraic sum of the instantaneous changes in potential (voltage drops) around any closed loop must be zero.

Kirchhoff’s current law implies that the same current passes through all elements in each circuit of Figure 3.13.

To apply Kirchhoff’s voltage law, we need to know the voltage drop across each element of the circuit. These voltage formulas are stated below (you can consult an introductory physics text for further details).

(a) According to Ohm’s law, the voltage drop ERacross a resistor is proportional to the current Ipassing through the resistor:

The proportionality constant Ris called the resistance.

(b) It can be shown using Faraday’s law and Lenz’s law that the voltage drop ELacross an inductor is proportional to the instantaneous rate of change of the current I:

The proportionality constant Lis called the inductance.

(c) The voltage drop ECacross a capacitor is proportional to the electrical charge qon the capacitor:

The constant Cis called the capacitance.

The common units and symbols used for electrical circuits are listed in Table 3.3.

E_C 1 Cq . ELLdI

dt . ERRI .

A voltage source is assumed to addvoltage or potential energy to the circuit. If we let E t denote the voltage supplied to the circuit at time t, then applying Kirchhoff’s voltage law to the RLcircuit in Figure 3.13(a) gives

(1) ELERE t .

Substituting into (1) the expressions for ELand ERgives (2)

Note that this equation is linear (compare Section 2.3), and upon writing it in standard form we obtain the integrating factor

which leads to the following general solution [see equation (8), Section 2.3, page 47]

(3)

For the RLcircuit, one is usually given the initial current I 0 as an initial condition.

An RLcircuit with a 1-Ωresistor and a 0.01-H inductor is driven by a voltage E t sin 100t V. If the initial inductor current is zero, determine the subsequent resistor and inductor voltages and the current.

From equation (3) and the integral tables, we find that the general solution to the linear equa-tion (2) is given by

sin 100tcos 100t

2 Ke^100t .

e^100tc^{100 e100tA}100 sin 100t100 cos 100tB

10,00010,000 Kd

IAtBe^100ta^{e100tsin 100t0.01 dtKb}

B A B

A IAtBe^Rt/^Lc^{eRt/LELAtB dtKd .}

m(t)e^AR/^LBdte^Rt/^L , LdI

dtRIEAtB . B A

B A Section 3.5 Electrical Circuits 119

TABLE 3.3 Common Units and Symbols Used With Electrical Circuits

Letter Symbol

Quantity Representation Units Representation

Voltage source E volt (V) Generator

Battery

Resistance R ohm ()

Inductance L henry (H)

Capacitance C farad (F)

Charge q coulomb (C)

Current I ampere (A)

Example 1

Solution

For I 0 0, we obtain so and the current is I t 0.5 sin 100tcos 100te^100t .

The inductor and resistor voltages are then given by

◆

Now we turn to the RCcircuit in Figure 3.13(b). Applying Kirchhoff’s voltage law yields RIq CE t .

The capacitor current, however, is the rate of change of its charge:Idq dt. So (4)

is the governing differential equation for the RCcircuit. The initial condition for a capacitor is its charge qat t0.

Suppose a capacitor of Cfarads holds an initial charge of Qcoulombs. To alter the charge, a constant voltage source of V volts is applied through a resistance of R ohms. Describe the capacitor charge for t 0.

Since E t Vis constant in equation (4), the latter is both separable and linear, and its general solution is easily derived:

The solution meeting the prescribed initial condition is

The capacitor charge changes exponentially from Qto CVas time increases. _◆

If we take V0 in Example 2, we see that the time constant—that is, the time required for the capacitor charge to drop to 1 etimes its initial value—is RC. Thus, a capacitor is a leaky energy-storage device; even the very high resistivity of the surrounding air can dissipate its charge, particularly on a humid day. Capacitors are used in cellular phones to store electrical energy from the battery while the phone is in a (more-or-less) idle receiving mode and then assist the battery in delivering energy during the transmitting mode.

The time constant for inductor current in the RLcircuit can be gleaned from Example 1 to be L R. An application of the RLcircuit is the spark plug of a combustion engine. If a voltage source establishes a nonzero current in an inductor and the source is suddenly disconnected, the rapid change of current produces a high dI dt and, in accordance with the formula , the inductor generates a voltage surge sufficient to cause a spark across the terminals—thus igniting the gasoline.

If an inductor and a capacitor bothappear in a circuit, the governing differential equation will be second order. We’ll return to RLCcircuits in Section 5.7.

ELLdI/dt

/

/

/

qAtBCV ^AQCVBe^t/^RC . qAtBCVKe^t/^RC .

B A

R dq dt q

C E

/

B

/