Model Validation and Comparison

Mechanics is the study of the motion of objects and the effect of forces acting on those objects. It is the foundation of several branches of physics and engineering. Newtonian,or classical, mechanicsdeals with the motion of ordinaryobjects—that is, objects that are large compared to an atom and slow moving compared with the speed of light. A model for Newtonian mechanics can be based on Newton’s laws of motion:^†

1. When a body is subject to no resultant external force, it moves with a constant velocity.

2. When a body is subject to one or more external forces, the time rate of change of the body’s momentum is equal to the vector sum of the external forces acting on it.

3. When one body interacts with a second body, the force of the first body on the second is equal in magnitude, but opposite in direction, to the force of the second body on the first.

Experimental results for more than two centuries verify that these laws are extremely use-ful for studying the motion of ordinary objects in an inertial reference frame—that is, a refer-ence frame in which an undisturbed body moves with a constant velocity. It is Newton’s second law, which applies only to inertial reference frames, that enables us to formulate the equations of motion for a moving body. We can express Newton’s second law by

dp

dt FAt, x, y_B ,

†For a discussion of Newton’s laws of motion, see Sears and Zemansky’s University Physics,12th ed., by H. D. Young, R. A. Freedman, J. R. Sandin, and A. L. Ford (Pearson Addison Wesley, San Francisco, 2008).

where is the resultant force on the body at time t, locationx, and velocity y, and is the momentum of the body at time t.The momentum is the product of the mass of the body and its velocity—that is,

—so we can express Newton’s second law as (1)

where is the acceleration of the body at time t.

Typically one substitutes for the velocity in (1) and obtains a second-order differential equation in the dependent variable x. However, in the present section, we will focus on situations where the force Fdoes not depend on x.This enables us to regard (1) as a first-order equation

(2) in .

To apply Newton’s laws of motion to a problem in mechanics, the following general procedure may be useful.

yAtB mdY

dt FAt,YB

ydx

/

^dt

ady

/

^dt

mdy

dt maFAt,x,yB , pAtBmy_AtB

pAtB FAt,x,y_B

Section 3.4 Newtonian Mechanics 109

Procedure for Newtonian Models

(a) Determineallrelevant forces acting on the object being studied. It is helpful to draw a simple diagram of the object that depicts these forces.

(b) Choose an appropriate axis or coordinate system in which to represent the motion of the object and the forces acting on it. Keep in mind that this coordinate system must be an inertial reference frame.

(c) Apply Newton’s second law as expressed in equation (2) to determine the equations of motion for the object.

In this section we express Newton’s second law in either of two systems of units: the U.S.

Customary System or the meter-kilogram-second (MKS) system. The various units in these systems are summarized in Table 3.2, along with approximate values for the gravitational acceleration (on the surface of Earth).

TABLE 3.2 Mechanical Units in the U.S. Customary and MKS Systems U.S. Customary

Unit System MKS System

Distance foot (ft) meter (m)

Mass slug kilogram (kg)

Time second (sec) second (sec)

Force pound (lb) newton (N)

g(Earth) 32 ft/sec² 9.81 m/sec²

An object of mass mis given an initial downward velocity y₀and allowed to fall under the influence of gravity. Assuming the gravitational force is constant and the force due to air resistance is proportional to the velocity of the object, determine the equation of motion for this object.

Two forces are acting on the object: a constant force due to the downward pull of gravity and a force due to air resistance that is proportional to the velocity of the object and acts in opposition to the motion of the object. Hence, the motion of the object will take place along a vertical axis. On this axis we choose the origin to be the point where the object was initially dropped and let denote the distance the object has fallen in time t(see Figure 3.7).

The forces acting on the object along this axis can be expressed as follows. The force due to gravity is

where gis the acceleration due to gravity at Earth’s surface (see Table 3.2). The force due to air resistance is

where is the proportionality constant^†and the negative sign is present because air resis-tance acts in opposition to the motion of the object. Hence, the net force acting on the object (see Figure 3.7) is

(3)

We now apply Newton’s second law by substituting (3) into (2) to obtain

Since the initial velocity of the object is y₀, a model for the velocity of the falling body is expressed by the initial value problem

(4)

where gand bare positive constants.

The model (4) is the same as the one we obtained in Section 2.1. Using separation of variables or the method of Section 2.3 for linear equations, we get

(5) Y_AtBmg

b a^Y0mg

bb^ebt

/

^m _.

mdy

dt mgby , y_A0By₀ , mdy

dt mgby .

FF₁F₂mgbyAtB . bA0B

F₂ by_AtB , F₁mg ,

xAtB

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

x(t)

t = 0

t

F ₂ = −b (t) F ₁ = mg

Figure 3.7 Forces on a falling object

Solution

†The units of bare lb-sec/ft in the U.S. system, and N-sec/m in the MKS system.

Example 1

Since we have taken x0 when t 0, we can determine the equation of motion of the object by integrating with respect to t.Thus, from (5) we obtain

and setting x0 when t0, we find

Hence, the equation of motion is

(6) _◆

In Figure 3.8, we have sketched the graphs of the velocity and the position as functions of t.

Observe that the velocity approaches the horizontal asymptote as and that the position asymptotically approaches the line

as The value of the horizontal asymptote for is called the limiting, or terminal, velocityof the object, and, in fact, is a solution of (4). Since the two forces are in balance, this is called an “equilibrium” solution.

Notice that the terminal velocity depends on the mass but not the initial velocity of the object; the velocity of every free-falling body approaches the limiting value . Heavier objects do,in the presence of friction, ultimately fall faster than lighter ones, but for a given object the terminal velocity is the same whether it initially is tossed upward or downward, or simply dropped from rest.

Now that we have obtained the equation of motion for a falling object with air resistance proportional to y, we can answer a variety of questions.

mg

/

^b

ymg

/

^bconstant y_AtB mg

/

^b

tSq.

x mg

b tm²g b² m

bY₀

xAtB ^yAtB ymg

/

^b tSq

xAtBmg b tm

b a^Y0mg

bbA1e^bt

/

^m_{B .}

cm

b a^y0 mg bb ^.

0 m

b a^y0 mg bb c ,

xAtB ^{yAtBdt mgb t mbay0 mgb bebt}

^/

^{mc ,}

ydx

/

^dt

Section 3.4 Newtonian Mechanics 111

x

Position Velocity

––– mg b

0 mg

––– b x = t −m ²

––– b ² + m ––– b ⁰ 0

0

g (t)

x(t)

t t

Figure 3.8 Graphs of the position and velocity of a falling object when y₀ 6 mg

/

^b

An object of mass 3 kg is released from rest 500 m above the ground and allowed to fall under the influence of gravity. Assume the gravitational force is constant, with g9.81 m/sec², and the force due to air resistance is proportional to the velocity of the object^†with proportionality constant b3 N-sec/m. Determine when the object will strike the ground.

We can use the model discussed in Example 1 with y₀0,m3,b3, and g9.81. From (6), the equation of motion in this case is

Because the object is released 500 m above the ground, we can determine when the object strikes the ground by setting and solving for t. Thus, we put

or

where we have rounded the computations to two decimal places. Unfortunately, this equation cannot be solved explicitly for t. We might try to approximate tusing Newton’s approximation method (see Appendix B), but in this case, it is not necessary. Since will be very small for t near , we simply ignore the term and obtain as our approximation

. ◆

A parachutist whose mass is 75 kg drops from a helicopter hovering 4000 m above the ground and falls toward the earth under the influence of gravity. Assume the gravitational force is constant. Assume also that the force due to air resistance is proportional to the velocity of the parachutist, with the proportionality constant b₁15 N-sec/m when the chute is closed and with constant b₂ 105 N-sec/m when the chute is open. If the chute does not open until 1 min after the parachutist leaves the helicopter, after how many seconds will she hit the ground?

We are interested only in when the parachutist will hit the ground, not where. Thus, we con-sider only the vertical component of her descent. For this, we need to use two equations: one to describe the motion before the chute opens and the other to apply after it opens. Before the chute opens, the model is the same as in Example 1 with y₀ 0, m 75 kg, b b₁ 15 N-sec/m, and g9.81 m/sec². If we let be the distance the parachutist has fallen in tsec and let , then substituting into equations (5) and (6), we have

^A49.05B A1e^0.2tB , y_1AtB ^A75B A9.81B

15

A

^1eA15

/

^75Bt

B

y₁dx₁

/

^dt

x₁AtB t51.97 sec

e^t 51.97 Ae^51.9710²²B

e^t te^t 509.81

9.81 51.97 , 500 A9.81Bt9.81 A9.81Be^t

xAtB500 xAtB ^A3B A9.81B

3 t ^A3B²A9.81B

A3B² A1e^3t

/

³_{B A9.81Bt A9.81B A1e}^t_{B .}

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

†The effects of more sophisticated air resistance models (such as a quadratic drag law) are analyzed numerically in Exercises 3.6, Problem 20.

Example 2

Solution

Example 3

Solution

and

Hence, after 1 min, when t60, the parachutist is falling at the rate

and has fallen

(In these and other computations for this problem, we round our answers to two decimal places.) Now when the chute opens, the parachutist is 4000 2697.75 or 1302.25 m above the ground and traveling at a velocity of 49.05 m/sec. To determine the equation of motion after the chute opens, let denote the position of the parachutist Tsec after the chute opens (so that Tt 60), taking at (see Figure 3.9). Further assume that the initial velocity of the parachutist after the chute opens is the same as the final velocity before it opens—that is, m/sec. Because the forces acting on the parachutist are the same as those acting on the object in Example 1, we can again use equations (5) and (6). With y₀49.05,m75,bb₂105, and g9.81, we find from (6) that

To determine when the parachutist will hit the ground, we set 1302.25, the height the parachutist was above the ground when her parachute opened. This gives

(7)

Again we cannot solve (7) explicitly for T. However, observe that is very small for T near 181.49, so we ignore the exponential term and obtain T181.49. Hence, the parachutist will strike the ground 181.49 sec after the parachute opens, or 241.49 sec after dropping from the helicopter. _◆

e^1.4T T4.28e^1.4T181.490 .

7.01T30.0330.03e^1.4T1302.25

x₂ATB 7.01T30.03A1e^1.4TB .

x₂ATB ^A75B A9.81B

105 T 75

105c^{49.05 A}75B A9.81B

105 d

A

¹e^A105

/

^75BT

B

x2¿A0Bx¿1A60B49.05 x₁A60B x₂A0B0

x₂ATB

x₁A60B A49.05B A60B A245.25B

A

^1e0.2A60B

B

2697.75 m . y_1A60B A49.05B

A

^1e0.2A60B

B

49.05 m/sec ,

49.05t245.25A1e^0.2tB . x₁AtB ^A75B A9.81B

15 t ^A75B²A9.81B

A15B²

A

^1eA15

/

^75Bt

B

Section 3.4 Newtonian Mechanics 113

x ₁ (t) t = 0

T = 0 1302.25

2697.75

Chute opens x ₂ (T)

x ₂ (0) = 0 atx ₁ (60)

Figure 3.9 The fall of the parachutist

In the computation for Tin equation (7), we found that the exponential was negligi-ble. Consequently, ignoring the corresponding exponential term in equation (5), we see that the parachutist’s velocity at impact is

which is the limiting velocity for her fall with the chute open.

In some situations the resistive drag force on an object is proportional to a power of other than 1. Then when the velocity is positive, Newton’s second law for a falling object gen-eralizes to

(8)

where mand ghave their usual interpretation and b 0 and the exponent rare experimental constants. [More generally, the drag force would be written as signy.] Express the solution to (8) for the case r2.

The (positive) equilibrium solution, with the forces in balance, is Other-wise, we write (8) as a separable equation that we can solve using partial fractions or the integral tables on the inside front cover:

and after some algebra,

Here is determined by the initial conditions. Again we see

that yapproaches the terminal velocity y₀as t→ ∞. ◆ c₃c₂ sign3^Ay0y_B

/

^Ay0y_B4

yy₀c₃e^2y0bt/^m c₃e^2y0bt/^m .

`y₀y

y₀y` c₂e^2y0bt/^m ,

^y2₀^{dyy2 2y1}0

ln`y<sub>0</sub>y y<sub>0</sub>y` b

mtc₁ , y¿bAy²₀y²_B

/

^m,

yy₀ 2mg

/

^b.

B b0y0^r A B

A mdy

dt mgby^r ,

0y0

mg

b₂ ^A75B A9.81B

105 7.01 m/sec ,

e^1.4T 114 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

Unless otherwise stated, in the following problems we assume that the gravitational force is constant with g 9.81 m /sec²in the MKS system and g 32ft/sec²in the U.S. Customary System.

1. An object of mass 5 kg is released from rest 1000 m above the ground and allowed to fall under the influ-ence of gravity. Assuming the force due to air resis-tance is proportional to the velocity of the object with proportionality constant b50 N-sec/m, deter-mine the equation of motion of the object. When will the object strike the ground?

2. A 400-lb object is released from rest 500 ft above the ground and allowed to fall under the influence of gravity. Assuming that the force in pounds due to air resistance is 10y, where y is the velocity of the object in ft/sec, determine the equation of motion of the object. When will the object hit the ground?

3. If the object in Problem 1 has a mass of 500 kg instead of 5 kg, when will it strike the ground? [Hint:

Here the exponential term is too large to ignore. Use Newton’s method to approximate the time twhen the object strikes the ground (see Appendix B).]

Example 4

Solution

In document A Systems level characterization and tradespace evaluation of a simulated airborne fourier transform infrared spectrometer for gas detection (Page 123-128)