Bedford Fowler Dynamics 5th Edition - Examples Ch12, Ch13 PDF

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CHAPTER

12 Introduction How do engineers design and construct the devices we use, from simple objects such as chairs and pencil sharpeners to complicated ones such as dams, cars, airplanes, and spacecraft? They must have a deep understanding of the physics underlying the design of such devices and must be able to use mathematical models to predict their behavior. Students of engineering begin to learn how to analyze and predict the behaviors of physical systems by studying mechanics.

! The motions of the bobsled and its crew—their positions, velocities, and accelerations—can be analyzed using the equations of dynamics. Engineers use dynamics to predict the motions of objects.

10

Chapter 12 Introduction

Active Example 12.1

Converting Units (! Related Problem 12.11) A man is riding a bicycle at a speed of 6 meters per second (m/s). How fast is he going in kilometers per hour (km/h)? Strategy One kilometer is 1000 meters and one hour is 60 minutes ! 60 seconds " 3600 seconds. We can use these unit conversions to determine his speed in km/h. Solution Convert meters to kilometers. Convert seconds to hours. 6 m/s " 6 m/s

1 km

! 1000 m " !

3600 s 1h

"

" 21.6 km/ h. Practice Problem A man is riding a bicycle at a speed of 10 feet per second (ft/s). How fast is he going in miles per hour (mi/h)? Answer: 6.82 mi/h.

Example 12.2

Converting Units of Pressure (! Related Problem 12.16) The pressure exerted at a point of the hull of the deep submersible vehicle is 3.00 * 106 Pa (pascals). A pascal is 1 newton per square meter. Determine the pressure in pounds per square foot. Strategy From Table 12.2, 1 pound = 4.448 newtons and1 foot = 0.3048 meters. With these unit conversions we can calculate the pressure in pounds per square foot. Solution The pressure (to three significant digits) is 3.00 * 106 N/m2 = 13.00 * 106 N/m22 a

0.3048 m 2 1 lb b ba 1 ft 4.448 N

= 62,700 lb/ft2.

Deep Submersible Vehicle.

Critical Thinking How could we have obtained this result in a more direct way? Notice from the table of unit conversions in the inside front cover that1 Pa = 0.0209 lb/ft2. Therefore, 3.00 * 106 N/m2 = 13.00 * 106 N/m22 a = 62,700 lb/ft2.

0.0209 lb/ft2 b 1 N/m2

12.1 Engineering and Mechanics

Example 12.3

Determining Units from an Equation (! Related Problem 12.20)

Suppose that in Einstein’s equation E = mc2, the mass m is in kilograms and the velocity of light c is in meters per second. (a) What are the SI units of E? (b) If the value of E in SI units is 20, what is its value in U.S. Customary base units?

Strategy (a) Since we know the units of the terms m and c, we can deduce the units of E from the given equation. (b) We can use the unit conversions for mass and length from Table 12.2 to convert E from SI units to U.S. Customary units.

Solution (a) From the equation for E, E = 1m kg21c m/s22, the SI units of E are kg-m2/s2. (b) From Table 12.2, 1 slug = 14.59 kg and 1 ft = 0.3048 m. Therefore, 1 kg-m2/s2 = 11 kg-m2/s22 a

2 1 slug 1 ft b ba 14.59 kg 0.3048 m

= 0.738 slug-ft2/s2. The value of E in U.S. Customary units is E = 120210.7382 = 14.8 slug-ft2/s2. Critical Thinking In part (a), how did we know that we could determine the units of E by determining the units of mc2? The dimensions, or units, of each term in an equation must be the same. For example, in the equationa + b = c, the dimensions of each of the terms a, b, and c must be the same. The equation is said to be dimensionally homogeneous. This requirement is expressed by the colloquial phrase “Don’t compare apples and oranges.”

11

16

Chapter 12 Introduction

RESULTS The gravitational force between two particles of mass m1 and m2 that are separated by a distance r is Gm1m2 , (12.1) F# r2 where G is the universal gravitational constant. The value of G in SI units is 6.67 ! 10"11 N-m2/ kg2.

Active Example 12.4

Newtonian gravitation.

When the earth is modeled as a homogeneous sphere of radius RE, the acceleration due to gravity at a distance r from the center is R2 (12.4) a # g 2E , r where g is the acceleration due to gravity at sea level.

Acceleration due to gravity of the earth.

W # mg , (12.6) where m is the mass of the object and g is the acceleration due to gravity at sea level.

Weight of an object at sea level.

Weight and Mass (! Related Problem 12.22) The C-clamp weighs 14 oz at sea level. [16 oz (ounces) # 1 lb.] The acceleration due to gravity at sea level is g # 32.2 ft/ s2. What is the mass of the C-clamp in slugs? Strategy We must first determine the weight of the C-clamp in pounds. Then we can use Eq. (12.6) to determine the mass in slugs. Solution

! 16 oz "# 0.875 lb.

Convert the weight from ounces to pounds.

W 0.875 lb # # 0.0272 slug. g 32.2 ft/s2

Use Eq. (12.6) to calculate the mass in slugs.

14 oz # 14 oz

m#

1 lb

Practice Problem The mass of the C-clamp is 0.397 kg. The acceleration due to gravity at sea level isg = 9.81 m/s2 . What is the weight of the C-clamp at sea level in newtons? Answer: 3.89 N.

12.2 Newtonian Gravitation

Example 12.5

Determining an Object’s Weight (! Related Problem 12.27)

When the Mars Exploration Rover was fully assembled, its mass was 180 kg. The acceleration due to gravity at the surface of Mars is3.68 m/s2 and the radius of Mars is 3390 km. (a) What was the rover’s weight when it was at sea level on Earth? (b) What is the rover’s weight on the surface of Mars? (c) The entry phase began when the spacecraft reached the Mars atmospheric entry interface point at 3522 km from the center of Mars. What was the rover’s weight at that point?

Mars Exploration Rover being assembled.

17

18

Chapter 12 Introduction

Strategy The rover’s weight at sea level on Earth is given by Eq. (12.6) withg = 9.81 m/s2. We can determine the weight on the surface of Mars by using Eq. (12.6) with the acceleration due to gravity equal to3.68 m/s2. To determine the rover’s weight as it began the entry phase, we can write an equation for Mars equivalent to Eq. (12.5).

Solution (a) The weight at sea level on Earth is W = mg = 1180 kg219.81 m/s22 = 1770 N 1397 lb2.

(b) Let g M = 3.68 m/s2 be the acceleration due to gravity at the surface of Mars. Then the weight of the rover on the surface of Mars is W = mg M

= 1180 kg213.68 m/s22

= 662 N 1149 lb2.

(c) Let RM = 3390 km be the radius of Mars. From Eq. (12.5), the rover’s weight when it is 3522 km above the center of Mars is W = mg M

R2M r2

= 1180 kg213.68 m/s22 = 614 N 1138 lb2.

13,390,000 m22

13,522,000 m22

Critical Thinking In part (c), how did we know that we could apply Eq. (12.5) to Mars? Equation (12.5) is applied to Earth based on modeling it as a homogeneous sphere. It can be applied to other celestial objects under the same assumption. The accuracy of the results depends on how aspherical and inhomogeneous the object is.

CHAPTER

13 Motion of a Point In this chapter we begin the study of motion. We are not yet concerned with the properties of objects or the causes of their motions—our objective is simply to describe and analyze the motion of a point in space. After defining the position, velocity, and acceleration of a point, we consider the simplest case, motion along a straight line. We then show how motion of a point along an arbitrary path, or trajectory, is expressed and analyzed using various coordinate systems.

! The lines show the paths followed by subatomic particles moving in a magnetic field. The particles with curved paths have both tangential and normal components of acceleration.

an

at

30

Chapter 13 Motion of a Point

Definite integrals can be used to determine the position. Here s0 is the velocity at time t0, and s is the velocity at time t. This result shows that the change in the position from time t0 to time t is equal to the area defined by the graph of the velocity from time t0 to time t.

L s0

s

dv !

L t0

t

v dt,

s ! s0 "

t v dt.

Lt0

v

Area ! s(t ) # s(t0) t0

t

t

When the Acceleration is Constant Suppose that the acceleration is a constant a ! a0. Equations (13.3)–(13.5) can be integrated to obtain these convenient results for the velocity v and position s at time t. Here v0 is the velocity at time t0, and s0 is the position at time t0.

Active Example 13.1

(13.11)

v ! v0 " a0(t # t0), 1 s ! s0 " v0(t # t0) " a0(t # t0)2, 2 2 2 " 2a (s # s ). v ! v0 0 0

(13.12) (13.13)

Acceleration that is a Function of Time (! Related Problem 13.12) The acceleration (in m/s2) of point P relative to point O is given as a function of time by a = 3t2, where t is in seconds. At t = 1 s, the position of P is s = 3 m, and at t = 2 s, the position of P is s = 7.5 m. What is the position of P at t = 3 s? O

P

s

s

Strategy Because the acceleration is given as a function of time, we can integrate it to obtain an equation for the velocity as a function of time. Then we can integrate the velocity to obtain an equation for the position as a function of time. The resulting equations will contain two unknown integration constants. We can evaluate them by using the given values of the position at t = 1 s and t = 2 s.

13.2 Straight-Line Motion

Solution Integrate the acceleration to determine the velocity as a function of time. A is an integration constant.

Integrate the velocity to determine the position as a function of time. B is an integration constant.

Use the known positions at t ! 1 s and at t ! 2 s to determine A and B, obtaining A ! 0.75 and B ! 2.

Determine the position at t ! 3 s.

dv ! 3t 2, dt v ! t 3 " A. a!

ds ! t 3 " A, dt 1 s ! t 4 " At " B. 4

v!

1 4 (1) " A(1) " B, 4 1 s t !2 s ! 7.5 ! (2)4 " A(2) " B. 4

s t !1 s ! 3 !

1 s ! t 4 " 0.75t " 2 : 4 1 s t !3 s ! (3)4 " 0.75(3) " 2 ! 24.5 m. 4

Practice Problem The acceleration (in ft/s2) of point P relative to point O is given as a function of time by a = 2t , where t is in seconds. Att = 3 s, the position and velocity of P are s = 30 ft and v = 14 ft/s. What are the position and velocity of P at t = 10 s?

O

P s

Answer: s = 389 ft, v = 105 ft/s.

s

31

32

Chapter 13 Motion of a Point

Example 13.2

Straight-Line Motion with Constant Acceleration (! Related Problem 13.1) Engineers testing a vehicle that will be dropped by parachute estimate that the vertical velocity of the vehicle when it reaches the ground will be 6m/s. If they drop the vehicle from the test rig shown, from what height h should they drop it to match the impact velocity of the parachute drop?

h

Strategy If the only significant force acting on an object near the earth’s surface is its weight, the acceleration of the object is approximately constant and equal to the acceleration due to gravity at sea level. Therefore, we can assume that the vehicle’s acceleration during its short fall is g = 9.81 m/s2. We can integrate Eqs. (13.3) and (13.4) to obtain the vehicle’s velocity and position as functions of time and then use them to determine the position of the vehicle when its velocity is 6 m/s. Solution Let t = 0 be the time at which the vehicle is dropped, and let s be the position of the bottom of the cushioning material beneath the vehicle relative to its position at t = 0 (Fig. a). The vehicle’s acceleration isa = 9.81 m/s2. From Eq. (13.4), dv = a = 9.81 m/s2. dt Integrating, we obtain v = 9.81t + A, where A is an integration constant. Because the vehicle is at rest when it is released, v = 0 at t = 0. Therefore, A = 0, and the vehicle’s velocity as a function of time is v = 9.81t m/s.

13.2 Straight-Line Motion

s

(a) The coordinate s measures the position of the bottom of the platform relative to its initial position.

We substitute this result into Eq. (13.3) to get ds = v = 9.81t dt and integrate, obtaining s = 4.91t2 + B. The position s = 0 when t = 0, so the integration constant B = 0, and the position as a function of time is s = 4.91t2. From our equation for the velocity as a function of time, the time necessary for the vehicle to reach 6 m/s as it falls is t =

6 m/s v = = 0.612 s. 9.81 m/s2 9.81 m/s2

Substituting this time into our equation for the position as a function of time yields the required height h: h = 4.91t2 = 4.9110.61222 = 1.83 m.

Critical Thinking Notice that we could have determined the height h from which the vehicle should be dropped in a simpler way by using Eq. (13.13), which relates the velocity to the position. v 2 = v 20 + 2a01s - s02:

16 m/s22 = 0 + 219.81 m/s221h - 02. Solving, we obtain h = 1.83 m. But it is essential to remember that Eqs. (13.11)–(13.13) apply only when the acceleration is constant, as it is in this example.

33

34

Chapter 13 Motion of a Point

Example 13.3

Graphical Solution of Straight-Line Motion (! Related Problem 13.26) The cheetah, Acinonyx jubatus, can run as fast as75 mi/h. If you assume that the animal’s acceleration is constant and that it reaches top speed in 4 s, what distance can the cheetah cover in 10 s?

Strategy The acceleration has a constant value for the first 4 s and is then zero. We can determine the distance traveled during each of these “phases” of the motion and sum them to obtain the total distance covered. We do so both analytically and graphically. Solution The top speed in terms of feet per second is 75 mi/h = 175 mi/h2 a

1h 5280 ft ba b = 110 ft/s. 1 mi 3600 s

First Method Let a0 be the acceleration during the first 4 s. We integrate Eq. (13.4) to get 0 L

v

dv =

L0

t

a0 dt,

v

t

0

0

c v d = a0 c t d , v - 0 = a01t - 02, obtaining the velocity as a function of time during the first 4 s: v = a0t ft/s. When t = 4 s, v = 110 ft/s; so a0 = 110/4 = 27.5 ft/s2. Therefore, the velocity during the first 4 s is v = 27.5t ft/s. Now we integrate Eq. (13.3), 0 L

s

ds =

L0

t

27.5t dt,

s

c s d = 27.5 c 0

t2 t d , 2 0

s - 0 = 27.5 a

t2 - 0b, 2

obtaining the position as a function of time during the first 4 s: s = 13.75t 2 ft.

13.2 Straight-Line Motion

At t = 4 s, the position is s = 13.751422 = 220 ft. From t = 4 s to t = 10 s, the velocity v = 110 ft/s. We write Eq. (13.3) as ds = v dt = 110 dt and integrate to determine the distance traveled during the second phase of the motion, L0

s

ds =

L4

10

110 dt,

s

10

0

4

c s d = 110c t d , s - 0 = 110110 - 42, obtaining s = 660 ft. The total distance the cheetah travels is 220 ft + 660 ft = 880 ft, or 293 yd, in 10 s. Second Method We draw a graph of the cheetah’s velocity as a function of time in Fig. a. The acceleration is constant during the first 4 s of motion, so the velocity is a linear function of time from v = 0 at t = 0 to v = 110 ft/s at t = 4 s. The velocity is constant during the last 6 s. The total distance covered is the sum of the areas during the two phases of motion: 1 2 14

s21110 ft/s2 + 16 s21110 ft/s2 = 220 ft + 660 ft = 880 ft.

v (ft/s)

Area equals the distance traveled from t ! 0 to t ! 10 s. 110

0 0

10

4

t (s)

(a) The cheetah’s velocity as a function of time.

Critical Thinking Notice that in the first method we used definite, rather than indefinite, integrals to determine the cheetah’s velocity and position as functions of time. You should rework the example using indefinite integrals and compare your results with ours. Whether to use definite or indefinite integrals is primarily a matter of taste, but you need to be familiar with both procedures.

35

Chapter 13 Motion of a Point

Active Example 13.4

Acceleration that is a Function of Velocity (! Related Problem 13.40) After deploying its drag parachute, the airplane’s acceleration (inm/s2 ) is a = -0.004v2, where v is the velocity in m/s. Determine the time required for the plane’s velocity to decrease from 80 m/s to 10 m/s. Strategy The airplane’s acceleration is known as a function of its velocity. Writing the acceleration asa = dv> dt, we can separate variables and integrate to determine the velocity as a function of time. Solution dv ! "0.004v2 : dt dv ! "0.004 dt. v2

Separate variables.

v

t

dv ! "0.004 dt, L80 v2 L0

Integrate, defining t ! 0 to be the time at which the velocity is 80 m/s. Here v is the velocity at time t.

1 v

! " "

v

t

!",

! "0.004 t

80

0

1 1 " # ! "0.004t. v 80 Solve for t in terms of the velocity. From this equation, we find that the time required for the volocity to decrease to 10 m/s is 21.9 s. The graph shows the airplane’s velocity as a function of time.

t ! 250

1

1

#v " 80 $ .

80 70 60 v (m/s)

44

50 40 30 20 10 0

0

5

10

15

20

21.9

25

30

t (s)

Practice Problem What distance does the airplane travel as its velocity decreases from 80 m/s to 10 m/s? Answer: 520 m.

45

13.3 Straight-Line Motion When the Acceleration Depends on Velocity or Position

Example 13.5

Gravitational (Position-Dependent) Acceleration (! Related Problem 13.62)

In terms of the distance s from the center of the earth, the magnitude of the acceleration due to gravity isgR2E>s 2, where RE is the radius of the earth. (See the discussion of gravity in Section 12.2.) If a spacecraft is a distances0 from the center of the earth, what outward velocityv0 must it be given to reach a specified distance h from the earth’s center? Strategy The acceleration is known as a function of the position s. We can apply the chain rule and separate variables, then integrate to determine the velocity as a function of s. Solution The acceleration due to gravity is toward the center of the earth: gRE2 a = - 2 . s Applying the chain rule results in gR E2 dv ds dv dv = = v = - 2 . a = dt ds dt ds s Separating variables, we obtain gRE2 v dv = - 2 ds. s We integrate this equation using the initial condition (v = v0 when s = s0 ) as the lower limits and the final condition (v = 0 whens = h ) as the upper limits: h

0

Lv0 c

v dv = -

Ls0

gR2E s2

ds,

v2 0 1 h d = gR2E c d , s s0 2 v0

0 -

v20 1 1 = gR E2 a - b. h 2 s0

Solving for v0, we obtain ...