MECÁNICA PARA INGENIERÍA

DINÁMICA

(SOLUCIONARIO) Quinta edición

Bedford 1 Fowler

----PEARSON

Prentice

Hall

®

Problem 12.1 The value of π is 3.1415962654. . . . . If Solution:

C is the circumference of a circle and r is its radius,

determine the value of r/C to four signiﬁcant digits. C = 2π r ⇒ r = 1 = 0.159154943.

C 2π

To four signiﬁcant digits we have r = 0.1592

C

Problem 12.2 The base of natural logarithms is e = Solution: The value of e is: e = 2.718281828

2.718281828 . . .

(a) To ﬁve signiﬁcant ﬁgures e = 2.7183

(a) Express e to ﬁve signiﬁcant digits. (b) e2 to ﬁve signiﬁcant ﬁgures is e2 = 7.3891

(b) Determine the value of e2 to ﬁve signiﬁcant digits. (c) Using the value from part (a) we ﬁnd e2 = 7.3892 which is

(c) Use the value of e you obtained in part (a) to deter-

not correct in the ﬁfth digit.

mine the value of e2 to ﬁve signiﬁcant digits.

[Part (c) demonstrates the hazard of using rounded-off

values in calculations.]

Problem 12.3 A machinist drills a circular hole in a Solution:

panel with a nominal radius r = 5 mm. The actual radius

of the hole is in the range r = 5 ± 0.01 mm. (a) To what (a) The radius is in the range r1 = 4.99 mm to r2 = 5.01 mm. These

number of signiﬁcant digits can you express the radius? numbers are not equal at the level of three signiﬁcant digits, but

(b) To what number of signiﬁcant digits can you express they are equal if they are rounded off to two signiﬁcant digits.

the area of the hole?

Two: r = 5.0 mm

5 mm

(b) The area of the hole is in the range from A1 = π r12 = 78.226 m2

to A2 = π r22 = 78.854 m2. These numbers are equal only if rounded

to one signiﬁcant digit:

One: A = 80 mm2

Problem 12.4 The opening in the soccer goal is 25 ft

wide and 8 ft high, so its area is 24 ft × 8 ft = 192 ft2.

What is its area in m2 to three signiﬁcant digits?

Solution: 1m 2

A = 192 ft2 3.281 ft = 17.8 m2

A = 17.8 m2

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Problem 12.5 The Burj Dubai, scheduled for comple-

tion in 2008, will be the world’s tallest building with a

height of 705 m. The area of its ground footprint will be

8000 m2. Convert its height and footprint area to U.S.

customary units to three signiﬁcant digits.

Solution:

h = 705 m 3.281 ft = 2.31 × 103 ft

1m

A = 8000 m2 3.218 ft 2

1m = 8.61 × 104 ft2

h = 2.31 × 103 ft, A = 8.61 × 104 ft2

Problem 12.6 Suppose that you have just purchased Solution: Convert the metric size n to inches, and compute the

a Ferrari F355 coupe and you want to know whether

you can use your set of SAE (U.S. Customary Units) percentage difference between the metric sized nut and the SAE

wrenches to work on it. You have wrenches with widths wrench. The results are:

w = 1/4 in, 1/2 in, 3/4 in, and 1 in, and the car has nuts

with dimensions n = 5 mm, 10 mm, 15 mm, 20 mm, 5 mm 1 inch = 0.19685.. in, 0.19685 − 0.25

and 25 mm. Deﬁning a wrench to ﬁt if w is no more 100

than 2% larger than n, which of your wrenches can you 25.4 mm 0.19685

use?

= −27.0%

n

10 mm 1 inch = 0.3937.. in, 0.3937 − 0.5 100 = −27.0%

25.4 mm 0.3937

15 mm 1 inch = 0.5905.. in, 0.5905 − 0.5 100 = +15.3%

25.4 mm 0.5905

20 mm 1 inch = 0.7874.. in, 0.7874 − 0.75 100 = +4.7%

25.4 mm 0.7874

25 mm 1 inch = 0.9843.. in, 0.9843 − 1.0 100 = −1.6%

25.4 mm 0.9843

A negative percentage implies that the metric nut is smaller than the

SAE wrench; a positive percentage means that the nut is larger then

the wrench. Thus within the deﬁnition of the 2% ﬁt, the 1 in wrench

will ﬁt the 25 mm nut. The other wrenches cannot be used.

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2

Problem 12.7 Suppose that the height of Mt. Everest Solution:

is known to be between 29,032 ft and 29,034 ft. Based

on this information, to how many signiﬁcant digits can (a) h1 = 29032 ft

you express the height (a) in feet? (b) in meters?

h2 = 29034 ft

The two heights are equal if rounded off to four signiﬁcant digits.

The ﬁfth digit is not meaningful.

Four: h = 29,030 ft

(b) In meters we have

h1 = 29032 ft 1m = 8848.52 m

3.281 ft

h2 = 29034 ft 1m = 8849.13 m

3.281 ft

These two heights are equal if rounded off to three signiﬁcant

digits. The fourth digit is not meaningful.

Three: h = 8850 m

Problem 12.8 The maglev (magnetic levitation) train

from Shanghai to the airport at Pudong reaches a speed

of 430 km/h. Determine its speed (a) in mi/h; (b) ft/s.

Solution:

(a) v = 430 km 0.6214 mi = 267 mi/h v = 267 mi/h

h 1 km

(b) v = 430 km 1000 m 1 ft 1 h = 392 ft/s

h 1 km 0.3048 m 3600 s

v = 392 ft/s

Problem 12.9 In the 2006 Winter Olympics, the men’s Solution:

15-km cross-country skiing race was won by Andrus

Veerpalu of Estonia in a time of 38 minutes, 1.3 seconds. (a) v = 15 km 60 min = 23.7 km/h v = 23.7 km/h

Determine his average speed (the distance traveled divi-

ded by the time required) to three signiﬁcant digits (a) in 38 + 1.3 min 1 h

km/h; (b) in mi/h. 60

(b) v = (23.7 km/h) 1 mi = 14.7 mi/h v = 14.7 mi/h

1.609 km

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Problem 12.10 The Porsche’s engine exerts 229 ft-lb

(foot-pounds) of torque at 4600 rpm. Determine the value

of the torque in N-m (Newton-meters).

Solution: 1 m = 310 N-m T = 310 N-m

3.281 ft

T = 229 ft-lb 1 N

0.2248 lb

Problem 12.11 The kinetic energy of the man in Active Solution:

Example 12.1 is deﬁned by 1 mv2, where m is his mass T = 1224 kg-m2/s2 1 slug 1 ft 2

2 14.59 kg

and v is his velocity. The man’s mass is 68 kg and he = 903 slug-ft2/s

0.3048 m

is moving at 6 m/s, so his kinetic energy is 1 (68 kg)

2

(6 m/s)2 = 1224 kg-m2/s2. What is his kinetic energy in

T = 903 slug-ft2/s

U.S. Customary units?

Problem 12.12 The acceleration due to gravity at sea Solution: Use Table 1.2. The result is:

level in SI units is g = 9.81 m/s2. By converting units,

use this value to determine the acceleration due to grav- m 1 ft ft ft

g = 9.81 s2 0.3048 m = 32.185 . . . s2 = 32.2 s2

ity at sea level in U.S. Customary units.

Problem 12.13 A furlong per fortnight is a facetious Solution: 1 furlong 3600 s 24 hr 14 day

unit of velocity, perhaps made up by a student as a 660 ft hr 1 day 1 fortnight

satirical comment on the bewildering variety of units v = 2 m/s 1 ft

engineers must deal with. A furlong is 660 ft (1/8 mile). 0.3048 m

A fortnight is 2 weeks (14 nights). If you walk to class

at 2 m/s, what is your speed in furlongs per fortnight to v = 12,000 furlongs

three signiﬁcant digits? fortnight

Problem 12.14 Determine the cross-sectional area of Solution:

the beam (a) in m2; (b) in in2.

A = (200 mm)2 − 2(80 mm)(120 mm) = 20800 mm2

y 1m 2

(a) A = 20800 mm2 1000 mm = 0.0208 m2 A = 0.0208 m2

40 mm

(b) A = 20800 mm2 1 in 2 A = 32.2 in2

120 mm x 25.4 mm = 32.2 in2

40 mm

40

mm

200 mm

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4

Problem 12.15 The cross-sectional area of the y

C12×30 American Standard Channel steel beam is A =

8.81 in2. What is its cross-sectional area in mm2? A

Solution:

A = 8.81 in2 25.4 mm 2 x

1 in = 5680 mm2

Problem 12.16 A pressure transducer measures a value Solution: Convert the units using Table 12.2 and the deﬁnition of

of 300 lb/in2. Determine the value of the pressure in

the Pascal unit. The result:

pascals. A pascal (Pa) is one newton per meter squared.

12 in 2 1 ft 2

lb 4.448 N

300 in2 1 lb 1 ft 0.3048 m

= 2.0683 . . . (106) N = 2.07(106) Pa

m2

Problem 12.17 A horsepower is 550 ft-lb/s. A watt is

1 N-m/s. Determine how many watts are generated by

the engines of the passenger jet if they are producing

7000 horsepower.

Solution: 1m 1 N = 5.22 × 106 W

3.281 ft 0.2248 lb

P = 7000 hp 550 ft-lb/s

1 hp

P = 5.22 × 106 W

Problem 12.18 Distributed loads on beams are expres- Solution: 1 m = 27.4 lb/ft w = 27.4 lb/ft

sed in units of force per unit length. If the value of a 3.281 ft

distributed load is 400 N/m, what is its value in lb/ft?. w = 400 N/m 0.2248 lb

1N

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Problem 12.19 The moment of inertia of the rectan- Solution:

gular area about the x axis is given by the equation

I = 1 bh3 . (a) I = 1 mm)(100 mm)3 = 66.7 × 106 mm4

3 (200

3

The dimensions of the area are b = 200 mm and h =

100 mm. Determine the value of I to four signiﬁcant (b) I = 66.7 × 106 mm4 1m 4

digits in terms of (a) mm4; (b) m4; (c) in4.

1000 mm = 66.7 × 10−6 m4

y 4

(c) I = 66.7 × 106 mm4 1 in = 160 in4

25.4 mm

h

x

b

Problem 12.20 In Example 12.3, instead of Einstein’s Solution:

equation consider the equation L = mc, where the mass

m is in kilograms and the velocity of light c is in meters (a) L = mc ⇒ Units(L) = kg-m/s

per second. (a) What are the SI units of L? (b) If the

value of L in SI units is 12, what is the value in U.S. (b) L = 12 kg-m/s 0.0685 slug 3.281 ft = 2.70 slug-ft/s

Customay base units? 1 kg 1m

L = 2.70 slug-ft/s

Problem 12.21 The equation Solution:

σ = My (a) My (N-m)m N

I σ= = = m2

I m4

is used in the mechanics of materials to determine

normal stresses in beams. My (2000 N-m)(0.1 m) 1 lb 0.3048 m 2

σ= =

(a) When this equation is expressed in terms of SI base (b) I 7 × 10−5 m4 4.448 N ft

units, M is in newton-meters (N-m), y is in meters

(m), and I is in meters to the fourth power (m4). = 59,700 lb

What are the SI units of σ ? ft2

(b) If M = 2000 N-m, y = 0.1 m, and I = 7 ×

10−5 m4, what is the value of σ in U.S. Customary

base units?

Problem 12.22 The acceleration due to gravity on the Solution:

surface of the moon is 1.62 m/s2. (a) What would be the

(a) The mass does not depend on location. The mass in kg is

mass of the C-clamp in Active Example 12.4 be on the 0.0272 slug 14.59 kg = 0.397 kg mass = 0.397 kg

1 slug

surface of the moon? (b) What would the weight of the

(b) The weight on the surface of the moon is

C-clamp in newtons be on the surface of the moon? W = mg = (0.397 kg)(1.62 m/s2) = 0.643 N W = 0.643 N

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6

Problem 12.23 The 1 ft × 1 ft × 1 ft cube of iron 1 ft

weighs 490 lb at sea level. Determine the weight in new- 1 ft 1 ft

tons of a 1 m × 1 m × 1 m cube of the same material

at sea level.

Solution: 490 lb

The weight density is γ = 1 ft3

The weight of the 1 m3 cube is:

W = γV = 490 lb (1 m)3 1 ft 3 1N = 77.0 kN

1 ft3 0.3048 m 0.2248 lb

Problem 12.24 The area of the Paciﬁc Ocean is Solution: The volume of the ocean is

64,186,000 square miles and its average depth is 12,925 ft. V = (64,186,000 mi2)(12,925 ft) 5,280 ft 2

Assume that the weight per unit volume of ocean water 1 mi = 2.312 × 1019 ft3

is 64 lb/ft3. Determine the mass of the Paciﬁc Ocean

(a) in slugs; (b) in kilograms.

(a) m = ρV = 64 lb/ft3 (2.312 × 1019f t3) = 4.60 × 1019 slugs

32.2 ft/s2

(b) m = (4.60 × 1019 slugs) 14.59 kg = 6.71 × 1020 kg

1 slug

Problem 12.25 The acceleration due to gravity at Solution: Use Eq. (12.3) a = GmE . Solve for the mass,

sea level is g = 9.81 m/s2. The radius of the earth R2

is 6370 km. The universal gravitational constant is m/s2)(6370 km)2 103 m 2

G = 6.67 × 10−11 N-m2/kg2. Use this information to km

gR2 (9.81

determine the mass of the earth. G

mE = = N-m2

kg2

6.67(10−11)

= 5.9679 . . . (1024) kg = 5.97(1024) kg

Problem 12.26 A person weighs 800 N sea level. Solution: Use Eq. (12.5).

The radius of the earth is 6372 km. What force is exerted

on the person by the gravitational attraction of the earth W = mg RE 2 WE g RE 2 6372 2

if he is in a space station in orbit 322 km above the r g RE + H 6372 + 322

surface of the earth? = = WE

= (800)(0.9519) = 262 N

Problem 12.27 The acceleration due to gravity on the Solution:

surface of the moon is 1.62 m/s2. The moon’s radius is

RM = 1738 km. (a) W = mgM = (10 kg)(1.26 m/s2) = 12.6 N W = 12.6 N

(a) What is the weight in newtons on the surface of (b) Adapting equation 1.4 we have aM = gM RM 2

the moon of an object that has a mass of 10 kg? r

. The force is

(b) Using the approach described in Example 12.5, de-

termine the force exerted on the object by the grav- then 2

ity of the moon if the object is located 1738 km

above the moon’s surface. F = maM = (10 kg)(1.62m/s2) 1738 km = 4.05 N

1738 km + 1738 km

F = 4.05 N

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Problem 12.28 If an object is near the surface of the Solution: Use a variation of Eq. (12.5).

earth, the variation of its weight with distance from the

center of the earth can often be neglected. The acceler- W = mg RE 2

ation due to gravity at sea level is g = 9.81 m/s2. The RE + h

radius of the earth is 6370 km. The weight of an object = 0.99 mg

at sea level is mg, where m is its mass. At what height

above the earth does the weight of the object decrease Solve for the radial height,

to 0.99 mg?

1

h = RE √ − 1 = (6370)(1.0050378 − 1.0)

0.99

= 32.09 . . . km = 32,100 m = 32.1 km

Problem 12.29 The planet Neptune has an equatorial

diameter of 49,532 km and its mass is 1.0247 × 1026 kg.

If the planet is modeled as a homogeneous sphere, what

is the acceleration due to gravity at its surface? (The uni-

versal gravitational constant is G = 6.67 × 10−11 h-m2/

kg2.)

Solution: W = G mN m = G mN m ⇒ gN = G mN

We have: rN2 r2 rN2

Note that the radius of Neptune is rN = 1 (49,532 km)

2

= 24,766 km

N-m2 1.0247 × 1026 kg

Thus gN = 6.67 × 10−11 kg2 (24766 km)2

× 1 km 2

1000 m = 11.1 m/s2

gN = 11.1 m/s2

Problem 12.30 At a point between the earth and the Solution: Let rEp be the distance from the Earth to the point where

moon, the magnitude of the force exerted on an object

by the earth’s gravity equals the magnitude of the force the gravitational accelerations are the same and let rMp be the distance

exerted on the object by the moon’s gravity. What is from the Moon to that point. Then, rEp + rMp = rEM = 383,000 km.

the distance from the center of the earth to that point The fact that the gravitational attractions by the Earth and the Moon

to three signiﬁcant digits? The distance from the center

of the earth to the center of the moon is 383,000 km, at this point are equal leads to the equation

and the radius of the earth is 6370 km. The radius of the

moon is 1738 km, and the acceleration due to gravity at gE RE 2 RM 2

its surface is 1.62 m/s2. rEp

= gM rMp ,

where rEM = 383,000 km. Substituting the correct numerical values

leads to the equation

m 6370 km 2 m 1738 km 2

9.81 s2 rEp s2 ,

= 1.62

rEM − rEp

where rEp is the only unknown. Solving, we get rEp = 344,770 km =

345,000 km.

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8

Problem 13.1 In Example 13.2, suppose that the vehi-

cle is dropped from a height h = 6m. (a) What is the

downward velocity 1 s after it is released? (b) What is

its downward velocity just before it reaches the ground?

Solution: The equations that govern the motion are:

a = −g = −9.81 m/s2

v = −gt

s = − 1 gt 2 + h h

2

s

(a) v = −gt = −(9.81 m/s2)(1 s) = −9.81 m/s.

The downward velocity is 9.81 m/s.

(b) We need to ﬁrst determine the time at which the vehicle hits the

ground

s = 0 = − 1 gt 2 + h ⇒ t = 2h = 2(6 m) = 1.106 s

2 g 9.81 m/s2

Now we can solve for the velocity

v = −gt = −(9.81 m/s2)(1.106 s) = −10.8 m/s.

The downward velocity is 10.8 m/s.

Problem 13.2 The milling machine is programmed so

that during the interval of time from t = 0 to t = 2 s,

the position of its head (in inches) is given as a function

of time by s = 4t − 2t3. What are the velocity (in in/s)

and acceleration (in in/s2) of the head at t = 1 s?

Solution: The motion is governed by the equations

s = (4 in/s)t − (2 in/s2)t2,

v = (4 in/s) − 2(2 in/s2)t,

a = −2(2 in/s2).

At t = 1 s, we have v = 0, a = −4 in/s2.

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Problem 13.3 In an experiment to estimate the accel- s

eration due to gravity, a student drops a ball at a distance sϭ0

of 1 m above the ﬂoor. His lab partner measures the time

it takes to fall and obtains an estimate of 0.46 s.

(a) What do they estimate the acceleration due to grav-

ity to be?

(b) Let s be the ball’s position relative to the ﬂoor.

Using the value of the acceleration due to gravity

that they obtained, and assuming that the ball is

released at t = 0, determine s (in m) as a function

of time.

Solution: The governing equations are

a = −g

v = −gt

s = − 1 gt 2 + h

2

(a) When the ball hits the ﬂoor we have

0 = − 1 gt 2 +h ⇒ g = 2h = 2(1 m) = 9.45 m/s2

2 t2 (0.46 s)2

g = 9.45 m/s2

(b) The distance s is then given by

s = − 1 (9.45 m/s2) + 1 m. s = −(4.73 m/s2)t2 + 1.0 m.

2

Problem 13.4 The boat’s position during the interval

of time from t = 2 s to t = 10 s is given by s = 4t +

1.6t2 − 0.08t3 m.

(a) Determine the boat’s velocity and acceleration at

t = 4 s.

(b) What is the boat’s maximum velocity during this

interval of time, and when does it occur?

Solution:

s = 4t + 1.6t2 − 0.08t3 a) v(4s) = 12.96 m/s2

v = ds = 4 + 3.2t − 0.24t2 ⇒ a(4s) = 1.28 m/s2

dt b) a = 3.2 − 0.48t = 0 ⇒ t = 6.67s

a = dv = 3.2 − 0.48t v(6.67s) = 14.67 m/s

dt

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10

Problem 13.5 The rocket starts from rest at t = 0 and

travels straight up. Its height above the ground as a

function of time can be approximated by s = bt2 + ct3,

where b and c are constants. At t = 10 s, the rocket’s

velocity and acceleration are v = 229 m/s and a = 28.2

m/s2. Determine the time at which the rocket reaches

supersonic speed (325 m/s). What is its altitude when

that occurs?

Solution: The governing equations are

s = bt2 + ct3,

s

v = 2bt + 3ct2,

a = 2b + 6ct.

Using the information that we have allows us to solve for the constants

b and c.

(229 m/s) = 2b(10 s) + 3c(10 s)2,

(28.2 m/s2) = 2b + 6c(10 s).

Solving these two equations, we ﬁnd b = 8.80 m/s2, c = 0.177 m/s3.

When the rocket hits supersonic speed we have

(325 m/s) = 2(8.80 m/s2)t + 3(0.177 m/s3)t2 ⇒ t = 13.2 s.

The altitude at this time is s = 1940 m.

s = (8.80 m/s2)(13.2 s)2 + (0.177 m/s3)(13.2 s)3

Problem 13.6 The position of a point during the inter-

1

val of time from t = 0 to t = 6 s is given by s = − 2 t 3 +

6t2 + 4t m.

(a) What is the maximum velocity during this interval

of time, and at what time does it occur?

(b) What is the acceleration when the velocity is a

maximum?

Solution: Maximum velocity occurs where a = dv = 0 (it could be a minimum)

dt

s = − 1 t 3 + 6t2 + 4t m

2 da

This occurs at t = 4 s. At this point = −3 so we have a maximum.

v = − 3 t 2 + 12t + 4 m/s

2 dt

(a) Max velocity is at t = 4 s. where v = 28 m/s and

(b) a = 0 m/s2

a = −3t + 12 m/s2

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Problem 13.7 The position of a point during the inter- This is indeed a maximum, since d2v = −6 < 0. The maximum

val of time from t = 0 to t = 3 seconds is s = 12 + dt2

5t2 − t3 m. velocity is

(a) What is the maximum velocity during this interval v = 10t − 3t2 t=1.667 = 8.33 m/s

of time, and at what time does it occur?

(b) The acceleration is dv when the velocity is a maximum.

(b) What is the acceleration when the velocity is a =0

maximum?

Solution:

(a) The velocity is ds = 10t − 3t2. The maximum occurs when

dt

dv

= 10 − 6t = 0, from which

dt

t = 10 = 1.667 seconds.

6

dt

Problem 13.8 The rotating crank causes the position P

of point P as a function of time to be s = 0.4 sin s

(2π t) m.

(a) Determine the velocity and acceleration of P at

t = 0.375 s.

(b) What is the maximum magnitude of the velocity

of P ?

(c) When the magnitude of the velocity of P is a

maximum, what is the acceleration of P ?

Solution:

s = 0.4 sin(2π t) a) v(0.375s) = −1.777 m/s

v = ds = 0.8π cos(2π t) ⇒ a(0.375) = −11.2 m/s2

dt b) vmax = 0.8π = 2.513 m/s2

a = dv = −1.6π2 sin(2π t) c) vmax ⇒ t = 0, nπ ⇒ a = 0

dt

Problem 13.9 For the mechanism in Problem 13.8, Solution:

draw graphs of the position s, velocity v, and acce-

leration a of point P as functions of time for 0 ≤ t ≤ 2 s.

Using your graphs, conﬁrm that the slope of the graph

of s is zero at times for which v is zero, and the slope

of the graph of v is zero at times for which a is zero.

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Problem 13.10 A seismograph measures the horizon-

tal motion of the ground during an earthquake. An engi-

neer analyzing the data determines that for a 10-s interval

of time beginning at t = 0, the position is approximated

by s = 100 cos(2π t) mm. What are (a) the maximum

velocity and (b) maximum acceleration of the ground

during the 10-s interval?

Solution: (b) The acceleration is

(a) The velocity is

ds d2s = −0.4π2 cos(2π t).

= −(2π )100 sin(2π t) mm/s = −0.2π sin(2π t) m/s. dt2

dt The acceleration maxima occur at

The velocity maxima occur at

dv = −0.4π2 cos(2π t) = 0, d3s = d2v = 0.8π3 sin(2π t) = 0,

dt dt3 dt2

from which from which 2π t = nπ , or t = n = 0, 1, 2, . . . K, where

,n

2

2π t = (2n − 1)π or t = (2n − 1) K ≤ 10 seconds.

, , 2

24 These acceleration maxima have the absolute value

n = 1, 2, 3, . . . M, where (2M − 1) ≤ 10 seconds.

4

These velocity maxima have the absolute value dv nπ = 0.4π2 = 3.95 m/s2.

dt

t=

2

ds = [0.2π ] = 0.628 m/s.

dt (2n−1)

t = 4

Problem 13.11 In an assembly operation, the robot’s s

arm moves along a straight horizontal line. During an

interval of time from t = 0 to t = 1 s, the position of the

arm is given by s = 30t2 − 20t3 mm. (a) Determine the

maximum velocity during this interval of time. (b) What

are the position and acceleration when the velocity is a

maximum?

Solution: v = (60) 1 − 60 1 mm/s

24

s = 30t2 − 20t3 mm

v = 15 mm/s

v = 60t − 60t2 mm/s (b) The position and acceleration at this time are

a = 60 − 120t mm/s2 s = 7.5 − 2.5 mm

da = −120 mm/s3 s = 5 mm

dt a = 0 mm/s2

(a) Maximum velocity occurs when dv = a = 0. This occurs at

dt

0 = 60 − 120t or t = 1/2 second. (since da/dt < 0, we have

a maximum). The velocity at this time is

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Problem 13.12 In Active Example 13.1, the accelera- Solution: The governing equations are

tion (in m/s2) of point P relative to point O is given as

a function of time by a = 3t2. Suppose that at t = 0 the a = (3 m/s4)t2

position and velocity of P are s = 5 m and v = 2 m/s. v = 1 m/s4 )t 3 + (2 m/s)

(3

Determine the position and velocity of P at t = 4 s. 3

OP s s = 1 (3 m/s4)t4 + (2 m/s)t + (5 m)

s 12

At t = 4 s, we have s = 77 m, v = 66 m/s.

Problem 13.13 The Porsche starts from rest at time

t = 0. During the ﬁrst 10 seconds of its motion, its

velocity in km/h is given as a function of time by v =

22.8t − 0.88t2, where t is in seconds. (a) What is the

car’s maximum acceleration in m/s2, and when does it

occur? (b) What distance in km does the car travel dur-

ing the 10 seconds?

Solution: First convert the numbers into meters and seconds

km 1000 m 1hr = 6.33 m/s

22.8

hr 1 km 3600 s

km 1000 m 1 hr = 0.244 m/s

0.88

hr 1 km 3600 s

The governing equations are then

s = 1 m/s)(t2/s) − 1 m/s)(t 3 /s2 ),

(6.33 (0.88

23

v = (6.33 m/s)(t/s) − (0.88 m/s)(t2/s2),

a = (6.33 m/s2) − 2(0.88 m/s)(t/s2),

d a = −2(0.88 m/s)(1/s2) = −1.76 m/s3.

dt

The maximum acceleration occurs at t = 0 (and decreases linearly

from its initial value).

amax = 6.33 m/s2 @ t = 0

In the ﬁrst 10 seconds the car travels a distance

1 km (10 s)2 1 km (10 s)3 1 hr

s = 22.8 − 0.88 3600 s

2 hr s3 hr s2

s = 0.235 km.

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to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.

14

Problem 13.14 The acceleration of a point is a = At t = 0, s = 40 m, thus C2 = 40. The position is

20t m/s2. When t = 0, s = 40 m and v = −10 m/s. s = 10 t3 − 10t + 40 m.

What are the position and velocity at t = 3 s?

3

Solution: The velocity is At t = 3 seconds,

v = a dt + C1,

where C1 is the constant of integration. Thus

v = 20t dt + C1 = 10t2 + C1. s = 10 t3 − 10t + 40 = 100 m.

3 t=3

At t = 0, v = −10 m/s, hence C1 = −10 and the velocity is v =

10t2 − 10 m/s. The position is The velocity at t = 3 seconds is

v = 10t2 − 10 t=3 = 80 m/s .

s = v dt + C2,

where C2 is the constant of integration.

s= (10t2 − 10) dt + C2 = 10 t3 − 10t + C2.

3

Problem 13.15 The acceleration of a point is a = The position is

60t − 36t2 m/s 2. When t = 0, s = 0 and v = 20 m/s.

What are position and velocity as a function of time?

Solution: The velocity is

v = a dt + C1 = (60t − 36t2) + C1 = 30t2 − 12t3 + C1. s = v dt + C2 = (30t2 − 12t3 + 20) + C2

At t = 0, v = 20 m/s, hence C1 = 20, and the velocity as a function = 10t3 − 3t4 + 20t + C2.

of time is At t = 0, s = 0, hence C2 = 0, and the position is

s = 10t3 − 3t4 + 20t m

v = 30t2 − 12t3 + 20 m/s .

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Problem 13.16 As a ﬁrst approximation, a bioengineer

studying the mechanics of bird ﬂight assumes that the

snow petrel takes off with constant acceleration. Video

measurements indicate that a bird requires a distance of

4.3 m to take off and is moving at 6.1 m/s when it does.

What is its acceleration?

Solution: The governing equations are

a = constant, v = at, s = 1 at2.

Using the information given, we have 2

6.1 m/s = at, 4.3 m = 1 at2.

2

Solving these two equations, we ﬁnd t = 1.41 s and a = 4.33 m/s2.

Problem 13.17 Progressively developing a more real-

istic model, the bioengineer next models the acceleration

of the snow petrel by an equation of the form a =

C(1 + sin ωt), where C and ω are constants. From video

measurements of a bird taking off, he estimates that

ω = 18/s and determines that the bird requires 1.42 s

to take off and is moving at 6.1 m/s when it does. What

is the constant C?

Solution: We ﬁnd an expression for the velocity by integrating the

acceleration

a = C(1 + sin ωt),

v = Ct + C (1 − cos ωt) = C t + 1 − 1 cos ωt .

ω ωω

Using the information given, we have

6.1 m/s = C 1.42 s + s − s cos[18(1.42)]

18 18

Solving this equation, we ﬁnd C = 4.28 m/s2.

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16