,SOLUTION MANUAL FOR Systems Dynamics and Controls 1st Edition Kelly
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, Chapter 1
SHORT ANSWER PROBLEMS
SA1.1 What are the English units of mass?
lb∙s2
Answer:
ft
SA1.2 What are the SI units of mass?
Answer: kg
SA1.3 Using the FLT system what are the dimensions of energy?
Answer: F ∙ L
SA1.4 Using the MLT system what are the dimensions of energy?
M∙L2
Answer:
T2
SA1.5 Using the FLT system what are the dimensions of resistance?
V F∙L/C F∙L∙T
Answer: Ω = = =
A C/T C2
SA1.6 What are the English units of resistance?
V N∙m/C N∙m∙s
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀: Ω = = =
A C/s C2
SA1.7 Using the MLT system what are the dimensions of mass flow rate?
M
Answer:
T
SA1.8 Using the MLT system what are the dimensions of concentration?
mol
Answer:
m3
SA1.9 Using the MLT system what are the dimensions of dynamic viscosity?
M
Answer:
L∙T
SA1.10 Using the MLT system what are the dimensions of impulse?
M∙L
Answer:
T
SA1.11 What are the SI units of dynamic viscosity?
kg
Answer:
m∙s
SA1.12 What is the definition of the unit impulse function as applied to a mechanical system?
© 2025 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible
website, in whole or in part.
,Answer: The unit impulse function is the mathematical representation of a unit impulse. That is a very
large force applied over a short period of time such that the integral of force over time is one. If the force
0 𝑡𝑡 ≠ 𝑎𝑎
is applied at time a then the unit impulse function is defined by 𝛿𝛿 (𝑡𝑡 − 𝑎𝑎) = � where
∞ 𝑡𝑡 = 𝑎𝑎
∞
∫0 𝛿𝛿 (𝑡𝑡 − 𝑎𝑎)𝑑𝑑𝑑𝑑 = 1
SA1.13 An electrical system has a voltage spike totaling 26 V∙s. Express this spike in mathematical form?
Answer: 𝑣𝑣 (𝑡𝑡) = 26𝛿𝛿 (𝑡𝑡)
SA1.14 A mechanical system is subject to an impulse of 0.4 N∙s. Express this in mathematical form?
Answer: 𝐼𝐼 (𝑡𝑡) = 0.4𝛿𝛿 (𝑡𝑡)
SA1.15 An electrical system is subject to the voltage input shown in Figure SP 1.15. Write a mathematical
form for the voltage.
Answer: 𝑣𝑣 (𝑡𝑡) = 100𝑡𝑡 [𝑢𝑢(𝑡𝑡) − 𝑢𝑢(−1)] + 100𝑢𝑢 (𝑡𝑡 − 1) = 100𝑡𝑡𝑡𝑡(𝑡𝑡) + 100(𝑡𝑡 − 1)𝑢𝑢(𝑡𝑡 − 1)
SA1.16 A mechanical system is subject to a force that is illustrated in Figure SP 1.16. Write a unified
mathematical expression for the force.
Answer: 𝐹𝐹 (𝑡𝑡) = 240𝑡𝑡 [𝑢𝑢(𝑡𝑡) − 𝑢𝑢(𝑡𝑡 − 0.05)] − 240(𝑡𝑡 − 0.1)[𝑢𝑢(𝑡𝑡 − 0.05) − 𝑢𝑢(𝑡𝑡 − 0.1)] =
240[𝑡𝑡𝑡𝑡(𝑡𝑡) − (𝑡𝑡 − 0.05)𝑢𝑢(𝑡𝑡 − 0.05) + (𝑡𝑡 − 0.1)𝑢𝑢(𝑡𝑡 − 0.1)]
SA1.17 What is the total impulse imparted to a mechanical system from the force shown in Figure SP
1.16?
0.05 0.1 1
Answer: ∫0 240𝑡𝑡𝑡𝑡𝑡𝑡 + ∫0.05 240(0.1 − 𝑡𝑡)𝑑𝑑𝑑𝑑 = (12)(0.1) = 0.6 N ∙ s
2
SA1.18 Linearize the function (1 + 0.2𝑥𝑥 )3/2 for small 𝑥𝑥.
3 1 3 1
Answer: (1 + 0.2𝑥𝑥 )3/2 = 1 + (0.2𝑥𝑥 ) + � � � � (0.2𝑥𝑥 )2 + ⋯ ≈ 1 + 0.3𝑥𝑥
2 2 2 2
1
SA1.19 Linearize the function (1 − 0.2𝑥𝑥 )4 for small 𝑥𝑥.
1
1 1 1 3
Answer: (1 − 0.2𝑥𝑥 )4 = 1 + (−0.2𝑥𝑥 ) + � � �− � (−0.2𝑥𝑥 )2 + ⋯ ≈ 1 − 0.05𝑥𝑥
4 2 4 4
SA1.20 Linearize the function (1 + 0.3𝑥𝑥 )−0.76 for small 𝑥𝑥.
1
Answer: (1 + 0.3𝑥𝑥 )−0.76 = 1 + (−0.76)(0.3𝑥𝑥 ) + (−0.76)(−1.76)(0.3𝑥𝑥 )2 + ⋯ ≈ 1 − 0.228𝑥𝑥
2
SA1.21 Linearize the function 𝑒𝑒 −2𝑥𝑥 for small 𝑥𝑥.
1
Answer: 𝑒𝑒 −2𝑥𝑥 = 1 + (−2𝑥𝑥 ) + (−2𝑥𝑥 )2 + ⋯ ≈ 1 − 2𝑥𝑥
2
SA1.22 Linearize the function sin 𝜃𝜃 cos 2𝜃𝜃 for small 𝜃𝜃.
Answer: sin 𝜃𝜃 cos 2𝜃𝜃 ≈ 𝜃𝜃 (1) = 𝜃𝜃
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website, in whole or in part.
,SA1.23 Linearize the function tan 0.3𝜃𝜃 for small 𝜃𝜃.
Answer: tan 0.3𝜃𝜃 ≈ 0.3𝜃𝜃
SA1.24 Linearize the function sin 𝜃𝜃 cos 2 𝜃𝜃 for small 𝜃𝜃.
Answer: sin 𝜃𝜃 cos 2 𝜃𝜃 ≈ 𝜃𝜃 (1)2 = 𝜃𝜃
SA1.25 Linearize the differential equation assuming small 𝜃𝜃
𝑑𝑑 2 𝜃𝜃 𝑑𝑑𝜃𝜃
2
+ cos 𝜃𝜃 + tan 𝜃𝜃 = 6 sin 2𝑡𝑡
𝑑𝑑𝑡𝑡 𝑑𝑑𝑑𝑑
𝑑𝑑 2 𝜃𝜃 𝑑𝑑𝜃𝜃
Answer: + + 𝜃𝜃 = 6 sin 2𝑡𝑡
𝑑𝑑𝑡𝑡 2 𝑑𝑑𝑑𝑑
SA1.26 Linearize the differential equation assuming small 𝜃𝜃
𝑑𝑑 2 𝜃𝜃 𝑑𝑑𝜃𝜃 1
+ 4𝑡𝑡 cos 𝜃𝜃 + 16 sin � 𝜃𝜃� cos(3𝜃𝜃) = 6 sin 2𝑡𝑡
𝑑𝑑𝑡𝑡 2 𝑑𝑑𝑑𝑑 2
𝑑𝑑 2 𝜃𝜃 𝑑𝑑𝜃𝜃
Answer: + 4𝑡𝑡 + 8𝜃𝜃 = 6 sin 2𝑡𝑡
𝑑𝑑𝑡𝑡 2 𝑑𝑑𝑑𝑑
SA1.27 Fill in the blanks from the choices in parentheses regarding the differential equation
𝑑𝑑 2 𝜃𝜃 𝑑𝑑𝜃𝜃
2
+ 4𝑡𝑡 + 12𝜃𝜃 = 6 sin 2𝑡𝑡
𝑑𝑑𝑡𝑡 𝑑𝑑𝑑𝑑
Answer: The equation is an ordinary differential equation. It has an independent variable of t and a
dependent variable of 𝜃𝜃. It is a linear, nonhomogeneous differential equation with variable coefficients.
SA1.28 Fill in the blanks from the choices in parentheses regarding the differential equation
𝑑𝑑 2 𝜃𝜃 𝑑𝑑𝜃𝜃
2
+ 4 cos 𝜃𝜃 + 12 sin 𝜃𝜃 = 0
𝑑𝑑𝑡𝑡 𝑑𝑑𝑑𝑑
Answer: The equation is an ordinary differential equation. It has an independent variable of 𝑡𝑡 and a
dependent variable of 𝜃𝜃 It is a linear, homogeneous differential equation with constant coefficients.
PROBLEMS
1.1 Newton’s law of cooling is used to calculate the rate at which heat is transferred by convection to a
solid body at a temperature T from a surrounding fluid at a temperature T∞. The formula is
𝑄𝑄̇ = ℎ𝐴𝐴(𝑇𝑇 − 𝑇𝑇∞ ) (𝑎𝑎)
where Q˙ is the rate of heat transfer (rate of energy transfer), A is the area over which heat is transferred
by convection, and h is the heat transfer coefficient (also called the film coefficient). Use Equation (a) to
determine the basic dimensions of h. Suggest appropriate units for h using the English system and the SI
system.
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website, in whole or in part.
,Solution
Equation (a) of the problem statement is used to solve for ℎ as
𝑄𝑄̇
ℎ= (𝑏𝑏)
𝐴𝐴(𝑇𝑇 − 𝑇𝑇∞ )
The Principle of Dimensional Homogeneity is used to determine the dimensions of the heat transfer
coefficient. Using the F-L-T system, the dimensions of the quantities in Equation (b) are
F∙L
�𝑄𝑄̇� = � � (𝑐𝑐)
T
2
[𝐴𝐴] = [L ] (𝑑𝑑)
[𝑇𝑇 − 𝑇𝑇∞ ] = [Θ] (𝑒𝑒)
From Equations (b)-(e)the dimensions of the heat transfer coefficient are
F∙L F
[ℎ] = � 2
�=� � (𝑓𝑓)
T∙Θ∙L T∙Θ∙L
N
Possible units for the heat transfer coefficient using the SI system are while possible units using the
m∙s∙K
lb
English system are .
ft∙s∙R
1.2 The Reynolds number is used as a measure of the ratio of inertia forces to the friction forces in the
flow of a fluid in a circular pipe. The Reynolds number (Re) is defined as
𝜌𝜌𝜌𝜌𝜌𝜌
𝑅𝑅𝑅𝑅 = (𝑎𝑎)
𝜇𝜇
where 𝜌𝜌 is the mass density of the fluid, V is the average velocity of the flow, D is the diameter of the
pipe, and μ is the dynamic viscosity of the fluid. Show that the Reynolds number is dimensionless.
Solution
The dimensions on the quantities on the right-hand side of Equation (a) are obtained using Table 1.2 as
M
[𝜌𝜌] = � 3 � (𝑏𝑏)
L
L
[𝑉𝑉 ] = � � (𝑐𝑐)
T
[𝐷𝐷] = [L] (𝑑𝑑)
M
[𝜇𝜇] = � � (𝑒𝑒)
L∙T
Substituting Equations (b)-(e) into Equation (a) leads to
M L
L3∙T∙L
[Re] = � � = [1] (𝑓𝑓)
M
L∙T
Equation (f) shows that the Reynolds number is dimensionless.
𝑑𝑑𝑑𝑑
1.3 The relationship between voltage and current in a capacitor is 𝑖𝑖 = 𝐶𝐶 . Use this relation to determine
𝑑𝑑𝑑𝑑
the basic dimensions of the capacitance C.
Solution
The capacitance of a capacitor is defined by
𝑖𝑖
𝐶𝐶 = (𝑎𝑎)
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
The dimension of 𝑖𝑖 is that of electric current, which is a basic dimension. The dimensions of electric
potential are obtained from Table 1.2 as
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website, in whole or in part.
, F∙L
[𝑣𝑣 ] = � � (𝑏𝑏)
i∙T
Thus the dimensions of time rate of change of electric potential are
𝑑𝑑𝑑𝑑 F∙L
� �=� � (𝑐𝑐)
𝑑𝑑𝑑𝑑 i ∙ T2
Use of Equation (c) in Equation (a) leads to
i i2 ∙ T 2
[𝐶𝐶 ] = � �=� � (𝑑𝑑)
F∙L F∙L
i ∙ T2
𝑘𝑘
1.4 The natural frequency 𝜔𝜔𝑛𝑛 of a mechanical system of mass m and stiffness k is calculated by 𝜔𝜔𝑛𝑛 = � .
𝑚𝑚
(a) What are the dimensions of𝜔𝜔𝑛𝑛 ? (b) A mass-spring system of mass 100 g has a natural frequency of 20
Hz. What is the stiffness of the system?
Solution
M
(a)The dimension of spring stiffness in the M-L-T system is � 2 �. Thence the dimensions of natural
T
frequency are
M 1/2
2 1
[𝜔𝜔𝑛𝑛 ] = �� T � � = � � (𝑎𝑎)
M 𝑇𝑇
(b) Converting Hz to r/s
cycles cycles 2π r r
𝜔𝜔𝑛𝑛 = 20 Hz = 20 = �20 �� � = 125.7 (𝑏𝑏)
s s s s
The stiffness is calculated form the natural frequency as
r 2 N
𝑘𝑘 = 𝑚𝑚𝜔𝜔𝑛𝑛2 = (0.1 kg) �125.7 � = 1.58 × 103 (𝑐𝑐)
s m
1.5 Carbon nanotubes are new materials that consist of carbon atoms and are of significant interest
because of their good conductivity properties, light weight, and high strength. (a) The density of a carbon
nanotube is 1300 kg/m3 . A carbon nanotube is a tube with a diameter of 1 carbon atom, d = 0.68 nm.
Determine the mass of a carbon nanotube whose length is 80 nm. (b) The elastic modulus of a carbon
nanotube is 𝐸𝐸 = 1.1 TPa. What is its elastic modulus in pounds per square inch? (c) If modeled as a fixed-
fixed beam, the fundamental frequency of the nanotube is
𝐸𝐸𝐸𝐸
𝜔𝜔 = 22.37�
𝜌𝜌𝜌𝜌𝐿𝐿4
𝜋𝜋
where 𝐼𝐼 = 𝑟𝑟 4 is the cross-sectional moment of inertia of the beam and A is its cross-sectional area.
4
Calculate the fundamental frequency in hertz of a nanotube of length 80 nm.
Solution
(a) The mass of the nanotube is
kg
𝑚𝑚 = 𝜌𝜌𝜌𝜌𝜌𝜌 = 𝜌𝜌(𝜋𝜋𝑟𝑟 2 )𝐿𝐿 = �1300 3 � π(0.34 × 10−9 m)2 (80 × 10−9 m)
m
= 3.78 × 10−23 kg (a)
(b) Conversion between TPa and psi leads to
N lb 1 m 2 1 ft 2 lb
𝐸𝐸 = 1.1 𝑇𝑇𝑇𝑇𝑇𝑇 = �1.1 × 1012 2 � �0.225 � � � � � = 1.60 × 108 2 (𝑏𝑏)
m N 3.28 ft 12 in in
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website, in whole or in part.
,(c) Calculation of the fundamental frequency yields
N 𝜋𝜋
𝐸𝐸𝐸𝐸 �1.1 × 1012 2 � (0.34 × 10−9 m)4
m 4
𝜔𝜔 = 22.37� = 22.37�
𝜌𝜌𝜌𝜌𝐿𝐿4 kg
�1300 3 � π(0.34 × 10−9 m)2 (80 × 10−9 m)4
m
10
r
= 1.74 × 10 (𝑐𝑐)
s
Converting to Hz gives
r 1 cycle
𝜔𝜔 = �1.74 × 1010 � � � = 2.78 × 109 Hz (𝑑𝑑)
s 2π r
1.6 During 24 hours of operation, a motor expends 900 kW∙hr of energy. Noting that a horsepower (hp) is
a unit of power such that 1 hp= 550 ft∙lb/s, determine the power delivered by the motor in horsepower.
Solution
The power of the motor is
900 kW ∙ hr N ∙ m 0.225 lb
𝑃𝑃 = = 37.5 kW = 37.5 × 103 W = �37.5 × 103 �� �
24 hr s N
ft ∙ lb
= 2.77 × 104 (𝑎𝑎)
s
Conversion to hp yields
ft ∙ lb 1 hp
P = 2.77 × 104 � � = 50.3 hp (𝑏𝑏)
s ft ∙ lb
550
s
1.7 The density of water is 1.94 slugs/ft 3 . Express the density of water in kilograms per cubic meter
(kg/m3 ).
Solution
The conversion of density from English units to SI units is
slugs 1 kg 3.28 ft 3 kg
𝜌𝜌 = �1.94 3
� � � � � = 1000 3 (𝑎𝑎)
ft 0.00685 slugs m m
1.8 A train is traveling at a speed of 180 km/hr and is decelerating at 6 m/s 2 . Assuming uniform
deceleration, how far will the train travel, in miles, before it stops?
Solution
The constant acceleration of the train is
m
𝑎𝑎 = −6 2 (𝑎𝑎)
s
The velocity is obtained using Equation (a) as
𝑣𝑣(𝑡𝑡) = −6𝑡𝑡 + 𝐶𝐶 (𝑏𝑏)
The constant of integration is evaluated by requiring
km 1000 m 1 hr m
𝑣𝑣(0) = �180 �� �� � = 50 (𝑐𝑐)
hr km 3600 s s
Substituting Equation (c) inti Equation (b) leads to
𝑣𝑣 (𝑡𝑡) = −6𝑡𝑡 + 50 (𝑑𝑑)
The train stops when the velocity is zero
0 = −6𝑡𝑡 + 50 ⟹ 𝑡𝑡 = 8.33 s (e)
The distance traveled before the train stops is obtained by integrating the velocity with respect to time and
evaluating it at 8.33 s
𝑥𝑥 (𝑡𝑡) = −3𝑡𝑡 2 + 50𝑡𝑡 (𝑓𝑓)
The distance traveled before the train stops is
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website, in whole or in part.
, 𝑥𝑥 (8.33) = −3(8.33)2 + 50(8.33) = 208.3 m (g)
1.9 The differential equation governing the angular velocity v of a shaft of mass moment of inertia J,
acted on by a torque T and attached to bearings of torsional damping coefficient 𝑐𝑐𝑡𝑡 , is
𝑑𝑑𝑑𝑑
𝐽𝐽 + 𝑐𝑐𝑡𝑡 𝜔𝜔 = 𝑇𝑇 (𝑎𝑎)
𝑑𝑑𝑑𝑑
Use Equation (a) to determine the appropriate dimensions of 𝑐𝑐𝑡𝑡 .
Solution
The principle of dimensional homogeneity states that every term in an equation must have the same
dimensions. In this case, it is the dimensions of 𝑇𝑇, or torque, which are [F ∙ L]. The dimensions of angular
velocity are
1
� �. Thus, the dimension of 𝑐𝑐𝑡𝑡 are
T
F∙L
[𝑐𝑐𝑡𝑡 ] = � � = [F ∙ L ∙ T] (𝑏𝑏)
1
T
1.10 In an armature-controlled dc servomotor, the torque applied to the armature due to a magnetic
coupling field is given by
𝑇𝑇 = 𝐾𝐾𝑎𝑎 𝑖𝑖𝑎𝑎 𝑖𝑖𝑓𝑓 (𝑎𝑎)
and the back electromotive force (voltage) generated as the shaft rotates through the magnetic field is
𝑣𝑣 = 𝐾𝐾𝑓𝑓 𝑖𝑖𝑓𝑓 𝜔𝜔 (𝑏𝑏)
where 𝑖𝑖𝑎𝑎 is the current in the armature circuit, 𝑖𝑖𝑓𝑓 is the current in the field circuit, and 𝜔𝜔 is the angular
velocity of the shaft. Show that the constants𝐾𝐾𝑎𝑎 and 𝐾𝐾𝑓𝑓 have the same dimensions. Referring to Table 1.2,
what physical quantity has the same dimensions as these constants?
Solution
Equation (a) is rearranged to
𝑇𝑇
𝐾𝐾𝑎𝑎 = (𝑐𝑐 )
𝑖𝑖𝑎𝑎 𝑖𝑖𝑓𝑓
The dimensions of 𝐾𝐾𝑎𝑎 are
F∙L
[𝐾𝐾𝑎𝑎 ] = � 2 � (𝑑𝑑)
𝑖𝑖
Equation (b) is rearranged as
𝑣𝑣
𝐾𝐾𝑓𝑓 = (𝑒𝑒)
𝑖𝑖𝑓𝑓 𝜔𝜔
The dimensions of 𝐾𝐾𝑓𝑓 are
F∙L
F∙L
�𝐾𝐾𝑓𝑓 � = � 𝑖𝑖 ∙ 𝑇𝑇 � = � 2 � (𝑓𝑓)
1 𝑖𝑖
𝑖𝑖 ∙
𝑡𝑡
It is clear from Equations (d) and (f) that the dimensions are the same. The quantity of inductance has the
same dimensions as 𝐾𝐾𝑎𝑎 and 𝐾𝐾𝑓𝑓 .
1.11 The rate of heat transfer by radiation from one body to another is given by
𝑄𝑄̇ = 𝜎𝜎𝜎𝜎𝜎𝜎�𝑇𝑇 4 − 𝑇𝑇𝑏𝑏4 � (𝑎𝑎)
where A is the area of heat transfer, 𝜖𝜖 is the dimensionless emissivity of the body, T is the temperature of
the body, Tb is the temperature of the radiating body, and the Stefan-Boltzmann constant is
Btu W
𝜎𝜎 = 1.73 × 10−7 2 4
= 5.67 × 10−8 2 4 (𝑏𝑏)
ft ∙ hr ∙ R m ∙K
© 2025 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible
website, in whole or in part.
, (a) What are the basic dimensions of 𝑄𝑄̇ ? (b) The differential equation governing the transient temperature
in a body due to radiation heat transfer is
𝑑𝑑𝑑𝑑
𝜌𝜌𝜌𝜌 + 𝜎𝜎𝜎𝜎𝜎𝜎�𝑇𝑇 4 − 𝑇𝑇𝑏𝑏4 � = 0 (𝑐𝑐)
𝑑𝑑𝑑𝑑
where 𝜌𝜌 is the mass density of the body and c is its specific heat. The body is in a steady state with a
uniform temperature defined by a temperature Ts when the temperature of the radiating body is suddenly
changed from Tbs to Tb1. Let T1 be perturbation in temperature from the steady state such that the
temperature T is
𝑇𝑇 = 𝑇𝑇𝑠𝑠 + 𝑇𝑇1 (𝑡𝑡) (𝑑𝑑)
Substitute Equation (d) into Equation (c), use the binomial theorem to expand the nonlinear term, and
derive a linearized differential equation to solve for T1(t).
Solution
(a)The dimensions of 𝑄𝑄̇ are determined from Equation (a)
F∙L F∙L
�𝑄𝑄̇� = � 2 � [ 𝐿𝐿2 ][ 4 ]
Θ = � � (𝑒𝑒)
𝐿𝐿 ∙ 𝑇𝑇 ∙ Θ4 𝑇𝑇
(b) Substitution of Equation (d) into Equation (c) leads to
𝑑𝑑
𝜌𝜌𝜌𝜌 (𝑇𝑇𝑠𝑠 + 𝑇𝑇1 ) + 𝜎𝜎𝜎𝜎𝜎𝜎 [(𝑇𝑇𝑠𝑠 + 𝑇𝑇1 )4 − (𝑇𝑇𝑏𝑏𝑏𝑏 + 𝑇𝑇𝑏𝑏1 )4 ] = 0 (𝑓𝑓)
𝑑𝑑𝑑𝑑
Simplifying leads to
𝑑𝑑𝑇𝑇1 𝑇𝑇1 4 4
𝑇𝑇𝑏𝑏1 4
𝜌𝜌𝜌𝜌 + 𝜎𝜎𝜎𝜎𝜎𝜎 �𝑇𝑇𝑠𝑠4 �1 + � − 𝑇𝑇𝑏𝑏𝑏𝑏 �1 + � �=0 (𝑔𝑔)
𝑑𝑑𝑑𝑑 𝑇𝑇𝑠𝑠 𝑇𝑇𝑏𝑏𝑏𝑏
Applying the binomial expansion, keeping only through linear terms leads to
𝑑𝑑𝑇𝑇1 𝑇𝑇1 4
𝑇𝑇𝑏𝑏1
𝜌𝜌𝜌𝜌 + 𝜎𝜎𝜎𝜎𝜎𝜎 �𝑇𝑇𝑠𝑠4 �4 � − 𝑇𝑇𝑏𝑏𝑏𝑏 �4 �� = 0 (ℎ)
𝑑𝑑𝑑𝑑 𝑇𝑇𝑠𝑠 𝑇𝑇𝑏𝑏𝑏𝑏
Simplifying Equation (h) and rearranging leads to
𝑑𝑑𝑇𝑇1 3
𝜌𝜌𝜌𝜌 + 4𝜎𝜎𝜎𝜎𝜎𝜎𝑇𝑇𝑠𝑠3 𝑇𝑇1 = 4𝜎𝜎𝜎𝜎𝜎𝜎𝑇𝑇𝑏𝑏𝑏𝑏 𝑇𝑇𝑏𝑏1 (𝑖𝑖)
𝑑𝑑𝑑𝑑
1.12 A nonlinear differential equation governing the motion of a mechanical system is
1 1
𝑚𝑚𝐿𝐿2 𝜃𝜃̈ + 𝑐𝑐𝐿𝐿2 𝜃𝜃̇ + 𝑘𝑘𝐿𝐿2 sin 𝜃𝜃 cos 𝜃𝜃 = 0 (𝑎𝑎)
3 4
Assuming small 𝜃𝜃, derive a linearized differential equation for the system.
Solution
The differential equation is linearized by using the small angle assumption which implies sin 𝜃𝜃 ≈ 𝜃𝜃 and
cos 𝜃𝜃 ≈ 1. Using these approximations in the differential equation leads to the linearized approximation
as
1 1
𝑚𝑚𝐿𝐿2 𝜃𝜃̈ + 𝑐𝑐𝐿𝐿2 𝜃𝜃̇ + 𝑘𝑘𝐿𝐿2 𝜃𝜃 = 0 (𝑏𝑏)
3 4
1.13 The differential equation governing the motion of the system of Figure P1.13 is
1 𝐿𝐿
𝑚𝑚𝐿𝐿2 𝜃𝜃̈ + 𝐿𝐿𝑦𝑦̈ sin 𝜃𝜃 + 𝐿𝐿𝑥𝑥̈ cos 𝜃𝜃 + 𝑚𝑚𝑚𝑚 sin 𝜃𝜃 = 0 (𝑎𝑎)
3 2
where x(t) and y(t) are known functions of time. Derive a linearized equation governing the motion of the
system for small 𝜃𝜃 .
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, Chapter 1
SHORT ANSWER PROBLEMS
SA1.1 What are the English units of mass?
lb∙s2
Answer:
ft
SA1.2 What are the SI units of mass?
Answer: kg
SA1.3 Using the FLT system what are the dimensions of energy?
Answer: F ∙ L
SA1.4 Using the MLT system what are the dimensions of energy?
M∙L2
Answer:
T2
SA1.5 Using the FLT system what are the dimensions of resistance?
V F∙L/C F∙L∙T
Answer: Ω = = =
A C/T C2
SA1.6 What are the English units of resistance?
V N∙m/C N∙m∙s
𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀𝐀: Ω = = =
A C/s C2
SA1.7 Using the MLT system what are the dimensions of mass flow rate?
M
Answer:
T
SA1.8 Using the MLT system what are the dimensions of concentration?
mol
Answer:
m3
SA1.9 Using the MLT system what are the dimensions of dynamic viscosity?
M
Answer:
L∙T
SA1.10 Using the MLT system what are the dimensions of impulse?
M∙L
Answer:
T
SA1.11 What are the SI units of dynamic viscosity?
kg
Answer:
m∙s
SA1.12 What is the definition of the unit impulse function as applied to a mechanical system?
© 2025 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible
website, in whole or in part.
,Answer: The unit impulse function is the mathematical representation of a unit impulse. That is a very
large force applied over a short period of time such that the integral of force over time is one. If the force
0 𝑡𝑡 ≠ 𝑎𝑎
is applied at time a then the unit impulse function is defined by 𝛿𝛿 (𝑡𝑡 − 𝑎𝑎) = � where
∞ 𝑡𝑡 = 𝑎𝑎
∞
∫0 𝛿𝛿 (𝑡𝑡 − 𝑎𝑎)𝑑𝑑𝑑𝑑 = 1
SA1.13 An electrical system has a voltage spike totaling 26 V∙s. Express this spike in mathematical form?
Answer: 𝑣𝑣 (𝑡𝑡) = 26𝛿𝛿 (𝑡𝑡)
SA1.14 A mechanical system is subject to an impulse of 0.4 N∙s. Express this in mathematical form?
Answer: 𝐼𝐼 (𝑡𝑡) = 0.4𝛿𝛿 (𝑡𝑡)
SA1.15 An electrical system is subject to the voltage input shown in Figure SP 1.15. Write a mathematical
form for the voltage.
Answer: 𝑣𝑣 (𝑡𝑡) = 100𝑡𝑡 [𝑢𝑢(𝑡𝑡) − 𝑢𝑢(−1)] + 100𝑢𝑢 (𝑡𝑡 − 1) = 100𝑡𝑡𝑡𝑡(𝑡𝑡) + 100(𝑡𝑡 − 1)𝑢𝑢(𝑡𝑡 − 1)
SA1.16 A mechanical system is subject to a force that is illustrated in Figure SP 1.16. Write a unified
mathematical expression for the force.
Answer: 𝐹𝐹 (𝑡𝑡) = 240𝑡𝑡 [𝑢𝑢(𝑡𝑡) − 𝑢𝑢(𝑡𝑡 − 0.05)] − 240(𝑡𝑡 − 0.1)[𝑢𝑢(𝑡𝑡 − 0.05) − 𝑢𝑢(𝑡𝑡 − 0.1)] =
240[𝑡𝑡𝑡𝑡(𝑡𝑡) − (𝑡𝑡 − 0.05)𝑢𝑢(𝑡𝑡 − 0.05) + (𝑡𝑡 − 0.1)𝑢𝑢(𝑡𝑡 − 0.1)]
SA1.17 What is the total impulse imparted to a mechanical system from the force shown in Figure SP
1.16?
0.05 0.1 1
Answer: ∫0 240𝑡𝑡𝑡𝑡𝑡𝑡 + ∫0.05 240(0.1 − 𝑡𝑡)𝑑𝑑𝑑𝑑 = (12)(0.1) = 0.6 N ∙ s
2
SA1.18 Linearize the function (1 + 0.2𝑥𝑥 )3/2 for small 𝑥𝑥.
3 1 3 1
Answer: (1 + 0.2𝑥𝑥 )3/2 = 1 + (0.2𝑥𝑥 ) + � � � � (0.2𝑥𝑥 )2 + ⋯ ≈ 1 + 0.3𝑥𝑥
2 2 2 2
1
SA1.19 Linearize the function (1 − 0.2𝑥𝑥 )4 for small 𝑥𝑥.
1
1 1 1 3
Answer: (1 − 0.2𝑥𝑥 )4 = 1 + (−0.2𝑥𝑥 ) + � � �− � (−0.2𝑥𝑥 )2 + ⋯ ≈ 1 − 0.05𝑥𝑥
4 2 4 4
SA1.20 Linearize the function (1 + 0.3𝑥𝑥 )−0.76 for small 𝑥𝑥.
1
Answer: (1 + 0.3𝑥𝑥 )−0.76 = 1 + (−0.76)(0.3𝑥𝑥 ) + (−0.76)(−1.76)(0.3𝑥𝑥 )2 + ⋯ ≈ 1 − 0.228𝑥𝑥
2
SA1.21 Linearize the function 𝑒𝑒 −2𝑥𝑥 for small 𝑥𝑥.
1
Answer: 𝑒𝑒 −2𝑥𝑥 = 1 + (−2𝑥𝑥 ) + (−2𝑥𝑥 )2 + ⋯ ≈ 1 − 2𝑥𝑥
2
SA1.22 Linearize the function sin 𝜃𝜃 cos 2𝜃𝜃 for small 𝜃𝜃.
Answer: sin 𝜃𝜃 cos 2𝜃𝜃 ≈ 𝜃𝜃 (1) = 𝜃𝜃
© 2025 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible
website, in whole or in part.
,SA1.23 Linearize the function tan 0.3𝜃𝜃 for small 𝜃𝜃.
Answer: tan 0.3𝜃𝜃 ≈ 0.3𝜃𝜃
SA1.24 Linearize the function sin 𝜃𝜃 cos 2 𝜃𝜃 for small 𝜃𝜃.
Answer: sin 𝜃𝜃 cos 2 𝜃𝜃 ≈ 𝜃𝜃 (1)2 = 𝜃𝜃
SA1.25 Linearize the differential equation assuming small 𝜃𝜃
𝑑𝑑 2 𝜃𝜃 𝑑𝑑𝜃𝜃
2
+ cos 𝜃𝜃 + tan 𝜃𝜃 = 6 sin 2𝑡𝑡
𝑑𝑑𝑡𝑡 𝑑𝑑𝑑𝑑
𝑑𝑑 2 𝜃𝜃 𝑑𝑑𝜃𝜃
Answer: + + 𝜃𝜃 = 6 sin 2𝑡𝑡
𝑑𝑑𝑡𝑡 2 𝑑𝑑𝑑𝑑
SA1.26 Linearize the differential equation assuming small 𝜃𝜃
𝑑𝑑 2 𝜃𝜃 𝑑𝑑𝜃𝜃 1
+ 4𝑡𝑡 cos 𝜃𝜃 + 16 sin � 𝜃𝜃� cos(3𝜃𝜃) = 6 sin 2𝑡𝑡
𝑑𝑑𝑡𝑡 2 𝑑𝑑𝑑𝑑 2
𝑑𝑑 2 𝜃𝜃 𝑑𝑑𝜃𝜃
Answer: + 4𝑡𝑡 + 8𝜃𝜃 = 6 sin 2𝑡𝑡
𝑑𝑑𝑡𝑡 2 𝑑𝑑𝑑𝑑
SA1.27 Fill in the blanks from the choices in parentheses regarding the differential equation
𝑑𝑑 2 𝜃𝜃 𝑑𝑑𝜃𝜃
2
+ 4𝑡𝑡 + 12𝜃𝜃 = 6 sin 2𝑡𝑡
𝑑𝑑𝑡𝑡 𝑑𝑑𝑑𝑑
Answer: The equation is an ordinary differential equation. It has an independent variable of t and a
dependent variable of 𝜃𝜃. It is a linear, nonhomogeneous differential equation with variable coefficients.
SA1.28 Fill in the blanks from the choices in parentheses regarding the differential equation
𝑑𝑑 2 𝜃𝜃 𝑑𝑑𝜃𝜃
2
+ 4 cos 𝜃𝜃 + 12 sin 𝜃𝜃 = 0
𝑑𝑑𝑡𝑡 𝑑𝑑𝑑𝑑
Answer: The equation is an ordinary differential equation. It has an independent variable of 𝑡𝑡 and a
dependent variable of 𝜃𝜃 It is a linear, homogeneous differential equation with constant coefficients.
PROBLEMS
1.1 Newton’s law of cooling is used to calculate the rate at which heat is transferred by convection to a
solid body at a temperature T from a surrounding fluid at a temperature T∞. The formula is
𝑄𝑄̇ = ℎ𝐴𝐴(𝑇𝑇 − 𝑇𝑇∞ ) (𝑎𝑎)
where Q˙ is the rate of heat transfer (rate of energy transfer), A is the area over which heat is transferred
by convection, and h is the heat transfer coefficient (also called the film coefficient). Use Equation (a) to
determine the basic dimensions of h. Suggest appropriate units for h using the English system and the SI
system.
© 2025 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible
website, in whole or in part.
,Solution
Equation (a) of the problem statement is used to solve for ℎ as
𝑄𝑄̇
ℎ= (𝑏𝑏)
𝐴𝐴(𝑇𝑇 − 𝑇𝑇∞ )
The Principle of Dimensional Homogeneity is used to determine the dimensions of the heat transfer
coefficient. Using the F-L-T system, the dimensions of the quantities in Equation (b) are
F∙L
�𝑄𝑄̇� = � � (𝑐𝑐)
T
2
[𝐴𝐴] = [L ] (𝑑𝑑)
[𝑇𝑇 − 𝑇𝑇∞ ] = [Θ] (𝑒𝑒)
From Equations (b)-(e)the dimensions of the heat transfer coefficient are
F∙L F
[ℎ] = � 2
�=� � (𝑓𝑓)
T∙Θ∙L T∙Θ∙L
N
Possible units for the heat transfer coefficient using the SI system are while possible units using the
m∙s∙K
lb
English system are .
ft∙s∙R
1.2 The Reynolds number is used as a measure of the ratio of inertia forces to the friction forces in the
flow of a fluid in a circular pipe. The Reynolds number (Re) is defined as
𝜌𝜌𝜌𝜌𝜌𝜌
𝑅𝑅𝑅𝑅 = (𝑎𝑎)
𝜇𝜇
where 𝜌𝜌 is the mass density of the fluid, V is the average velocity of the flow, D is the diameter of the
pipe, and μ is the dynamic viscosity of the fluid. Show that the Reynolds number is dimensionless.
Solution
The dimensions on the quantities on the right-hand side of Equation (a) are obtained using Table 1.2 as
M
[𝜌𝜌] = � 3 � (𝑏𝑏)
L
L
[𝑉𝑉 ] = � � (𝑐𝑐)
T
[𝐷𝐷] = [L] (𝑑𝑑)
M
[𝜇𝜇] = � � (𝑒𝑒)
L∙T
Substituting Equations (b)-(e) into Equation (a) leads to
M L
L3∙T∙L
[Re] = � � = [1] (𝑓𝑓)
M
L∙T
Equation (f) shows that the Reynolds number is dimensionless.
𝑑𝑑𝑑𝑑
1.3 The relationship between voltage and current in a capacitor is 𝑖𝑖 = 𝐶𝐶 . Use this relation to determine
𝑑𝑑𝑑𝑑
the basic dimensions of the capacitance C.
Solution
The capacitance of a capacitor is defined by
𝑖𝑖
𝐶𝐶 = (𝑎𝑎)
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
The dimension of 𝑖𝑖 is that of electric current, which is a basic dimension. The dimensions of electric
potential are obtained from Table 1.2 as
© 2025 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible
website, in whole or in part.
, F∙L
[𝑣𝑣 ] = � � (𝑏𝑏)
i∙T
Thus the dimensions of time rate of change of electric potential are
𝑑𝑑𝑑𝑑 F∙L
� �=� � (𝑐𝑐)
𝑑𝑑𝑑𝑑 i ∙ T2
Use of Equation (c) in Equation (a) leads to
i i2 ∙ T 2
[𝐶𝐶 ] = � �=� � (𝑑𝑑)
F∙L F∙L
i ∙ T2
𝑘𝑘
1.4 The natural frequency 𝜔𝜔𝑛𝑛 of a mechanical system of mass m and stiffness k is calculated by 𝜔𝜔𝑛𝑛 = � .
𝑚𝑚
(a) What are the dimensions of𝜔𝜔𝑛𝑛 ? (b) A mass-spring system of mass 100 g has a natural frequency of 20
Hz. What is the stiffness of the system?
Solution
M
(a)The dimension of spring stiffness in the M-L-T system is � 2 �. Thence the dimensions of natural
T
frequency are
M 1/2
2 1
[𝜔𝜔𝑛𝑛 ] = �� T � � = � � (𝑎𝑎)
M 𝑇𝑇
(b) Converting Hz to r/s
cycles cycles 2π r r
𝜔𝜔𝑛𝑛 = 20 Hz = 20 = �20 �� � = 125.7 (𝑏𝑏)
s s s s
The stiffness is calculated form the natural frequency as
r 2 N
𝑘𝑘 = 𝑚𝑚𝜔𝜔𝑛𝑛2 = (0.1 kg) �125.7 � = 1.58 × 103 (𝑐𝑐)
s m
1.5 Carbon nanotubes are new materials that consist of carbon atoms and are of significant interest
because of their good conductivity properties, light weight, and high strength. (a) The density of a carbon
nanotube is 1300 kg/m3 . A carbon nanotube is a tube with a diameter of 1 carbon atom, d = 0.68 nm.
Determine the mass of a carbon nanotube whose length is 80 nm. (b) The elastic modulus of a carbon
nanotube is 𝐸𝐸 = 1.1 TPa. What is its elastic modulus in pounds per square inch? (c) If modeled as a fixed-
fixed beam, the fundamental frequency of the nanotube is
𝐸𝐸𝐸𝐸
𝜔𝜔 = 22.37�
𝜌𝜌𝜌𝜌𝐿𝐿4
𝜋𝜋
where 𝐼𝐼 = 𝑟𝑟 4 is the cross-sectional moment of inertia of the beam and A is its cross-sectional area.
4
Calculate the fundamental frequency in hertz of a nanotube of length 80 nm.
Solution
(a) The mass of the nanotube is
kg
𝑚𝑚 = 𝜌𝜌𝜌𝜌𝜌𝜌 = 𝜌𝜌(𝜋𝜋𝑟𝑟 2 )𝐿𝐿 = �1300 3 � π(0.34 × 10−9 m)2 (80 × 10−9 m)
m
= 3.78 × 10−23 kg (a)
(b) Conversion between TPa and psi leads to
N lb 1 m 2 1 ft 2 lb
𝐸𝐸 = 1.1 𝑇𝑇𝑇𝑇𝑇𝑇 = �1.1 × 1012 2 � �0.225 � � � � � = 1.60 × 108 2 (𝑏𝑏)
m N 3.28 ft 12 in in
© 2025 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible
website, in whole or in part.
,(c) Calculation of the fundamental frequency yields
N 𝜋𝜋
𝐸𝐸𝐸𝐸 �1.1 × 1012 2 � (0.34 × 10−9 m)4
m 4
𝜔𝜔 = 22.37� = 22.37�
𝜌𝜌𝜌𝜌𝐿𝐿4 kg
�1300 3 � π(0.34 × 10−9 m)2 (80 × 10−9 m)4
m
10
r
= 1.74 × 10 (𝑐𝑐)
s
Converting to Hz gives
r 1 cycle
𝜔𝜔 = �1.74 × 1010 � � � = 2.78 × 109 Hz (𝑑𝑑)
s 2π r
1.6 During 24 hours of operation, a motor expends 900 kW∙hr of energy. Noting that a horsepower (hp) is
a unit of power such that 1 hp= 550 ft∙lb/s, determine the power delivered by the motor in horsepower.
Solution
The power of the motor is
900 kW ∙ hr N ∙ m 0.225 lb
𝑃𝑃 = = 37.5 kW = 37.5 × 103 W = �37.5 × 103 �� �
24 hr s N
ft ∙ lb
= 2.77 × 104 (𝑎𝑎)
s
Conversion to hp yields
ft ∙ lb 1 hp
P = 2.77 × 104 � � = 50.3 hp (𝑏𝑏)
s ft ∙ lb
550
s
1.7 The density of water is 1.94 slugs/ft 3 . Express the density of water in kilograms per cubic meter
(kg/m3 ).
Solution
The conversion of density from English units to SI units is
slugs 1 kg 3.28 ft 3 kg
𝜌𝜌 = �1.94 3
� � � � � = 1000 3 (𝑎𝑎)
ft 0.00685 slugs m m
1.8 A train is traveling at a speed of 180 km/hr and is decelerating at 6 m/s 2 . Assuming uniform
deceleration, how far will the train travel, in miles, before it stops?
Solution
The constant acceleration of the train is
m
𝑎𝑎 = −6 2 (𝑎𝑎)
s
The velocity is obtained using Equation (a) as
𝑣𝑣(𝑡𝑡) = −6𝑡𝑡 + 𝐶𝐶 (𝑏𝑏)
The constant of integration is evaluated by requiring
km 1000 m 1 hr m
𝑣𝑣(0) = �180 �� �� � = 50 (𝑐𝑐)
hr km 3600 s s
Substituting Equation (c) inti Equation (b) leads to
𝑣𝑣 (𝑡𝑡) = −6𝑡𝑡 + 50 (𝑑𝑑)
The train stops when the velocity is zero
0 = −6𝑡𝑡 + 50 ⟹ 𝑡𝑡 = 8.33 s (e)
The distance traveled before the train stops is obtained by integrating the velocity with respect to time and
evaluating it at 8.33 s
𝑥𝑥 (𝑡𝑡) = −3𝑡𝑡 2 + 50𝑡𝑡 (𝑓𝑓)
The distance traveled before the train stops is
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website, in whole or in part.
, 𝑥𝑥 (8.33) = −3(8.33)2 + 50(8.33) = 208.3 m (g)
1.9 The differential equation governing the angular velocity v of a shaft of mass moment of inertia J,
acted on by a torque T and attached to bearings of torsional damping coefficient 𝑐𝑐𝑡𝑡 , is
𝑑𝑑𝑑𝑑
𝐽𝐽 + 𝑐𝑐𝑡𝑡 𝜔𝜔 = 𝑇𝑇 (𝑎𝑎)
𝑑𝑑𝑑𝑑
Use Equation (a) to determine the appropriate dimensions of 𝑐𝑐𝑡𝑡 .
Solution
The principle of dimensional homogeneity states that every term in an equation must have the same
dimensions. In this case, it is the dimensions of 𝑇𝑇, or torque, which are [F ∙ L]. The dimensions of angular
velocity are
1
� �. Thus, the dimension of 𝑐𝑐𝑡𝑡 are
T
F∙L
[𝑐𝑐𝑡𝑡 ] = � � = [F ∙ L ∙ T] (𝑏𝑏)
1
T
1.10 In an armature-controlled dc servomotor, the torque applied to the armature due to a magnetic
coupling field is given by
𝑇𝑇 = 𝐾𝐾𝑎𝑎 𝑖𝑖𝑎𝑎 𝑖𝑖𝑓𝑓 (𝑎𝑎)
and the back electromotive force (voltage) generated as the shaft rotates through the magnetic field is
𝑣𝑣 = 𝐾𝐾𝑓𝑓 𝑖𝑖𝑓𝑓 𝜔𝜔 (𝑏𝑏)
where 𝑖𝑖𝑎𝑎 is the current in the armature circuit, 𝑖𝑖𝑓𝑓 is the current in the field circuit, and 𝜔𝜔 is the angular
velocity of the shaft. Show that the constants𝐾𝐾𝑎𝑎 and 𝐾𝐾𝑓𝑓 have the same dimensions. Referring to Table 1.2,
what physical quantity has the same dimensions as these constants?
Solution
Equation (a) is rearranged to
𝑇𝑇
𝐾𝐾𝑎𝑎 = (𝑐𝑐 )
𝑖𝑖𝑎𝑎 𝑖𝑖𝑓𝑓
The dimensions of 𝐾𝐾𝑎𝑎 are
F∙L
[𝐾𝐾𝑎𝑎 ] = � 2 � (𝑑𝑑)
𝑖𝑖
Equation (b) is rearranged as
𝑣𝑣
𝐾𝐾𝑓𝑓 = (𝑒𝑒)
𝑖𝑖𝑓𝑓 𝜔𝜔
The dimensions of 𝐾𝐾𝑓𝑓 are
F∙L
F∙L
�𝐾𝐾𝑓𝑓 � = � 𝑖𝑖 ∙ 𝑇𝑇 � = � 2 � (𝑓𝑓)
1 𝑖𝑖
𝑖𝑖 ∙
𝑡𝑡
It is clear from Equations (d) and (f) that the dimensions are the same. The quantity of inductance has the
same dimensions as 𝐾𝐾𝑎𝑎 and 𝐾𝐾𝑓𝑓 .
1.11 The rate of heat transfer by radiation from one body to another is given by
𝑄𝑄̇ = 𝜎𝜎𝜎𝜎𝜎𝜎�𝑇𝑇 4 − 𝑇𝑇𝑏𝑏4 � (𝑎𝑎)
where A is the area of heat transfer, 𝜖𝜖 is the dimensionless emissivity of the body, T is the temperature of
the body, Tb is the temperature of the radiating body, and the Stefan-Boltzmann constant is
Btu W
𝜎𝜎 = 1.73 × 10−7 2 4
= 5.67 × 10−8 2 4 (𝑏𝑏)
ft ∙ hr ∙ R m ∙K
© 2025 Cengage. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible
website, in whole or in part.
, (a) What are the basic dimensions of 𝑄𝑄̇ ? (b) The differential equation governing the transient temperature
in a body due to radiation heat transfer is
𝑑𝑑𝑑𝑑
𝜌𝜌𝜌𝜌 + 𝜎𝜎𝜎𝜎𝜎𝜎�𝑇𝑇 4 − 𝑇𝑇𝑏𝑏4 � = 0 (𝑐𝑐)
𝑑𝑑𝑑𝑑
where 𝜌𝜌 is the mass density of the body and c is its specific heat. The body is in a steady state with a
uniform temperature defined by a temperature Ts when the temperature of the radiating body is suddenly
changed from Tbs to Tb1. Let T1 be perturbation in temperature from the steady state such that the
temperature T is
𝑇𝑇 = 𝑇𝑇𝑠𝑠 + 𝑇𝑇1 (𝑡𝑡) (𝑑𝑑)
Substitute Equation (d) into Equation (c), use the binomial theorem to expand the nonlinear term, and
derive a linearized differential equation to solve for T1(t).
Solution
(a)The dimensions of 𝑄𝑄̇ are determined from Equation (a)
F∙L F∙L
�𝑄𝑄̇� = � 2 � [ 𝐿𝐿2 ][ 4 ]
Θ = � � (𝑒𝑒)
𝐿𝐿 ∙ 𝑇𝑇 ∙ Θ4 𝑇𝑇
(b) Substitution of Equation (d) into Equation (c) leads to
𝑑𝑑
𝜌𝜌𝜌𝜌 (𝑇𝑇𝑠𝑠 + 𝑇𝑇1 ) + 𝜎𝜎𝜎𝜎𝜎𝜎 [(𝑇𝑇𝑠𝑠 + 𝑇𝑇1 )4 − (𝑇𝑇𝑏𝑏𝑏𝑏 + 𝑇𝑇𝑏𝑏1 )4 ] = 0 (𝑓𝑓)
𝑑𝑑𝑑𝑑
Simplifying leads to
𝑑𝑑𝑇𝑇1 𝑇𝑇1 4 4
𝑇𝑇𝑏𝑏1 4
𝜌𝜌𝜌𝜌 + 𝜎𝜎𝜎𝜎𝜎𝜎 �𝑇𝑇𝑠𝑠4 �1 + � − 𝑇𝑇𝑏𝑏𝑏𝑏 �1 + � �=0 (𝑔𝑔)
𝑑𝑑𝑑𝑑 𝑇𝑇𝑠𝑠 𝑇𝑇𝑏𝑏𝑏𝑏
Applying the binomial expansion, keeping only through linear terms leads to
𝑑𝑑𝑇𝑇1 𝑇𝑇1 4
𝑇𝑇𝑏𝑏1
𝜌𝜌𝜌𝜌 + 𝜎𝜎𝜎𝜎𝜎𝜎 �𝑇𝑇𝑠𝑠4 �4 � − 𝑇𝑇𝑏𝑏𝑏𝑏 �4 �� = 0 (ℎ)
𝑑𝑑𝑑𝑑 𝑇𝑇𝑠𝑠 𝑇𝑇𝑏𝑏𝑏𝑏
Simplifying Equation (h) and rearranging leads to
𝑑𝑑𝑇𝑇1 3
𝜌𝜌𝜌𝜌 + 4𝜎𝜎𝜎𝜎𝜎𝜎𝑇𝑇𝑠𝑠3 𝑇𝑇1 = 4𝜎𝜎𝜎𝜎𝜎𝜎𝑇𝑇𝑏𝑏𝑏𝑏 𝑇𝑇𝑏𝑏1 (𝑖𝑖)
𝑑𝑑𝑑𝑑
1.12 A nonlinear differential equation governing the motion of a mechanical system is
1 1
𝑚𝑚𝐿𝐿2 𝜃𝜃̈ + 𝑐𝑐𝐿𝐿2 𝜃𝜃̇ + 𝑘𝑘𝐿𝐿2 sin 𝜃𝜃 cos 𝜃𝜃 = 0 (𝑎𝑎)
3 4
Assuming small 𝜃𝜃, derive a linearized differential equation for the system.
Solution
The differential equation is linearized by using the small angle assumption which implies sin 𝜃𝜃 ≈ 𝜃𝜃 and
cos 𝜃𝜃 ≈ 1. Using these approximations in the differential equation leads to the linearized approximation
as
1 1
𝑚𝑚𝐿𝐿2 𝜃𝜃̈ + 𝑐𝑐𝐿𝐿2 𝜃𝜃̇ + 𝑘𝑘𝐿𝐿2 𝜃𝜃 = 0 (𝑏𝑏)
3 4
1.13 The differential equation governing the motion of the system of Figure P1.13 is
1 𝐿𝐿
𝑚𝑚𝐿𝐿2 𝜃𝜃̈ + 𝐿𝐿𝑦𝑦̈ sin 𝜃𝜃 + 𝐿𝐿𝑥𝑥̈ cos 𝜃𝜃 + 𝑚𝑚𝑚𝑚 sin 𝜃𝜃 = 0 (𝑎𝑎)
3 2
where x(t) and y(t) are known functions of time. Derive a linearized equation governing the motion of the
system for small 𝜃𝜃 .
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website, in whole or in part.
