Solutions Manual for Systems Dynamics and Controls 1st Edition by S. Graham Kelly

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?

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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.