Chapter 4
Dynamics, Kinematics and Mechanical Vibrations
Bachelors Program in Mechnical Engineering
Fundamentals of Mechanical Engineering
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Bachelor
Dynamics, Kinematics and Mechanical Vibrations
Introduction
Mechanical engineering is concerned not only with whether a component can carry a load, but also with how it moves, accelerates, vibrates, and interacts with neighboring components. Kinematics describes motion without considering its causes. Dynamics relates motion to forces and moments. Mechanical vibration studies oscillatory motion about an equilibrium position and is essential for the safe design of machines, vehicles, structures, and rotating equipment.
This chapter develops particle and rigid-body analysis using force–acceleration, work–energy, and impulse–momentum methods. It then introduces mechanisms, ideal vibration models, resonance, damping, and balancing.
Learning Objectives
By the end of this chapter, the student should be able to:
Describe particle motion using displacement, velocity, and acceleration.
Apply Newton’s laws to particles and rigid bodies.
Select and use force–acceleration, work–energy, and impulse–momentum methods.
Distinguish translation, rotation, and general plane motion.
Explain the function and classification of mechanisms.
Determine natural frequency, damping behavior, and resonance conditions.
Explain the principles of static and dynamic balancing.
1. Kinematics of Particles
1.1 Position, displacement, velocity, and acceleration
The position of a particle is specified relative to a reference frame. For one-dimensional motion, position is represented by x(t). Displacement between two instants is
∆x = x₂ − x₁
where x₁ and x₂ are positions in metres. Displacement is a vector quantity and may be positive or negative according to the selected coordinate direction.
Average velocity and average acceleration are
vavg = ∆x/∆t
aavg = ∆v/∆t
Instantaneous velocity and acceleration are defined by
v = dx/dt
a = dv/dt = d²x/dt²
The SI units are m/s for velocity and m/s² for acceleration. These definitions apply to any differentiable motion.
For constant acceleration,
v = v₀ + at
x = x₀ + v₀t + ½at²
v² = v₀² + 2a(x − x₀)
Here v₀ and x₀ are the initial velocity and position, t is time, and a is constant acceleration. These equations must not be used when acceleration varies with time unless the variation is first integrated.
1.2 Curvilinear motion
For motion along a curved path, acceleration has tangential and normal components:
a = at et + an en
at = dv/dt
an = v²/ρ
where e_t and e_n are unit vectors tangent and normal to the path, and ρ is the local radius of curvature. The tangential component changes the speed; the normal component changes the direction of velocity. For circular motion of radius r, ρ = r.
For uniform circular motion, speed is constant, so a_t = 0 and
an = ω²r = v²/r
where ω is angular speed in rad/s.
2. Dynamics: Force–Acceleration Method
2.1 Newton’s second law
For a particle of mass m, the vector form of Newton’s second law is
∑F = ma
The resultant force and acceleration must be expressed in the same coordinate system. Mass is measured in kilograms, force in newtons, and acceleration in m/s². Weight is a force given by
W = mg
where g ≈ 9.81 m/s² near Earth’s surface.
A reliable solution begins with a free-body diagram. The diagram should isolate the body and show all external forces, including weight, normal reactions, friction, applied forces, and cable tensions. Internal forces are not included in a free-body diagram of the isolated body.
For planar motion, resolve Newton’s law into perpendicular axes:
∑Fx = max
∑Fy = may
The friction force is often modeled as
Fs ≤ μsN
for static contact, and
Fk = μkN
for sliding contact. Here N is the normal reaction and μ_s and μ_k are coefficients of static and kinetic friction. These are engineering idealizations; actual friction depends on materials, speed, lubrication, temperature, and surface condition.
Worked Example 1: Inclined plane
A 10 kg block slides down a 30° incline. The coefficient of kinetic friction is 0.20. Determine its acceleration, neglecting air resistance.
Choose the positive direction down the slope. The component of weight down the slope is mg sin 30°, while friction acts up the slope. The normal reaction is N = mg cos 30°.
mg sin 30° − μk mg cos 30° = ma
Cancel m:
a = g(sin 30° − μk cos 30°)
Substitution gives
a = 9.81(0.500 − 0.20 × 0.866) = 3.21 m/s²
The block accelerates down the slope at approximately 3.21 m/s². Notice that the mass cancels because both gravitational and friction forces are proportional to mass under this model.
3. Dynamics of Rigid Bodies
A rigid body is an ideal body in which the distance between any two particles remains constant. Real bodies deform, but the rigid-body approximation is suitable when deformation is small compared with the dimensions relevant to motion.
3.1 Translation
In pure translation, every line fixed in the body remains parallel to itself. The equations are
∑F = maG
where G is the centre of mass. For planar translation, the body has no angular acceleration and
∑MG = 0
when moments are taken about the centre of mass.
3.2 Rotation about a fixed axis
For a rigid body rotating about a fixed axis through O,
∑MO = IO α
where I_O is the mass moment of inertia about the axis and α is angular acceleration. The rotational kinematic relations for constant α are
ω = ω₀ + αt
θ = θ₀ + ω₀t + ½αt²
ω² = ω₀² + 2α(θ − θ₀)
Angular displacement is measured in radians. The corresponding point velocity and normal acceleration are
v = rω
an = rω²
and the tangential acceleration is a_t = rα.
The moment of inertia depends on mass distribution. For a slender rod of length L about its centre, perpendicular to the rod,
IG = 1/12 mL²
About an axis through one end parallel to that axis,
IO = IG + md²
where d is the distance between axes. This is the parallel-axis theorem.
3.3 General plane motion
General plane motion combines translation of the centre of mass and rotation about the centre of mass:
∑F = maG
∑MG = IGα
For two points A and B on the same rigid body,
aB = aA + α × rB/A − ω²rB/A
This vector relation is most conveniently applied using components or a carefully drawn rotating coordinate system. It assumes planar rigid-body motion and uses the cross-product interpretation of the angular terms.
4. Work–Energy and Impulse–Momentum Methods
4.1 Work and kinetic energy
The work of a force along a path is
U₁→₂ = ∫₁² F · dr
For a constant force parallel to displacement, U = Fs. Work is measured in joules. The kinetic energy of a particle is
T = ½mv²
The work–energy equation is
T₁ + U₁→₂ = T₂
For a rigid body undergoing translation and rotation,
T = ½mvG² + ½IGω²
This method is especially efficient when forces vary with position or when time is not required. It does not directly provide the time history unless combined with kinematics.
Potential energy for elevation and a linear spring is
Vg = mgh
Vs = ½kx²
where h is elevation, k is spring stiffness in N/m, and x is spring deformation in metres. For conservative systems,
T₁ + V₁ = T₂ + V₂
If nonconservative work is present, it must be included separately.
Worked Example 2: Spring-driven motion
A 2 kg collar is released from rest after compressing a spring by 0.10 m. The spring stiffness is 800 N/m. Neglect friction. Find the collar speed when the spring reaches its natural length.
Initially, the energy is spring potential; finally, it is kinetic energy:
½kx² = ½mv²
Therefore,
v = x√(k/m) = 0.10√(800/2) = 2.00 m/s
The answer is dimensionally consistent because k/m has units s⁻² and its square root has units s⁻¹.
4.2 Linear impulse and momentum
Linear momentum is
p = mv
The impulse–momentum equation is
∫t₁t₂ ∑F dt = mv₂ − mv₁
For a constant resultant force, this becomes F∆t = m(v₂ − v₁). Impulse is measured in N·s, equivalent to kg·m/s. This method is particularly valuable for impact, collision, and short-duration force events.
Angular momentum about a fixed point O is
HO = IOω
and the angular impulse–momentum equation is
∫t₁t₂ ∑MO dt = HO2 − HO1
These equations require consistent sign conventions and careful treatment of external impulses and moments.
5. Mechanisms and Machine Motion
A mechanism is a set of links connected by joints so that the motion of one link produces a constrained motion in another. A machine is a mechanism or group of mechanisms that transmits forces and performs useful work.
Important terms include:
Link: a resistant body or group of bodies forming a member of a mechanism.
Kinematic pair: two links joined so that relative motion is constrained.
Revolute pair: a pin joint permitting rotation.
Prismatic pair: a sliding joint permitting translation.
Ground link: the fixed reference link.
Input link: the link receiving motion or power.
Output link: the link delivering the desired motion.
Common mechanisms include the four-bar linkage, slider–crank mechanism, cam and follower, and gear train. A mechanism is analyzed through position, velocity, and acceleration relationships. In a slider–crank engine, rotation of the crank produces reciprocating motion of the piston. In a cam mechanism, a prescribed follower motion is generated by the cam profile.
The transmission-angle concept is important in four-bar mechanisms. A very small transmission angle can create large joint forces and poor force transmission. Mechanism design therefore considers motion range, singular positions, manufacturability, wear, lubrication, and guarding.
Slider-crank mechanism showing crank, connecting rod, and slider
Figure: Slider-crank mechanism showing crank, connecting rod, and slider
Figure: Principal components of a slider–crank mechanism.
6. Mechanical Vibrations
Vibration is oscillatory motion about an equilibrium position. Vibration can be useful, as in compactors and ultrasonic equipment, or harmful, as in turbines, pumps, bridges, and vehicle suspensions.
A vibration model is described by its degrees of freedom, the minimum number of independent coordinates required to specify its configuration. A single-degree-of-freedom (SDOF) system commonly contains a mass, spring, and damper.
6.1 Undamped free vibration
For a mass m connected to a linear spring of stiffness k, measured from static equilibrium,
mẍ + kx = 0
The undamped natural circular frequency is
ωn = √(k/m)
and the natural frequency in hertz is
fn = ωn/(2π)
The general motion is
x(t) = X cos(ωnt − φ)
where X is amplitude and φ is phase angle. The model assumes a linear spring, constant mass, no damping, and small enough motion that the properties remain constant.
6.2 Damped free vibration
With viscous damping c, the equation becomes
mẍ + cẋ + kx = 0
The critical damping coefficient is
cc = 2√(km)
and the damping ratio is
ζ = c/cc
For 0 < ζ < 1, the system is underdamped and oscillates with damped natural frequency
ωd = ωn√(1 − ζ²)
For ζ = 1, the system is critically damped and returns to equilibrium without oscillation in the shortest idealized time. For ζ > 1, it is overdamped and returns without oscillation more slowly. Real damping may arise from friction, fluid resistance, material hysteresis, joints, and electromagnetic effects; viscous damping is a useful approximation rather than a universal law.
6.3 Forced vibration and resonance
For harmonic excitation F₀ sin ωt,
mẍ + cẋ + kx = F₀ sin(ωt)
The steady-state amplitude is
X = (F₀/k) / √[(1 − r²)² + (2ζr)²]
where r = ω/ω_n is the frequency ratio. This expression assumes a linear SDOF system, harmonic forcing, and viscous damping.
When the forcing frequency approaches the natural frequency, amplitude can become large. This phenomenon is resonance. Damping limits the peak response, but it does not eliminate the need to avoid hazardous operating speeds. The phase difference between force and displacement also changes rapidly near resonance.
Engineering controls include changing mass or stiffness, adding damping, shifting operating speed, isolating the machine from its foundation, and using a tuned mass absorber. A vibration survey should measure amplitude, frequency, phase, and operating condition rather than relying on amplitude alone.
7. Balancing of Rotating Machinery
An unbalanced rotating mass produces a centrifugal force. For a mass m rotating at radius r with angular speed ω,
Fu = mrω²
This force increases with the square of speed. Static balancing requires the resultant rotating force to be zero. For a single-plane rotor, this means the vector sum of unbalance products mr is zero.
Dynamic balancing additionally requires the resultant couple to be zero. It is necessary when masses lie in different axial planes. Balancing machines determine correction mass magnitude and angular position. Good practice includes accurate mass measurement, controlled tolerances, secure attachment of correction weights, and verification at the intended operating speed.
8. Applied Case: Vibration in a Motor–Pump Assembly
Background
Consider an illustrative industrial case in which a motor drives a centrifugal pump through a flexible coupling. Operators report increasing vibration after maintenance.
Problem
The vibration is strongest at the rotational frequency and increases noticeably with speed. Possible causes include rotor imbalance, coupling misalignment, looseness, resonance, or bearing damage.
Analysis
A technician first records vibration amplitude and frequency in horizontal, vertical, and axial directions. A dominant once-per-revolution component suggests imbalance or misalignment, but frequency evidence alone is not conclusive. The machine is inspected for loose foundation bolts, soft foot, bent shafts, coupling condition, and bearing temperature. Alignment is checked using laser equipment, and phase measurements are compared between motor and pump.
Suppose the rotating speed is 1,800 rpm. The rotational frequency is
f = 1800/60 = 30 Hz
If the vibration spectrum shows a strong component at 30 Hz and its amplitude scales approximately with speed squared, imbalance is plausible because F_u = mrω². If a structural natural frequency is also near 30 Hz, resonance may amplify the response. A coast-down test can identify a critical speed by observing the amplitude peak as speed decreases.
Approach and solution
The foundation bolts are tightened, soft foot is corrected, shaft alignment is restored, and the rotor is dynamically balanced. A follow-up measurement confirms that the 30 Hz amplitude has fallen substantially and that no damaging bearing-frequency components are present.
Outcome and learning points
The case illustrates that vibration diagnosis should combine physical inspection, time-domain measurements, frequency spectra, phase information, and operating history. Replacing a bearing without identifying the excitation mechanism may provide only temporary improvement. Preventive maintenance should establish baseline vibration signatures and alarm limits appropriate to the machine and applicable standards.
9. Limitations, Safety, and Professional Practice
Analytical models are simplified representations. Rigid-body theory neglects deformation; Coulomb friction may not represent lubricated contacts; linear springs and viscous dampers may fail at large amplitudes; and SDOF vibration models may overlook important modes. Engineers must state assumptions and validate predictions with measurement or higher-fidelity analysis.
Rotating machinery requires guarding against ejected parts, entanglement, and stored energy. Lockout–tagout procedures must be followed before inspection or adjustment. Balancing and vibration tests should use calibrated instruments and appropriate operating procedures. Relevant machinery-safety, guarding, and vibration standards should be consulted for the specific industry and jurisdiction.
Computer-aided engineering, finite-element modal analysis, digital sensors, and machine-learning-based condition monitoring are increasingly useful. However, AI-based fault classification must be trained on representative data, checked for false alarms and missed detections, and interpreted by qualified engineers. Automated recommendations should not replace physical safety controls or engineering judgment.
Best Practices for Analysis
Define the system boundary and coordinate directions.
Draw a complete free-body diagram before writing equations.
Check units and signs at every stage.
Select the method suited to the question: force–acceleration for forces and accelerations, energy for speed and displacement, and impulse–momentum for short-duration events.
State initial conditions and model assumptions.
Distinguish rotational speed from frequency and use radians consistently in angular equations.
For vibration problems, identify mass, stiffness, damping, excitation frequency, and degrees of freedom.
Compare calculations with physical measurements and safety requirements.
Chapter Summary
Kinematics describes particle and rigid-body motion through position, velocity, acceleration, angular velocity, and angular acceleration. Dynamics relates motion to force and moment using Newton’s laws. The force–acceleration method is direct and provides detailed force information; work–energy efficiently relates speed and displacement; and impulse–momentum is suited to impacts and short-duration loading.
Rigid-body motion may be translational, rotational, or general plane motion. Mechanisms constrain motion through links and joints and are fundamental to engines, pumps, robotics, and manufacturing equipment. Mechanical vibration is modeled using mass, stiffness, and damping. Natural frequency describes free oscillation, damping controls decay and peak response, and resonance occurs when excitation approaches a natural frequency. Rotating imbalance creates forces proportional to mrω², making static and dynamic balancing essential for safe and reliable machinery.
Key Terms
Kinematics: Study of motion without considering its causes.
Dynamics: Study of motion in relation to forces and moments.
Displacement: Change in position of a particle.
Velocity: Time rate of change of position.
Acceleration: Time rate of change of velocity.
Free-body diagram: Diagram showing all external forces and moments on an isolated body.
Rigid body: Ideal body whose internal distances remain constant.
Mass moment of inertia: Measure of resistance to angular acceleration.
Work: Energy transferred by a force through displacement.
Kinetic energy: Energy associated with motion.
Impulse: Integral of force over a time interval.
Mechanism: Connected links that produce constrained relative motion.
Degree of freedom: Number of independent coordinates needed to describe motion.
Natural frequency: Frequency of free vibration of a system.
Damping: Energy dissipation that reduces vibration amplitude.
Resonance: Large response caused by excitation near a natural frequency.
Static balancing: Removal of resultant rotating force.
Dynamic balancing: Removal of both resultant rotating force and couple.
Critical speed: Operating speed associated with a significant vibration response.
Review Questions
5 Short Answer Questions
Distinguish kinematics from dynamics.
What information should appear on a free-body diagram?
State the conditions under which the constant-acceleration equations may be used.
Define natural frequency and damping ratio.
Why is dynamic balancing required for some rotors but not others?
5 Long Answer Questions
Explain particle motion in Cartesian and tangential–normal coordinates, including the physical meaning of each acceleration component.
Compare the force–acceleration, work–energy, and impulse–momentum methods, including the types of problems for which each is most efficient.
Derive the equation of motion for an undamped mass–spring system and interpret its solution.
Explain the operation and applications of a slider–crank mechanism.
Discuss resonance, its risks, and engineering methods for reducing resonant response.
3 Analytical / Application Questions
A 5 kg block is pulled up a 20° incline by a 40 N force parallel to the incline. If μ_k = 0.15, calculate its acceleration. Use g = 9.81 m/s².
A rotor has an unbalance mass of 0.020 kg at a radius of 0.10 m and rotates at 3,000 rpm. Calculate the centrifugal force and explain how it changes if speed doubles.
A machine is modeled as an SDOF system with m = 50 kg and k = 20,000 N/m. Calculate its natural frequency in rad/s and hertz, then explain why an operating speed near that frequency may be unsafe.
Multiple Choice Questions with Answers
Question 1
Which equation defines instantaneous acceleration in one-dimensional motion?
A. a = dx/dt
B. a = dv/dt
C. a = x/t
D. a = v²/t
Correct Answer: B. a = dv/dt
Question 2
A 10 kg block is acted on by a resultant force of 50 N. What is its acceleration?
A. 0.2 m/s²
B. 5 m/s²
C. 50 m/s²
D. 500 m/s²
Correct Answer: B. 5 m/s²
Question 3
Which method is generally most convenient for finding speed as a function of displacement when time is not required?
A. Work–energy
B. Impulse–momentum
C. Dimensional analysis
D. Static balancing
Correct Answer: A. Work–energy
Question 4
For a rigid body rotating about a fixed axis, the governing moment equation is:
A. ∑M = ma
B. ∑M = Iα
C. ∑M = mv
D. ∑M = kx
Correct Answer: B. ∑M = Iα
Question 5
The undamped natural circular frequency of a mass–spring system is:
A. √(m/k)
B. k/m
C. √(k/m)
D. 2π√(k/m)
Correct Answer: C. √(k/m)
Question 6
What is the main characteristic of resonance?
A. Zero displacement at all frequencies
B. Large response near a natural frequency
C. Constant acceleration independent of frequency
D. Removal of all damping
Correct Answer: B. Large response near a natural frequency
Question 7
For a viscously damped system, critical damping is represented by which condition?
A. ζ = 0
B. ζ < 0
C. ζ = 1
D. ζ → ∞
Correct Answer: C. ζ = 1
Question 8
A rotor’s centrifugal unbalance force is proportional to:
A. mr/ω²
B. mrω²
C. m/(rω)
D. r/(mω²)
Correct Answer: B. mrω²
Question 9
Which joint permits relative rotation between two links?
A. Prismatic pair
B. Revolute pair
C. Welded joint
D. Fixed support only
Correct Answer: B. Revolute pair
Question 10
A 2 kg mass is attached to a spring of stiffness 800 N/m. Its natural circular frequency is approximately:
A. 10 rad/s
B. 20 rad/s
C. 40 rad/s
D. 800 rad/s
Correct Answer: C. 40 rad/s
Fundamentals of Mechanical Engineering
Chapter 4
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