Engineering Dynamics

Introduction and Context

Engineering dynamics studies motion and the forces, moments, energy, and momentum that produce it. In mechatronics, dynamics connects mechanical structure to actuators, sensors, controllers, and algorithms. A robot arm accelerates because motor torque overcomes inertia and resistance; a conveyor starts and stops according to force and impulse; a rotating spindle vibrates because mass distribution is not balanced.

Dynamics is commonly divided into:

• Kinematics: description of motion without considering its causes.
• Kinetics: relationship between motion and applied forces or moments.

Three complementary methods are used throughout engineering:

1. Force–acceleration: directly relates resultant force or moment to acceleration.
2. Work–energy: relates forces and displacement to changes in kinetic energy.
3. Impulse–momentum: relates forces acting over time to changes in momentum.

The best method depends on the available information. Force–acceleration is usually effective when acceleration is required at an instant. Work–energy is useful when displacement and speed are known but time is not. Impulse–momentum is especially useful for impacts, short-duration actuator forces, and changes in velocity.

Learning Objectives

After studying this chapter, you should be able to:

1. Describe particle and rigid-body motion using appropriate coordinates.
2. Relate position, velocity, and acceleration in Cartesian, normal–tangential, and polar descriptions.
3. Apply Newton’s second law to translating and rotating systems.
4. Use work–energy and impulse–momentum methods for particles and rigid bodies.
5. Analyze planar rigid-body motion and rolling.
6. Explain balancing, rotational inertia, transient response, and their relevance to actuators and robots.
7. Select assumptions, coordinate systems, and solution methods responsibly.

1. Kinematics of Particles

A particle is an idealization in which the dimensions and rotation of a body are unimportant for the motion being analyzed. Its position is represented by a vector r(t). Velocity and acceleration are defined by differentiation:

Here, r is position in metres, v is velocity in metres per second, and a is acceleration in metres per second squared. These definitions apply to any differentiable particle trajectory.

Cartesian coordinates

For planar motion,

The unit vectors i and j are fixed Cartesian directions. Thus, the component equations can be integrated independently when acceleration is known. For constant acceleration,

The last equation is valid for constant vector acceleration and is often used in one-dimensional motion. Symbols r₀ and v₀ denote initial position and velocity, and t is time.

Curvilinear coordinates

For motion along a known curved path, normal–tangential coordinates are convenient:

Here, t and n are unit vectors tangent and normal to the path, v is speed, and ρ is the local radius of curvature. The normal acceleration always points toward the centre of curvature. Tangential acceleration changes speed; normal acceleration changes direction.

For planar polar coordinates,

The terms ṙ and r̈ are radial velocity and acceleration; θ̇ and θ̈ are angular velocity and angular acceleration. The term 2ṙθ̇ is sometimes called the Coriolis component. These equations assume planar motion and orthonormal polar unit vectors.

Angular motion

For rotation about a fixed axis,

where ω is angular velocity in radians per second and α is angular acceleration in radians per second squared. Since the radian is dimensionless, ω has units s⁻¹, although rad/s is customary.

For a point at radius r on a rotating body,

The tangential and normal accelerations are perpendicular. The formulas apply to rotation about a fixed axis and a point fixed in the rotating body.

Figure: Tangential and inward normal acceleration for a point on a rotating disk.

2. Kinetics of Particles

Newton’s second law

The fundamental equation for a particle of constant mass is

where ΣF is the vector resultant of external forces, m is mass in kilograms, and a is acceleration. In Cartesian components,

A free-body diagram must show the isolated body, all relevant external forces, dimensions needed for moments, and the chosen axes. Internal forces do not appear when the complete body is isolated.

Common forces include weight W = mg, normal reaction, friction, spring force, damping force, and actuator force. The gravitational acceleration g is approximately 9.81 m/s² near Earth’s surface. Kinetic friction is often modelled as F_k = μ_kN, but this is an approximation and is unreliable when friction depends strongly on speed, temperature, lubrication, or direction.

Equation of motion in normal–tangential form

For a particle moving on a path,

The normal equation describes the force required to curve the path; it does not imply that a separate “centripetal force” exists. The inward resultant of real forces produces normal acceleration.

D’Alembert’s principle

An inertial force F_i = −ma can be introduced to form a dynamic equilibrium equation:

This is a mathematical rearrangement of Newton’s law, not an additional physical force. It is useful in some graphical and structural formulations but must not be confused with an externally applied force.

3. Work and Energy

The differential work done by a force is

For a variable force, total work is

Work has units of joules. A force perpendicular to displacement does no instantaneous work. The kinetic energy of a particle is

The work–energy theorem is

or equivalently, the net work by all external forces equals the change in kinetic energy. It applies to a particle between two states, regardless of whether the forces are conservative.

For a conservative force, potential energy V can be defined such that

Therefore, if only conservative forces do work,

Near Earth’s surface, gravitational potential energy is V_g = mgy. For a linear spring,

where k is stiffness in N/m and x is deformation from the unstretched position. With nonconservative work such as friction or motor losses,

where U_nc is work by nonconservative forces.

Worked example 1: actuator-driven translation

A 12 kg carriage moves on a horizontal guide. A motor applies a constant 90 N force over 2.0 m. Friction opposing motion is 18 N. The carriage starts at 1.5 m/s. Find its final speed.

The net work is

U = (90 − 18)(2.0) = 144 J.

Initial kinetic energy is T₁ = ½(12)(1.5²) = 13.5 J. Hence,

T₂ = 13.5 + 144 = 157.5 J.

Thus,

v₂ = √(2T₂/m) = √(315/12) = 5.12 m/s.

The result assumes constant force, negligible vertical motion, and friction represented adequately by a constant opposing force.

4. Impulse and Momentum

Linear momentum is

For constant mass, Newton’s law gives

Integrating over time gives the impulse–momentum equation:

Impulse has units N·s, equivalent to kg·m/s. The equation is valuable when force varies rapidly, as in impact or a solenoid stroke. If external impulse is negligible in a direction, momentum in that direction is conserved.

For two colliding bodies, the coefficient of restitution is

For one-dimensional impact,

with signs chosen consistently along the line of impact. The value is normally between 0 and 1 for passive impacts: e = 0 represents perfectly inelastic contact and e = 1 perfectly elastic contact. Real impacts also dissipate energy through deformation, sound, and heat.

5. Planar Rigid-Body Dynamics

A rigid body is an idealization in which distances between particles remain constant. It may translate, rotate, or undergo general plane motion. For a body of mass m and mass-centre acceleration a_G,

The rotational equation about the mass centre is

where I_G is the mass moment of inertia about the axis through the centre of mass. For planar motion, these equations are independent and together provide three scalar equations: two force equations and one moment equation.

The moment of a force about point O is

In two dimensions, its magnitude is M_O = Fd, where d is the perpendicular distance from O to the force line of action. Sign conventions must be stated, for example counterclockwise positive.

For rotation about a fixed stationary axis O,

The parallel-axis theorem is

where d is the distance between parallel axes through G and O. Mass moment of inertia has units kg·m². Some useful values are:

Rigid-body work–energy

The kinetic energy of a planar rigid body is

This separates translation of the mass centre from rotation about the mass centre. For fixed-axis rotation, it reduces to

The rotational work of a torque is

For constant torque, U = M(θ₂ − θ₁). Mechanical power transmitted by a torque is

with power in watts when torque is in N·m and angular velocity in rad/s.

Rolling without slipping

For a wheel of radius R rolling without slipping,

These are kinematic constraints, not universal laws. They require zero relative velocity at the contact point. If slipping occurs, friction and energy relations must be modelled differently.

Worked example 2: motor acceleration of a drum

A drum has I = 0.080 kg·m². A motor applies 4.0 N·m, while a resisting torque of 0.80 N·m acts opposite rotation. Find angular acceleration.

The net torque is 3.20 N·m. Therefore,

α = ΣM/I = 3.20/0.080 = 40 rad/s².

If the torque remains constant for 0.50 s from rest, ω = αt = 20 rad/s. The ideal model neglects motor electrical dynamics, bearing losses beyond the specified torque, shaft flexibility, and changes in inertia.

6. Balancing and Rotating Systems

Unbalance occurs when the mass centre does not lie on the intended rotation axis or when rotating masses generate unequal dynamic effects. A rotating mass m at eccentricity e produces an inertial force of magnitude

This force increases with the square of speed. Static balancing requires the resultant centrifugal force to be zero. Dynamic balancing additionally requires the resultant moment of centrifugal forces to be zero; it generally requires correction in two planes.

For a reciprocating mechanism, inertia forces arise because piston acceleration is not purely sinusoidal at finite connecting-rod length. Counterweights can reduce vibration but may introduce bearing loads and transmitted forces. Balancing is therefore a design compromise involving vibration, strength, mass, cost, and available space.

A practical balancing procedure is:

1. Identify rotating and reciprocating masses.
2. Establish the shaft axis and rotational speed range.
3. Determine eccentricity and phase angles.
4. Resolve unbalance into force and couple components.
5. Specify correction masses and planes.
6. Verify residual vibration experimentally across the operating range.

7. Transient Mechanical Response

Actuators and mechanisms rarely change speed instantaneously. A simple rotational model is

where J is equivalent rotational inertia, b is viscous damping coefficient, T_L is load torque, and T_m is motor torque. This equation assumes a rigid shaft, constant inertia, viscous damping, and a lumped load.

For constant motor torque and zero load torque, the undamped model gives constant angular acceleration. With viscous damping and constant T_m, the steady speed is

and the time constant is

The first-order response from rest is

This simplified response is useful for preliminary sizing, but real servomotors also involve winding resistance, inductance, back electromotive force, Coulomb friction, saturation, gearbox compliance, sensor delay, and feedback control.

For a translational mass–spring–damper system,

where m is mass, c damping coefficient, k stiffness, and F(t) applied force. The undamped natural frequency is

and the damping ratio is

These parameters help predict overshoot, settling, resonance, and vibration sensitivity. A mechatronic controller should not be tuned from a rigid-body model alone if flexible modes lie within the control bandwidth.

8. Applied Case: Pick-and-Place Robot Axis

Background

Consider an illustrative single-axis robot joint driving a link through a reduction gearbox. The link and payload have an equivalent inertia of 0.18 kg·m² at the motor shaft. The motor can provide 2.4 N·m peak torque. Reflected friction and load torque together are estimated as 0.30 N·m.

Problem

Estimate peak acceleration and determine whether a commanded change from 0 to 30 rad/s in 0.20 s is feasible under this simplified model.

Analysis

The available net torque is

T_net = 2.4 − 0.30 = 2.10 N·m.

Using the rotational equation,

α = T_net/J = 2.10/0.18 = 11.67 rad/s².

The time required to reach 30 rad/s from rest is

t = Δω/α = 30/11.67 = 2.57 s.

Thus the command is not feasible with the stated peak torque and inertia. The required net acceleration is 30/0.20 = 150 rad/s², requiring net torque

T_req = Jα = 0.18(150) = 27 N·m.

Including the estimated resisting torque, the motor would need approximately 27.3 N·m, far above its 2.4 N·m capability.

Engineering response

Possible changes include reducing reflected inertia, lowering payload, increasing gear ratio while checking reflected motor inertia and efficiency, using a higher-torque actuator, relaxing the motion profile, or using a trapezoidal/ S-curve trajectory that limits jerk. A controller cannot create torque that the actuator and power electronics cannot supply.

Outcome and learning points

The simple dynamic model quickly identifies an infeasible command before hardware testing. However, final validation should include gearbox efficiency, current and voltage limits, thermal duty cycle, structural flexibility, backlash, encoder resolution, emergency stopping, and worst-case payload. Safety factors and independent motion limits are essential because an underestimated load can cause collision, overheating, or loss of control.

9. Modelling Practice, Limitations, and Professional Responsibility

A credible dynamics model begins with a clearly stated system boundary. The analyst should identify inputs, outputs, degrees of freedom, constraints, initial conditions, and neglected effects. Units must be checked at every stage: kg·m/s² is a newton, N·m is a torque unit, and kg·m² is a rotational inertia unit.

Common modelling errors include:

• treating a rigid body as a particle when rotation affects energy or reactions;
• omitting reaction forces or the torque of an actuator;
• using v = Rω when rolling slip exists;
• mixing degrees and radians in angular calculations;
• using static friction as a fixed force in all conditions;
• ignoring reflected inertia and gearbox efficiency;
• assuming a motor can deliver peak torque continuously;
• applying conservation of energy when dissipative work is significant;
• selecting a time step or sensor bandwidth that cannot resolve impacts or vibration.

Computer algebra, multibody simulation, finite-element models, and digital twins are valuable for verification, but they do not replace free-body diagrams or physical reasoning. AI-assisted design tools can suggest trajectories or identify anomalous vibration, yet engineers remain responsible for validating training data, model assumptions, uncertainty, traceability, and safety-critical decisions. Experimental correlation should include parameter identification and uncertainty bounds rather than relying on a visually plausible simulation.

In industrial practice, guarding, emergency stops, torque limits, risk assessment, and relevant machinery and electrical safety standards must be addressed. A dynamically correct model is not by itself a safe machine.

Chapter Summary

Dynamics provides the mathematical foundation for predicting motion in mechatronic mechanisms. Kinematics describes position, velocity, and acceleration using Cartesian, normal–tangential, polar, and angular coordinates. Kinetics relates motion to causes through ΣF = ma and ΣM = Iα. Work–energy connects force and displacement to changes in kinetic energy, while impulse–momentum connects force over time to changes in momentum.

Planar rigid-body analysis separates translation of the centre of mass from rotation about it. Mass moments of inertia, the parallel-axis theorem, rolling constraints, and torque–power relations are central to actuator and robot design. Rotating unbalance produces forces proportional to ω², making balancing increasingly important at high speed. Lumped transient models reveal how inertia, damping, stiffness, load torque, and actuator limits influence acceleration, resonance, and settling.

Good engineering practice requires explicit assumptions, dimensional consistency, free-body diagrams, model validation, and recognition of actuator, structural, sensing, and safety limitations.

Key Terms

Particle: Idealized body whose dimensions and rotation are neglected.

Rigid body: Body whose interparticle distances remain constant.

Kinematics: Description of motion without its causes.

Kinetics: Study of forces and moments producing motion.

Velocity: Time rate of change of position.

Acceleration: Time rate of change of velocity.

Normal acceleration: Inward acceleration changing the direction of velocity.

Tangential acceleration: Acceleration changing speed along a path.

Free-body diagram: Isolated-body diagram showing external forces and moments.

Mass moment of inertia: Measure of resistance to angular acceleration.

Torque: Rotational effect of a force about an axis.

Work: Energy transferred by a force through displacement.

Kinetic energy: Energy associated with motion.

Impulse: Time integral of force.

Linear momentum: Product of mass and velocity.

Coefficient of restitution: Ratio of relative separation speed to approach speed.

Rolling without slipping: Contact condition satisfying v_G = Rω.

Unbalance: Uneven mass distribution relative to a rotation axis.

Transient response: Time-dependent response before steady operation.

Damping: Mechanism that dissipates mechanical energy.

Review Questions

Short Answer Questions

1. Distinguish kinematics from kinetics.
2. What physical effects are represented by tangential and normal acceleration?
3. State the work–energy theorem for a particle.
4. Why is the parallel-axis theorem useful?
5. What assumptions are required for rolling without slipping?

Long Answer Questions

1. Explain how a free-body diagram is constructed for a translating and rotating rigid body.
2. Compare force–acceleration, work–energy, and impulse–momentum methods, including a suitable application for each.
3. Derive the kinetic energy expression for a planar rigid body as translation of its mass centre plus rotation about its mass centre.
4. Explain static and dynamic balancing and discuss why rotating unbalance becomes more severe as speed increases.
5. Discuss the assumptions and limitations of a lumped motor–load transient model.

Analytical / Application Questions

1. A 5 kg slider is pulled by 40 N over 1.5 m against a constant 8 N resistance, starting from rest. Determine its final speed using work–energy.
2. A wheel has mass 6 kg, radius 0.20 m, and is a solid disk. Determine its central-axis inertia and the torque required to produce 12 rad/s² angular acceleration, neglecting resistance.
3. A robot joint has equivalent inertia 0.10 kg·m² and resisting torque 0.25 N·m. If its motor supplies 1.25 N·m, determine angular acceleration and the ideal time to reach 10 rad/s from rest.

Multiple Choice Questions with Answers

Question 1

Which equation directly relates the resultant external force on a constant-mass particle to its acceleration?

A. ΣF = ma

B. ΣM = Iα

C. T = ½mv²

D. p = mv

Correct Answer: A. ΣF = ma

Question 2

For a particle moving along a curved path, which expression gives normal acceleration?

A. dv/dt

B. v²/ρ

C. rα

D. 2ṙθ̇

Correct Answer: B. v²/ρ

Question 3

A 10 N force acts through 3 m in its direction. What work does it perform?

A. 3.33 J

B. 13 J

C. 30 J

D. 300 J

Correct Answer: C. 30 J

Question 4

Which method is generally most appropriate for a short-duration impact when the force varies rapidly with time?

A. Impulse–momentum

B. Static equilibrium

C. Geometric similarity

D. Steady heat balance

Correct Answer: A. Impulse–momentum

Question 5

The mass moment of inertia of a solid disk about its central axis is:

A. mR²

B. ½mR²

C. 1/12mR²

D. 2/5mR²

Correct Answer: B. ½mR²

Question 6

A drum with inertia 0.20 kg·m² experiences a net torque of 2.0 N·m. Its angular acceleration is:

A. 0.10 rad/s²

B. 0.40 rad/s²

C. 4.0 rad/s²

D. 10 rad/s²

Correct Answer: D. 10 rad/s²

Question 7

For rolling without slipping, the centre velocity of a wheel of radius R is:

A. v_G = R/ω

B. v_G = Rω

C. v_G = ω/R

D. v_G = Rω²

Correct Answer: B. v_G = Rω

Question 8

If rotating unbalance is modelled by F_u = meω², doubling angular speed changes the unbalance force by a factor of:

A. 2

B. 4

C. ½

D. ¼

Correct Answer: B. 4

Question 9

For a rotational model Jω̇ + bω = T_m, the steady speed for constant motor torque is:

A. J/T_m

B. T_m/J

C. T_m/b

D. b/T_m

Correct Answer: C. T_m/b

Question 10

Which statement about a free-body diagram is correct?

A. It shows only internal forces.

B. It replaces all forces with kinetic energy.

C. It isolates a body and shows relevant external forces and moments.

D. It is required only for static systems.

Correct Answer: C. It isolates a body and shows relevant external forces and moments.