Acceleration Analysis of Mechanisms

Introduction

Velocity analysis describes how fast links and points in a mechanism move. Acceleration analysis goes further: it determines how rapidly the velocity changes in magnitude and direction. This information is essential for calculating inertia forces, bearing loads, motor torque, vibration, noise, fatigue, and dynamic balancing requirements.

In a mechanism, acceleration may arise because a point moves along a curved path, because a rotating link changes its angular speed, or because a sliding point is carried by a rotating link. The principal components are radial (normal), tangential, and Coriolis acceleration. These components are vector quantities and must be combined with correct directions.

Acceleration analysis is commonly performed by:

1. Analytical methods, using vector equations and differentiation.
2. Graphical methods, using acceleration diagrams or polygons drawn to scale.
3. Computer-based methods, using multibody simulation, numerical differentiation, or symbolic kinematics.

The analytical method gives numerical precision and is convenient for programming. The graphical method gives a strong physical interpretation and remains useful for checking calculations and teaching mechanism geometry.

Learning objectives

After studying this chapter, the reader should be able to:

• distinguish radial, tangential, Coriolis, and relative acceleration;
• formulate acceleration equations for rotating and translating links;
• derive the acceleration relation for a rigid link;
• construct and interpret acceleration polygons;
• analyse slider-crank and four-bar mechanisms;
• calculate accelerations using consistent sign and vector conventions;
• identify assumptions, sources of error, and practical implications of acceleration results.

1. Fundamental ideas

Acceleration is the time rate of change of velocity:

where 𝐚 is acceleration in m/s² and 𝐯 is velocity in m/s. Since velocity is a vector, acceleration can result from a change in either speed or direction.

For a particle moving in plane polar coordinates, the position is specified by radial distance r and angular position θ. Its acceleration is

Here, ṙ and r̈ are radial velocity and radial acceleration, θ̇ = ω is angular velocity in rad/s, θ̈ = α is angular acceleration in rad/s², 𝐞ᵣ is the outward radial unit vector, and 𝐞θ is the transverse unit vector in the direction of increasing θ. This expression assumes planar motion and an inertial reference frame.

The terms have clear meanings:

• −rθ̇²𝐞ᵣ is the inward normal or centripetal acceleration caused by rotation.
• r̈𝐞ᵣ is acceleration caused by change in radial motion.
• rθ̈𝐞θ is tangential acceleration caused by angular acceleration.
• 2ṙθ̇𝐞θ is Coriolis acceleration, present when radial sliding occurs in a rotating frame.

For a fixed-radius point, ṙ = 0 and r̈ = 0. Therefore,

The radial and tangential components are perpendicular, so the magnitude is

This formula applies only to a point fixed on a rotating link with instantaneous radius r.

Angular acceleration and units

Angular velocity ω is measured in rad/s and angular acceleration α in rad/s². The radian is dimensionless in SI analysis, although it is retained to clarify rotational quantities. For a point at radius r:

where aₙ and aₜ are in m/s². The normal component points toward the centre of rotation; the tangential component is perpendicular to the radius and follows the sense of α.

2. Acceleration of a rigid link

Consider two points A and B on the same rigid link. The relative position of B from A is 𝐫ᴮ⁄ᴬ. Differentiating the velocity relation gives

For a rigid link rotating instantaneously with angular velocity ω and angular acceleration α,

In planar engineering notation this becomes

where

and

The relative normal acceleration is directed from B toward A and has magnitude ω²AB. The relative tangential acceleration is perpendicular to AB and has magnitude αAB.

If the angular velocity of the link is known but its angular acceleration is unknown, the normal component can be drawn immediately, while the tangential component is drawn perpendicular to the link with unknown magnitude. This is the foundation of acceleration polygons.

Vector-component form

With x and y axes fixed in an inertial frame, let the link AB make angle φ with the x-axis. Then

and

Thus,

These equations are useful for numerical solution of four-bar and other constrained mechanisms.

3. Acceleration of a sliding point on a rotating link

Suppose a slider P moves along a slot in link OA, while OA rotates with angular velocity ω and angular acceleration α. Let u be the distance OP measured along the slot, and let u̇ and ü be the relative sliding velocity and acceleration. The absolute acceleration of P is

For a fixed origin O, 𝐚ᴼ = 0. The terms are:

• α × 𝐫ᴾ⁄ᴼ: tangential acceleration due to angular acceleration;
• ω × (ω × 𝐫ᴾ⁄ᴼ): radial acceleration toward O;
• 2ω × 𝐯ᴾ⁄ᴼ: Coriolis acceleration;
• 𝐚ᴾ⁄ᴼ: relative acceleration along the slot.

The Coriolis component has magnitude

where vᵣ = u̇ is the relative sliding speed. Its direction is obtained by rotating the relative velocity vector through 90° in the direction of the angular velocity. A useful practical rule is: rotate the relative velocity vector by 90° in the sense of rotation of the slotted link, then multiply its magnitude by 2ω.

Coriolis acceleration is not present merely because a link rotates. It requires relative motion along or across the rotating member. Omitting it in a slotted-link mechanism can produce a substantial error in the acceleration of the slider.

4. Graphical acceleration analysis

An acceleration diagram represents acceleration vectors with a selected scale, such as 1 cm = 2 m/s². It is not the mechanism itself; it is a vector construction that obeys the acceleration equations.

General construction procedure

1. Draw the mechanism accurately in its specified configuration.
2. Select a positive angular direction and coordinate system.
3. Determine known angular velocities, angular accelerations, and point accelerations.
4. Choose an acceleration scale and a pole Oₐ.
5. Draw known absolute acceleration vectors from the pole.
6. For each rigid link, draw the normal component parallel to the link and directed toward the instantaneous centre or toward the first point of the link.
7. From the end of the normal component, draw a line perpendicular to the link to represent the unknown tangential component.
8. For sliders, draw the permitted acceleration direction, including guide constraints.
9. The intersection of construction lines gives the unknown point acceleration.
10. Measure vector lengths and convert them using the selected scale.
11. Check the result analytically or by resolving the polygon into x and y components.

The polygon must preserve vector order. For example, the equation 𝐚ᴮ = 𝐚ᴬ + 𝐚ᴮ⁄ᴬ means that the vector from the acceleration pole to B is obtained by adding the vector to A and then the relative vector from A to B.

Figure: Schematic acceleration polygon for a rigid link. The normal component is drawn first and the tangential component is perpendicular to the link.

Graphical accuracy

Graphical solutions are affected by drawing scale, line thickness, inaccurate mechanism geometry, and measurement of small vector differences. A large drawing scale and sharp construction lines improve accuracy. When a result is safety-critical, the graphical solution should be verified by component equations or simulation.

5. Slider-crank acceleration analysis

A slider-crank mechanism consists of a crank OA, connecting rod AB, and slider B constrained to move along a line. It is used in engines, compressors, pumps, and reciprocating machinery.

Let r be crank radius, l connecting-rod length, θ crank angle measured from the line of stroke, and ω crank angular velocity. For constant crank speed, α = 0. Taking the line of stroke as x and choosing a suitable origin, the slider position may be written as

This expression assumes that the crank and rod geometry remains physically possible, l ≥ r|sinθ|, and that θ changes uniformly with time.

Differentiating with respect to time gives slider velocity. A useful acceleration expression is

The exact form depends on the selected positive direction and coordinate origin; therefore the sign must be interpreted from the stated convention. For a long connecting rod, r/l is small, and the common approximation is

The first term is the primary acceleration, associated with crank rotation. The second is the secondary acceleration caused by connecting-rod obliquity. This approximation is useful when r/l is sufficiently small, but it should not replace the exact relation when precision is required.

Worked example 1: rotating crank pin

A crank of radius 0.05 m rotates at 20 rad/s with constant speed. Determine the radial and tangential acceleration of its crank pin.

Since α = 0,

Thus the crank pin has 20 m/s² directed toward the crank centre. The result is physically consistent: constant angular speed does not mean zero acceleration, because the velocity direction continually changes.

Worked example 2: approximate slider acceleration

A slider-crank has r = 0.04 m, l = 0.16 m, ω = 25 rad/s, and θ = 60°. Estimate the slider acceleration using the long-rod approximation.

Here, cos60° = 0.5 and cos120° = −0.5.

The negative sign indicates acceleration opposite to the selected positive x-direction. The primary contribution is −12.5 m/s² and the secondary correction is +3.125 m/s². The approximation should be checked against an exact calculation if the result is used for component design.

6. Four-bar mechanism

A four-bar mechanism contains four links connected by four turning pairs. Let the fixed link be AD, input link AB, coupler BC, and output link CD. The vector loop is

Differentiating once gives the velocity equation, and differentiating twice gives

because the fixed-link acceleration is zero when A and D are fixed. Each link acceleration is resolved into normal and tangential parts. The known input-link acceleration is calculated first. The coupler and output-link unknown angular accelerations are then found by resolving the vector equation into two scalar equations.

For link AB of length l₁ and angle θ₁,

The same form applies to links BC and CD. A graphical acceleration polygon is constructed by placing these vectors head to tail. The solution is valid only for the specified configuration; another assembly position generally has different velocities and accelerations.

7. Sign conventions and interpretation

Many errors in mechanism analysis are not arithmetic errors but direction errors. A reliable procedure is:

• define x and y axes before writing equations;
• define positive angular velocity and angular acceleration;
• distinguish a vector from its magnitude;
• draw normal acceleration toward the centre of curvature;
• draw tangential acceleration perpendicular to the link;
• state whether slider displacement is positive to the right, left, upward, or downward;
• retain signs until the final interpretation.

Angular velocity may be clockwise or counterclockwise. In a planar vector treatment, changing the sign of ω changes the direction of velocity, but the normal acceleration term depends on ω² and therefore retains its inward direction. Tangential acceleration changes direction when α changes sign.

8. Applied case: acceleration in a reciprocating compressor

Background

Consider an illustrative single-cylinder reciprocating compressor operating at 1,200 rpm. Its crank radius is 40 mm and connecting-rod length is 160 mm. The crank angle is 60° from the inner-dead-centre line. The crank speed is assumed constant.

Problem

The designer must estimate piston acceleration to assess inertia force and bearing loading. Excessive acceleration may increase vibration and cyclic stress.

Analysis

The angular speed is

Using the long-rod approximation,

With r = 0.04 m, l = 0.16 m, θ = 60°, cosθ = 0.5 and cos2θ = −0.5,

If the reciprocating mass is 2.0 kg, the corresponding inertia-force magnitude, neglecting gas force and friction, is approximately

This is an illustrative estimate, not a complete dynamic design calculation. Gas pressure, connecting-rod angular inertia, bearing clearance, crankshaft flexibility, friction, and transient speed variation must also be considered.

Outcome and learning points

The secondary acceleration reduces the magnitude of acceleration at this crank position relative to the primary component alone. At other crank angles, the secondary term may increase it. The case demonstrates why connecting-rod length affects vibration and why speed is especially important: acceleration varies with ω², so doubling speed ideally increases acceleration and inertia force by a factor of four.

9. Limitations, technology, and professional practice

Acceleration analysis in introductory mechanism theory commonly assumes rigid links, ideal joints, planar motion, constant dimensions, negligible clearance, and an inertial fixed frame. Real mechanisms depart from these assumptions. Flexible links can have vibration modes; clearances create impacts; lubrication affects friction; and manufacturing tolerances alter the geometry.

Computer-aided engineering tools can calculate position, velocity, and acceleration over a complete cycle. However, numerical differentiation of measured position data can amplify noise. Smoothing, appropriate sampling, and independent checks are therefore necessary. Symbolic differentiation is often more reliable for idealized geometry, while multibody dynamics is useful when masses, forces, compliance, and contact are included.

Professional best practice includes:

• checking units and orders of magnitude;
• comparing graphical and analytical results;
• testing limiting cases, such as α = 0 or ṙ = 0;
• documenting coordinate and sign conventions;
• reporting whether values are exact, approximate, measured, or simulated;
• using safety factors and validated models in design decisions.

Acceleration results directly influence guarding, structural integrity, noise, and worker safety. A calculation should not be used as the sole basis for releasing a machine without considering dynamic loads, standards, testing, and failure modes.

Chapter Summary

Acceleration analysis determines how the velocity of mechanism points changes in magnitude and direction. A point on a rotating link has normal acceleration rω² toward the centre and tangential acceleration rα perpendicular to the radius. A rigid-link relation expresses the acceleration of one point as the acceleration of another plus relative normal and tangential components. A sliding point on a rotating link additionally has relative acceleration and Coriolis acceleration of magnitude 2ωvᵣ.

Graphical acceleration polygons provide a direct vector solution: known absolute and normal components are drawn first, while unknown tangential and constrained components are represented by construction lines. Analytical component equations provide precise numerical results and are especially suitable for four-bar and slider-crank mechanisms. Slider-crank acceleration contains primary and secondary terms, while four-bar analysis requires a differentiated vector loop. Correct directions, sign conventions, assumptions, and validation are essential. In practical machines, acceleration estimates support inertia-force calculations, vibration control, fatigue assessment, and safe design.

Key Terms

Acceleration: Rate of change of velocity with time.

Angular velocity: Rate of change of angular position, measured in rad/s.

Angular acceleration: Rate of change of angular velocity, measured in rad/s².

Normal acceleration: Component directed toward the centre of curvature; for rotation its magnitude is rω².

Radial acceleration: Acceleration along a radial direction, commonly used interchangeably with normal acceleration in fixed-radius rotation.

Tangential acceleration: Component perpendicular to the radius, with magnitude rα for a rotating point.

Relative acceleration: Acceleration of one point measured relative to another point on a link or mechanism.

Rigid link: Link whose distance between any two material points remains constant.

Coriolis acceleration: Additional acceleration 2ωvᵣ of a point sliding relative to a rotating member.

Acceleration polygon: Vector diagram used to solve acceleration relations graphically.

Slider-crank: Mechanism converting rotary motion into reciprocating motion or vice versa.

Four-bar mechanism: Closed chain of four links connected by four lower pairs.

Primary acceleration: Main slider-crank acceleration term proportional to rω².

Secondary acceleration: Slider-crank correction caused by connecting-rod obliquity.

Instantaneous centre: Point about which a link has instantaneous pure rotation.

Vector loop: Position equation expressing closure of a mechanism chain.

Inertia force: Force associated with mass acceleration, commonly represented in dynamic analysis as ma.

Relative sliding velocity: Velocity of a point along a slot or guide relative to the rotating member.

Review Questions

Short Answer Questions

1. Why can a body moving at constant speed still have non-zero acceleration?
2. State the magnitudes and directions of normal and tangential acceleration for a point on a rotating link.
3. Under what condition does Coriolis acceleration occur?
4. What is the purpose of an acceleration polygon?
5. Why must sign conventions be stated before solving a mechanism problem?

Long Answer Questions

1. Derive the acceleration relation between two points on a rigid planar link and explain each component.
2. Explain the construction procedure for a graphical acceleration polygon.
3. Derive the acceleration expression for a point sliding in a rotating slot and discuss the Coriolis term.
4. Explain primary and secondary acceleration in a slider-crank mechanism.
5. Describe analytical acceleration analysis of a four-bar mechanism using a differentiated vector loop.

Analytical / Application Questions

1. A point is 0.12 m from the centre of a link rotating at 30 rad/s with angular acceleration 8 rad/s². Calculate its normal and tangential accelerations and their resultant magnitude.
2. A slider moves along a radial slot at 3 m/s relative to a link rotating at 10 rad/s. Determine the magnitude of its Coriolis acceleration and state how its direction is obtained.
3. For a slider-crank with r = 0.05 m, l = 0.20 m, ω = 40 rad/s, and θ = 0°, use the long-rod approximation to calculate the slider acceleration and identify the primary and secondary contributions.

Multiple Choice Questions with Answers

Question 1

A point at radius r on a link rotating with angular velocity ω has normal acceleration of magnitude

A. rω

B. rω²

C. rα

D. 2rω

Correct Answer: B. rω²

Question 2

Tangential acceleration of a fixed-radius point on a rotating link is

A. rω²

B. ω²/r

C. rα

D. 2ωṙ

Correct Answer: C. rα

Question 3

Coriolis acceleration occurs when a point

A. is fixed on a rotating link

B. slides relative to a rotating member

C. moves in a straight guide fixed to ground

D. has zero relative velocity

Correct Answer: B. slides relative to a rotating member

Question 4

The direction of normal acceleration of a point on a rotating rigid link is

A. away from the centre of rotation

B. parallel to the tangential direction

C. toward the centre of rotation

D. opposite to angular acceleration

Correct Answer: C. toward the centre of rotation

Question 5

If a crank rotates at constant angular velocity, its crank-pin tangential acceleration is

A. zero

B. rω²

C. 2rω

D. r/ω²

Correct Answer: A. zero

Question 6

A link rotates at 10 rad/s while a point slides at 2 m/s relative to it. The Coriolis acceleration magnitude is

A. 5 m/s²

B. 20 m/s²

C. 40 m/s²

D. 200 m/s²

Correct Answer: C. 40 m/s²

Question 7

In an acceleration polygon for a rigid link, the relative normal acceleration is drawn

A. perpendicular to the link and away from the first point

B. along the link toward the first point

C. parallel to the ground only

D. in the direction of angular acceleration only

Correct Answer: B. along the link toward the first point

Question 8

For a slider-crank, the secondary acceleration is primarily caused by

A. zero crank speed

B. connecting-rod obliquity

C. bearing friction alone

D. constant piston velocity

Correct Answer: B. connecting-rod obliquity

Question 9

If rotational speed is doubled while geometry remains unchanged, acceleration terms proportional to ω² become

A. half as large

B. twice as large

C. four times as large

D. unchanged

Correct Answer: C. four times as large

Question 10

Which method is especially useful for checking the direction and composition of mechanism accelerations?

A. An acceleration polygon

B. A mass-only calculation

C. A static force diagram alone

D. A dimensional drawing without vectors

Correct Answer: A. An acceleration polygon