Planar Kinetics of Rigid Bodies

Introduction

Planar kinetics examines the motion of rigid bodies when all particles move in planes parallel to a fixed reference plane. Unlike planar kinematics, which describes motion without considering its causes, kinetics relates motion to applied forces and moments. The principal analytical tools are the equations of force–acceleration, work–energy, impulse–momentum, and angular momentum.

These methods are central to the analysis of gears, flywheels, wheels, rolling vehicles, robotic links, crank mechanisms, conveyors, and impact systems. A rigid body may translate, rotate, or undergo general plane motion consisting of simultaneous translation and rotation. Correct analysis depends on selecting an appropriate free-body diagram, identifying the mass properties, and choosing equations that match the physical situation.

Learning Objectives

By the end of this chapter, the student should be able to:

1. Formulate planar equations of motion for translating, rotating, and general-plane-motion rigid bodies.
2. Determine mass moments of inertia and use the parallel-axis theorem.
3. Apply work–energy and impulse–momentum methods to rigid bodies.
4. Analyze eccentric rotation, rolling, connected bodies, and impact.
5. Explain dynamic balance and identify sources of shaking forces and moments.
6. Select an efficient solution method and assess the assumptions behind the result.

1. Kinetic Description of a Rigid Body

A rigid body is an idealized body in which the distance between every pair of particles remains constant. In engineering analysis, deformation may be neglected when it is small compared with the overall motion. A planar rigid-body motion has three degrees of freedom: two translational coordinates and one angular coordinate.

For a body of mass m, the mass center G is especially useful. The translational acceleration of G is denoted by a_G, and the angular velocity and angular acceleration are ω and α. The mass moment of inertia about an axis perpendicular to the plane through G is I_G.

The total kinetic energy is

where v_G is the speed of the mass center. The first term represents translation of the mass center, and the second represents rotation about the mass center.

For a point P fixed in the body,

and

Here r_P/G is the position vector from G to P. The equations apply to a rigid body undergoing plane motion, with vectors interpreted in three-dimensional form even though the motion is planar.

2. Mass Moment of Inertia

The mass moment of inertia measures resistance to angular acceleration. For a continuous body,

where r is the perpendicular distance from the mass element dm to the axis through O. The SI unit is kg·m².

If the centroidal moment of inertia I_G is known, the parallel-axis theorem gives

where d is the perpendicular distance between the parallel axes through G and O. This theorem applies only to parallel axes, with one axis passing through the mass center.

Common results include:

The radius of gyration k is defined by

Thus, k = √(I_G/m). It is the distance from the axis at which the entire mass could be concentrated without changing the moment of inertia.

3. Force–Acceleration Equations

For a rigid body in planar motion, Newton–Euler equations are

The first two equations describe translation of the mass center. The third describes rotation about the mass center. Moments are positive or negative according to the selected sign convention, usually counterclockwise positive.

A moment equation about another point O fixed in an inertial frame can be written as

If O is fixed in the inertial frame, this equation is valid. A frequently used special case occurs when O is a fixed pin and the body rotates about O:

This form is not generally valid when O is accelerating. For a point that is instantaneously stationary but not fixed, additional care is required; an instantaneous center is useful for velocity analysis but should not automatically be used as a moment center in the kinetic equation.

Free-body diagrams

A free-body diagram should show the body isolated, all applied forces, support reactions, weight, applied couples, and relevant dimensions. Friction must be included when it is physically present, even if its work is zero during ideal rolling. Unknown reactions can often be eliminated by taking moments about a convenient point, but the complete force equations are still needed when the acceleration of the mass center is required.

4. General Plane Motion and Special Cases

Translation

In pure translation, ω = 0 and α = 0. Every point has the same acceleration, so

The rotational equation is satisfied because the net moment about G is zero.

Rotation about a fixed axis

If a body rotates about a fixed point O, every point follows a circular path about O. For a point at radius r,

where a_t is tangential acceleration toward the direction of increasing angular speed and a_n is inward normal acceleration.

General plane motion

In general motion, the mass center translates while the body rotates. The three Newton–Euler equations should normally be used. An alternative is to use a translating-and-rotating reference description, but the mass-center formulation is usually the clearest and least error-prone.

5. Work–Energy Method

The work–energy equation for a rigid body is

where T₁ and T₂ are initial and final kinetic energies, and U_1→2 is the work of external forces and couples between the two positions. Equivalently,

For planar motion,

The work of a force F is

For a constant force acting through displacement s at angle θ,

The work of a constant couple M through angular displacement Δθ is

with angles measured in radians. Weight is conservative, so its work may be represented by a change in gravitational potential energy. An ideal fixed-axis bearing does no work because its point of application has zero displacement. Static friction at a stationary contact in pure rolling also does no work in the ideal rigid-body model; it may nevertheless be essential for enforcing the rolling constraint.

The work–energy method is scalar and is especially efficient when reactions are unknown, the motion is between two positions, and time is not required.

Worked Example 1: Rolling disk down an incline

A solid disk of mass 10 kg and radius 0.20 m rolls without slipping from rest through a vertical drop of 1.5 m. Determine its speed at the lower position. Neglect resistance.

For a solid disk, I_G = 1/2 mr². Pure rolling gives v_G = rω. Conservation of mechanical energy gives

Substitution yields

Therefore,

and

The mass cancels, as expected for ideal rolling bodies of the same shape. The angular speed is ω = v/r = 22.2 rad/s.

6. Impulse–Momentum Method

Impulse–momentum is useful when forces act over a time interval, particularly during impact or rapidly changing motion. The linear impulse–momentum equation is

The angular impulse–momentum equation about G is

If a force or moment is approximately constant,

and

Impulse has units N·s, equivalent to kg·m/s. Angular impulse has units N·m·s, equivalent to kg·m²/s.

During a short impact, ordinary weight and other small forces may be neglected compared with the large contact impulse. This is an approximation that must be justified by comparing force magnitudes and durations. Impulse–momentum determines changes in velocity and angular velocity, but it does not by itself determine energy loss.

7. Angular Momentum

The angular momentum of a rigid body about G is

The moment equation in angular-momentum form is

For a fixed point O,

and, when O is fixed in an inertial frame,

If the external impulsive moment about O is zero, angular momentum about O is conserved:

This principle is often useful for a body striking a pin or for coupled rotating components. Conservation of angular momentum does not imply conservation of kinetic energy unless the interaction is also perfectly elastic.

8. Rolling and Connected Bodies

For rolling without slipping on a stationary surface,

and, while contact is maintained,

These relations apply along the direction of motion and assume a rigid circular rolling surface. Rolling without slipping is a kinematic constraint, not an energy law. It requires sufficient static friction; if the required friction exceeds the available limit, slipping occurs.

The friction condition is commonly expressed as

where μ_s is the coefficient of static friction and N is the normal force. Equality indicates impending slip.

For connected bodies, first draw a free-body diagram of each body. Apply Newton’s equations separately and add the geometric constraint. For an inextensible cable over an ideal fixed pulley, the connected points have equal speed magnitudes and equal acceleration magnitudes. If the pulley has rotational inertia, the cable tensions on its two sides need not be equal.

Worked Example 2: Pulley with rotational inertia

Two masses are connected by a cable over a pulley of radius r. The pulley has moment of inertia I_O and angular acceleration α. If the masses accelerate with magnitude a, no slip between cable and pulley gives a = rα. Let the tensions be T₁ and T₂. For the pulley,

Thus,

Therefore, tension is unequal unless I_O = 0. Treating the pulley as massless when its rotational inertia is significant produces an incorrect acceleration and tension distribution.

9. Eccentric Rotation

Eccentric rotation occurs when a body rotates about an axis that does not pass through its mass center. The mass center then travels in a circular path of radius e, where e is the eccentricity. If the angular speed is constant, the mass center has centripetal acceleration

The required resultant force is

This force rotates with the shaft and produces vibration in bearings and supports. If angular speed varies, the mass center also has tangential acceleration αe, and the resultant force contains both radial and tangential components.

The rotating unbalance of a point mass m_u at radius e is often represented by the product m_u e. Several rotating masses are balanced when the vector sum of their centrifugal-force effects is zero. Static balance requires

as a vector equation. For dynamic balance, both the force and the couple effects must vanish:

and

where z_i is the axial location relative to a reference plane. These conditions assume a rigid rotor and a common angular speed. In practice, balancing machines measure vibration and calculate correction masses in one or more planes.

10. Impact of Rigid Bodies

Impact is a short-duration interaction involving large contact forces. The coefficient of restitution e is defined along the line of impact as

For two bodies A and B moving along one line,

with signs selected consistently along the line of impact. For ordinary passive impacts, 0 ≤ e ≤ 1. Perfectly plastic impact has e = 0, while a perfectly elastic impact has e = 1.

The linear momentum equation for an isolated two-body system is

Together with restitution, it provides two equations for two unknown final velocities in a one-dimensional impact. Kinetic energy is conserved only for a perfectly elastic impact. Inelastic impact converts part of the kinetic energy into deformation, heat, sound, and vibration.

For bodies with rotation, angular momentum about a suitable fixed point may be combined with the restitution relation. The line of impact and contact geometry must be identified carefully. Friction during impact can create tangential impulses and alter angular velocities.

11. A Systematic Solution Procedure

A reliable workflow is:

1. Define the system and state whether the body is treated as rigid.
2. Draw a free-body diagram and a kinetic diagram if useful.
3. Choose coordinates and a sign convention.
4. Identify mass properties, including I_G or I_O.
5. Add geometric constraints such as cable length or no-slip rolling.
6. Select the method: force–acceleration for instantaneous forces, work–energy for position-to-position motion, impulse–momentum for short-duration events, and angular momentum when external moments are conveniently eliminated.
7. Write equations symbolically before inserting numbers.
8. Check units, signs, limiting cases, and whether support reactions are physically plausible.

12. Illustrative Case: Balancing a High-Speed Rotor

Background

Consider an illustrative industrial rotor used in a machine tool. A small mass has been added during repair, but its axial position is not known precisely. At operating speed, the machine exhibits increasing bearing vibration.

Problem

The rotor has two correction planes. Measurements indicate a once-per-revolution vibration whose amplitude increases approximately with the square of rotational speed. The engineering team must decide whether the problem is static unbalance, couple unbalance, or another fault.

Analysis

A rotating eccentric mass produces a centrifugal force proportional to ω². If the rotor has effective unbalance U = m_u e,

The observed speed dependence supports an unbalance hypothesis, but it does not identify the axial distribution. A single correction plane can remove the resultant force, but if unbalance exists in separated planes, a residual rotating couple may remain. Therefore, two-plane balancing is appropriate.

The balancing procedure measures vibration amplitude and phase at a trial speed, adds a known trial mass, and observes the change. From these data, the required correction masses and angular locations are calculated. The rotor is then tested at operating speed, subject to guarding and approved overspeed procedures.

Outcome

After correction, both the once-per-revolution force and couple are reduced. Bearing loads decrease, vibration falls, and service life improves. If vibration remained, other causes—misalignment, bent shafts, looseness, resonance, or bearing defects—would require investigation.

Learning points

1. Dynamic balance requires control of both force and moment effects.
2. Vibration amplitude alone is insufficient; phase and axial location matter.
3. A simple rigid-body model is valuable but must be validated against measurements.
4. High-speed testing requires machine guarding, risk assessment, and adherence to applicable safety standards.

13. Limitations, Safety, and Contemporary Practice

Rigid-body kinetics neglects elastic deformation, joint clearance, damping, frictional heating, and material failure. These simplifications are appropriate when deformations are small and the time scale does not excite structural flexibility. For high-speed machinery, flexible-body dynamics and finite-element analysis may be necessary.

Computer-aided engineering software can integrate equations of motion, model multibody mechanisms, and estimate reaction forces. However, software does not replace free-body diagrams or physical validation. Incorrect constraints, mass properties, contact definitions, or units can produce precise but meaningless results. Sensor-based condition monitoring and digital twins increasingly combine measured acceleration with dynamic models, but uncertainty in sensor calibration and model parameters must be reported.

Safety is integral to kinetics. Rotating equipment must be guarded; impact tests require containment; stored energy in springs, flywheels, and elevated bodies must be isolated before maintenance. Engineers must distinguish an ideal mathematical prediction from a permissible operating condition and must follow relevant occupational, machinery, and laboratory regulations.

Chapter Summary

Planar rigid-body kinetics relates forces and moments to translation and rotation. The mass-center equations are ∑F = ma_G and ∑M_G = I_Gα, while the kinetic energy is the sum of translational and rotational terms. The parallel-axis theorem permits moments of inertia to be transferred between parallel axes.

The force–acceleration method is appropriate for instantaneous dynamics and reaction forces. Work–energy is efficient for motion between positions and does not require time. Impulse–momentum is suited to short-duration force interactions, and angular momentum is particularly useful when external moments about a point vanish. Rolling requires both kinematic constraints and sufficient friction. Connected bodies require separate free-body diagrams and compatible geometric constraints. Eccentric rotation generates speed-dependent shaking forces, while dynamic balance requires elimination of both resultant force and resultant couple. Impact analysis combines momentum principles with restitution, and kinetic energy is generally not conserved in inelastic collisions.

Sound engineering practice requires clear assumptions, correct mass properties, consistent units, physical checks, and attention to safety and model limitations.

Key Terms

Rigid body: Idealized body whose internal distances remain constant.

Planar motion: Motion in which all particles move in planes parallel to one reference plane.

Mass center: Point representing the translational motion of a body’s mass distribution.

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

Radius of gyration: Distance k satisfying I = mk².

Parallel-axis theorem: Relation I_O = I_G + md² for parallel axes.

General plane motion: Simultaneous translation and rotation of a rigid body.

Force–acceleration method: Newton–Euler analysis using resultant forces and moments.

Work–energy method: Method relating external work to change in kinetic energy.

Impulse: Time integral of force.

Angular impulse: Time integral of moment.

Angular momentum: Rotational momentum, commonly H_G = I_Gω.

Rolling without slipping: Motion satisfying v_G = rω.

Eccentric rotation: Rotation about an axis not passing through the mass center.

Static balance: Condition in which the resultant centrifugal force is zero.

Dynamic balance: Condition in which both centrifugal force and couple effects are zero.

Impact: Short-duration interaction involving large contact forces.

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

Line of impact: Direction along which impact impulse acts in a frictionless contact model.

Kinetic energy: Energy associated with translational and rotational motion.

Review Questions

Short Answer Questions

1. Why is the mass center used in the translational equations of rigid-body motion?
2. State the parallel-axis theorem and identify its applicability condition.
3. What is the difference between rolling without slipping and zero frictional force?
4. When is the impulse–momentum method preferable to the work–energy method?
5. Why are cable tensions generally unequal when a pulley has rotational inertia?

Long Answer Questions

1. Derive the kinetic energy expression for a rigid body undergoing general plane motion.
2. Explain the Newton–Euler equations for planar motion and discuss the restrictions on taking moments about a moving point.
3. Develop the equations for a disk rolling down an incline and explain the role of static friction.
4. Explain static and dynamic balance of a rotor, including the physical meaning of the force and couple conditions.
5. Describe the analysis of a one-dimensional impact using conservation of linear momentum and the coefficient of restitution.

Analytical / Application Questions

1. A 12 kg solid cylinder of radius 0.15 m rolls without slipping from rest through a vertical drop of 0.80 m. Determine its final speed and angular speed, neglecting resistance.
2. A 4 kg disk rotates at 30 rad/s about an axis parallel to, but 0.10 m from, its centroidal axis. Explain how to calculate the bearing force caused by eccentric rotation if the angular speed is constant, and identify the required mass property.
3. Two bodies collide along a straight line. Explain why linear momentum may be conserved while kinetic energy decreases, and identify the additional relation needed to determine the final velocities.

Multiple Choice Questions with Answers

Question 1

For a rigid body undergoing general plane motion, which equation describes rotation about its mass center?

A. ∑F_x = ma_Gx

B. ∑M_G = I_Gα

C. v_G = rω

D. T = mgh

Correct Answer: B. ∑M_G = I_Gα

Question 2

The parallel-axis theorem is correctly written as:

A. I_O = I_G − md²

B. I_O = I_G + md²

C. I_O = md

D. I_O = I_G/d²

Correct Answer: B. I_O = I_G + md²

Question 3

A solid disk rolls without slipping. Which relation applies?

A. v_G = ω/r

B. v_G = rω

C. v_G = r/ω

D. v_G = 0

Correct Answer: B. v_G = rω

Question 4

The kinetic energy of a planar rigid body is:

A. 1/2 mv_G² only

B. 1/2 I_Gω² only

C. 1/2 mv_G² + 1/2 I_Gω²

D. mgv_G

Correct Answer: C. 1/2 mv_G² + 1/2 I_Gω²

Question 5

Which method is generally most suitable for a short-duration collision?

A. Work–energy only

B. Impulse–momentum

C. Static equilibrium

D. Dimensional analysis only

Correct Answer: B. Impulse–momentum

Question 6

For a constant-speed eccentric rotation with eccentricity e, the mass-center acceleration magnitude is:

A. ωe

B. αe

C. ω²e

D. e/ω²

Correct Answer: C. ω²e

Question 7

If a pulley has nonzero rotational inertia, the tensions on its two sides are generally:

A. Always equal

B. Always zero

C. Unequal when the pulley angularly accelerates

D. Independent of angular acceleration

Correct Answer: C. Unequal when the pulley angularly accelerates

Question 8

For a perfectly plastic impact, the coefficient of restitution is:

A. 0

B. 0.5

C. 1

D. Greater than 1

Correct Answer: A. 0

Question 9

Which condition is required for ideal dynamic balance of a rotor?

A. Only zero mass

B. Zero resultant force and zero resultant couple

C. Constant linear velocity

D. No bearing supports

Correct Answer: B. Zero resultant force and zero resultant couple

Question 10

A 2 kg body moving at 3 m/s has translational kinetic energy equal to:

A. 3 J

B. 6 J

C. 9 J

D. 18 J

Correct Answer: C. 9 J