Dynamics · Chapter 8 of 10 · Intermediate
Plane Kinetics of Rigid Bodies
A rigid body needs two equations, not one: forces drive the mass center, and moments drive the spin. Mass moment of inertia is the rotational counterpart of mass.
Readiness check
This chapter combines forces, moments, and the kinematics of Chapter 7. Tick only what you can do closed-notes.
- Apply ΣF = ma to a particle.
- Take moments of forces about a point.
- Recall mass moment of inertia and the parallel-axis theorem.
- Relate aG = αr for rolling.
- Draw a free body for a rigid body.
The core idea
For a rigid body, the resultant force equals mass times the mass-center acceleration, and the resultant moment about the mass center equals the moment of inertia times the angular acceleration.
ΣF = m aGΣMG = IG αIO = IG + md²A particle has only translation, so one equation suffices. A rigid body can also spin, so it needs a second equation for rotation. The mass moment of inertia IG measures resistance to angular acceleration the way mass measures resistance to linear acceleration. The two equations are solved together.
The skills, taught in order
Rigid-body kinetics is two equations applied to three motion types. Five skills cover the equations, inertia, and each motion case.
8.1 The equations of motion
The pair ΣF = m aG and ΣMG = IG α holds for any plane motion. Forces are summed as in statics, but they now equal m aG; moments about the mass center equal IG α. Always draw the free body first.
8.2 Mass moment of inertia
IG depends on how mass is distributed about the axis: mass far from the axis counts more (it enters as r²). The parallel-axis theorem shifts the axis: IO = IG + md². The radius of gyration k, with I = mk², expresses inertia as an equivalent radius.
| Shape (centroidal axis) | IG |
|---|---|
| Slender rod, length L | (1/12)mL² |
| Solid disk or cylinder, radius r | ½mr² |
| Thin hoop or ring, radius r | mr² |
| Solid sphere, radius r | (2/5)mr² |
8.3 Translation
If the body does not rotate (α = 0), the moment equation reduces to ΣMG = 0, and ΣF = m aG governs. This is statics plus an inertia term, useful for accelerating vehicles and load shifts that change wheel reactions.
8.4 Fixed-axis rotation
When the body turns about a fixed axis O, it is often simplest to take moments there: ΣMO = IO α, with IO from the parallel-axis theorem. The pin reaction drops out of the moment equation, a major convenience for flywheels, pulleys, and pendulums.
8.5 General plane motion
The full case has both translation and rotation, so both equations are needed, linked by kinematics. Rolling without slipping is the classic example, where aG = αr ties the two together and friction provides the moment.
| Motion | Equations used |
|---|---|
| Translation | ΣF = m aG, ΣMG = 0 |
| Fixed-axis rotation | ΣMO = IO α |
| General plane motion | ΣF = m aG and ΣMG = IG α |
Engineering connection: flywheels and rotors, vehicle traction and weight transfer, gear and pulley drives, and any rolling or spinning machine part.
Worked example 1: spinning up a flywheel
A solid-disk flywheel of mass 40 kg and radius 0.3 m is driven by a constant torque of 24 N·m about its central axis. Find the angular acceleration and the time to reach 1200 rev/min from rest.
- ProblemFind α and the spin-up time for the flywheel in Figure 1.
- Given / findM = 40 kg, R = 0.3 m, T = 24 N·m, target 1200 rev/min, from rest. Find α and t.
- AssumptionsUniform solid disk, frictionless bearing, constant torque.
- ModelFixed-axis rotation about the center: ΣMO = IO α with IO = ½MR²; then constant-α kinematics for the time.
- EquationsIO = ½MR² α = T/IO t = ω/α
- SolveIO = ½ × 40 × 0.3² = 1.8 kg·m². α = 24/1.8 = 13.3 rad/s². Target ω = 1200 × 2π/60 = 125.7 rad/s, so t = 125.7/13.3 = 9.42 s.
- CheckUnits: N·m / (kg·m²) = 1/s², correct for α. A heavier or larger disk would raise IO and slow the spin-up, as intuition expects from a more massive flywheel.
- ConclusionTaking moments at the fixed axis kept the bearing reaction out of the math. The flywheel's large inertia is exactly what makes it useful for smoothing out speed fluctuations.
Worked example 2: a cylinder rolling down an incline
A uniform solid cylinder (mass 20 kg) rolls without slipping down a 25° incline. Find the acceleration of its center, the friction force, and the minimum friction coefficient that prevents slipping.
- ProblemFind aG, the friction force, and the minimum μs for the cylinder in Figure 2.
- Given / findm = 20 kg, θ = 25°, solid cylinder (IG = ½mr²), rolling without slipping. Find aG, f, μs,min.
- AssumptionsUniform cylinder, rolling without slipping so aG = αr, friction below the slip limit.
- ModelTwo equations: along the incline ΣF = maG, and moments about G with friction providing the torque; close with the rolling constraint.
- Equationsmg sinθ − f = m aG f r = IG α = ½mr²(aG/r) aG = (2/3)g sinθ
- SolveEliminating f gives aG = (2/3)g sinθ = (2/3)(9.81)(0.423) = 2.76 m/s². Then f = ½m aG = ½(20)(2.76) = 27.6 N. With N = mg cosθ = 177.8 N, the minimum coefficient is μs = f/N = 0.155.
- CheckA frictionless block would accelerate at g sinθ = 4.15 m/s²; the cylinder is slower because some effort spins it up, exactly as energy methods would predict. The required μ is modest, so most surfaces sustain rolling.
- ConclusionTwo equations and one constraint resolve the motion. The (2/3) factor is specific to a solid cylinder; a hoop (I = mr²) would give aG = ½g sinθ, slower still, because more of its mass is far from the axis.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Using only ΣF = ma | Rotation ignored for a spinning body | "Does the body rotate?" | Add the moment equation ΣMG = IG α. |
| Wrong moment of inertia | Inertia of a different shape or axis used | "Is this I for the actual shape and axis?" | Use the correct IG and shift it with the parallel-axis theorem. |
| Moments about the wrong point | Pin reaction left in the equation | "Did I take moments about G or the fixed axis?" | Use ΣMG = IG α, or ΣMO = IO α at a fixed axis. |
| Assuming maximum friction in rolling | f set to μN while rolling | "Is it rolling or slipping?" | In rolling, friction is whatever the motion needs (≤ μsN), found from the equations. |
Practice ladder
A 2 m slender rod of mass 6 kg pivots about one end. Find its moment of inertia about that pivot.
Show answer
About the center IG = (1/12)mL² = (1/12)(6)(4) = 2 kg·m²; the parallel-axis theorem gives IO = IG + m(L/2)² = 2 + 6(1)² = 8 kg·m². Equivalently (1/3)mL² = 8 kg·m².
For the Worked Example 2 incline, what acceleration would a thin hoop have instead of the solid cylinder?
Show answer
With IG = mr², the analysis gives aG = ½g sinθ = ½(9.81)(0.423) = 2.07 m/s², slower than the cylinder's 2.76. More mass at the rim means more inertia to spin up, so it rolls down more reluctantly, the classic race won by the solid cylinder.
The flywheel of Worked Example 1 now has a 50 N·m friction torque opposing it once the 24 N·m drive is removed at 125.7 rad/s. How long until it stops?
Show answer
α = −T/I = −50/1.8 = −27.8 rad/s². Time to stop = ω/|α| = 125.7/27.8 = 4.52 s. The same I that resists spin-up also resists stopping; a large flywheel coasts a long time under small friction.
Pick a real rotating or rolling part (a bench grinder, a yo-yo, a vehicle wheel under braking). Identify the motion type, write the relevant equations, and estimate an angular acceleration or required torque.
What good work looks like
The motion type identified, a free body drawn, the right I used, both equations (or the fixed-axis form) written, and a result checked against a limiting case.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Analyse one real rigid-body motion with both equations (or the fixed-axis form), justify the moment of inertia used, and end with an angular acceleration or torque plus a check.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. Write the two equations of motion for a rigid body.
ΣF = m aG and ΣMG = IG α.
2. State the parallel-axis theorem.
IO = IG + md², shifting the axis a distance d from the mass center.
3. Give IG for a solid disk and a slender rod.
Disk ½mr²; rod (1/12)mL² about its center.
4. Why take moments about a fixed axis when one exists?
The pin reaction has no moment there, so it drops out: ΣMO = IO α.
5. In rolling without slipping, how big is the friction?
Whatever the motion requires, up to μsN; it is found from the equations, not set to μN.
Textbook mapping
| Item | Mapping |
|---|---|
| Primary source | Meriam and Kraige, Engineering Mechanics: Dynamics (7th ed), Chapter 6, Section A (Force, Mass, and Acceleration) |
| Cross-reference | Hibbeler, Dynamics, Ch. 17 · Beer and Johnston, Ch. 16 |
| Core topics | 8.1 Equations of motion · 8.2 Moment of inertia · 8.3 Translation · 8.4 Fixed-axis rotation · 8.5 General plane motion |
| Engineering connection | Flywheels, rotors, vehicle traction, and gear and pulley drives. |
| Read next | Chapter 9: Rigid-Body Energy and Momentum. |