Kinetics of Rigid Bodies: Force and Acceleration

Learning Objectives

  • Apply Newton's Second Law and Euler's Equations to general plane motion.
  • Understand and utilize D'Alembert's Principle and the equivalence of Free-Body and Kinetic Diagrams.
  • Calculate the mass moment of inertia for various shapes using the parallel axis theorem.
  • Analyze the dynamics of rolling bodies, distinguishing between rolling with and without slipping.

The kinetics of rigid bodies relates the forces and couple moments acting on a body to its resulting translational and rotational acceleration. Unlike particle kinetics, which only considers forces and linear acceleration, rigid body kinetics must also account for the moments of forces and angular acceleration.

Equations of General Plane Motion

The motion of a rigid body in a plane is governed by three independent scalar equations: two for translation of the mass center and one for rotation about the mass center.

Fundamental Equations of General Plane Motion

Scalar equations governing the translational and rotational motion of a rigid body in a plane.

∑Fx=m(aG)x\sum F_x = m(a_G)_x∑Fy=m(aG)y\sum F_y = m(a_G)_y∑MG=IGα\sum M_G = I_G \alpha

Variables

SymbolDescriptionUnit
∑Fx,∑Fy\sum F_x, \sum F_ySum of external forces in x and y directionsN
mmMass of the rigid bodykg
(aG)x,(aG)y(a_G)_x, (a_G)_yAcceleration components of the center of mass Gm/s²
∑MG\sum M_GSum of external moments about the center of mass GN·m
IGI_GMass moment of inertia about the center of mass Gkg·m²
α\alphaAngular accelerationrad/s²

D'Alembert's Principle and Equivalence

A powerful way to conceptualize and solve rigid body dynamics problems is by establishing an equivalence between the external forces acting on the body and the effective forces driving its motion.

FBD=KD Equivalence

The fundamental equation ∑F=maG\sum \mathbf{F} = m\mathbf{a}_G and ∑MG=IGα\sum \mathbf{M}_G = I_G \mathbf{\alpha} can be visualized as an equality between two diagrams:

  • Free-Body Diagram (FBD): Shows all external forces (applied forces, weight, reaction forces, friction) and applied couple moments acting on the isolated body.
  • Kinetic Diagram (KD): Shows the "effective" forces (maGm\mathbf{a}_G) applied at the mass center GG, and the "effective" couple moment (IGαI_G \mathbf{\alpha}). The vector maGm\mathbf{a}_G must be oriented in the actual direction of acceleration, and IGαI_G \mathbf{\alpha} in the direction of angular acceleration.

The System is Equivalent: The system of external forces and moments on the FBD is equivalent to the system of effective forces and moments on the KD. You can take moments about any point PP, not just the mass center.

Applying the FBD=KD Method

  1. Establish a Coordinate System: Define your x−yx-y inertial axes and a positive direction for rotation (often counter-clockwise).
  2. Draw the Free-Body Diagram (FBD): Isolate the body and draw all applied forces, reaction forces at supports, and weight.
  3. Draw the Kinetic Diagram (KD): Sketch the body and place the kinetic vector maGm\mathbf{a}_G at the center of mass, and the kinetic couple IGαI_G\alpha.
  4. Apply Equations of Motion: Equate the sum of forces on the FBD to the kinetic vectors on the KD: ∑Fx=m(aG)x\sum F_x = m(a_G)_x and ∑Fy=m(aG)y\sum F_y = m(a_G)_y.
  5. Apply Moment Equation: Equate the moments about a convenient point PP on the FBD to the moments of the kinetic vectors about the same point PP on the KD.

Moment Equivalence Equation

Equates the sum of external moments to the sum of kinetic moments about any arbitrary point P.

∑MP (from FBD)=∑(Mk)P (from KD)\sum M_P \text{ (from FBD)} = \sum (M_k)_P \text{ (from KD)}

Variables

SymbolDescriptionUnit
∑MP\sum M_PSum of moments of all external forces and couples about point PN·m
∑(Mk)P\sum (M_k)_PSum of the moments of the kinetic vectors (m\mathbf{a}_G and I_G\mathbf{\alpha}) about point PN·m

D'Alembert's Principle Usage

Similar to particles, we can rewrite the equations as ∑F−maG=0\sum \mathbf{F} - m\mathbf{a}_G = 0 and ∑MG−IGα=0\sum \mathbf{M}_G - I_G \mathbf{\alpha} = 0. Here, −maG-m\mathbf{a}_G is the inertia force and −IGα-I_G \mathbf{\alpha} is the inertia couple. If these inertia effects are applied to the FBD, the body can be treated as being in a state of dynamic equilibrium. While mathematically identical to the FBD=KD method, the FBD=KD approach is generally preferred as it avoids sign errors associated with "reverse" inertia forces.

Mass Moment of Inertia (II)

The mass moment of inertia is a measure of a rigid body's resistance to angular acceleration about a given axis.

Mass Moment vs. Area Moment of Inertia

Do not confuse the mass moment of inertia (Im=∫r2dmI_m = \int r^2 dm, units: kg⋅m2kg \cdot m^2) used in dynamics with the area moment of inertia (Ix=∫y2dAI_x = \int y^2 dA, units: m4m^4 or in4in^4) used in mechanics of materials to calculate bending stresses. While mathematically similar, they represent entirely different physical properties.

Mass Moment of Inertia Integral

Defines the mass moment of inertia as the integral of the second moment of mass.

I=∫r2 dmI = \int r^2 \, dm

Variables

SymbolDescriptionUnit
IIMass moment of inertiakg·m²
rrPerpendicular distance from the axis of rotation to the mass element dmm
dmdmDifferential mass elementkg

Mass Moment of Inertia for Common Shapes

Interactive Simulation

Interact with the simulation below to explore mass moment of inertia for different shapes.

Mass Moment of Inertia & Parallel Axis Theorem

Shape & Forces

Parallel Axis Theorem

Centroidal Moment of Inertia:
IG=12mR2=0.5×8×12=4.000 kgâ‹…m2I_G = \frac{1}{2} m R^2 = 0.5 \times 8 \times 1^2 = 4.000\text{ kg}\cdot\text{m}^2
Parallel Axis Translation:
IO=IG+md2I_O = I_G + m d^2
IO=4.000+(8)(0.63)2=7.125 kgâ‹…m2I_O = 4.000 + (8)(0.63)^2 = 7.125\text{ kg}\cdot\text{m}^2
Separation d:0.63 m
Ang. Acceleration α:0.42 rad/s²
Spin Speed ω:0.0 rad/s
Parallel Axis Theorem Rotating WorkspaceGd = 0.63mAxis O
💡 Drag Axis O to translate the axis of rotation.
G (Centroid / Center of Mass)
O (Axis of Rotation / Pivot)
Distance d between G and O

Dynamics of Rolling

Rolling wheels, disks, and cylinders represent a common application of general plane motion. Analyzing rolling involves determining whether the body is slipping at its point of contact with the ground.

Rolling Motion Conditions

Assumption of Rolling Without Slipping

When solving rolling problems without explicitly knowing if slip occurs, assume rolling without slipping (aG=αra_G = \alpha r), solve for the required friction force (FfF_f), and check if Ff≤μsNF_f \le \mu_s N. If the inequality is violated, the assumption was wrong, the object slips, and you must re-solve using Ff=μkNF_f = \mu_k N.

Rotation about a Fixed Axis

For rotation about a fixed point OO (not necessarily the center of mass), the equations of motion can be simplified using the parallel axis theorem.

Fixed Axis Rotation Moment Equation

Relates the sum of moments about a fixed point to the angular acceleration.

∑MO=IOα\sum M_O = I_O \alpha

Variables

SymbolDescriptionUnit
∑MO\sum M_OSum of external moments about fixed axis ON·m
IOI_OMass moment of inertia about fixed axis Okg·m²
α\alphaAngular accelerationrad/s²

Parallel Axis Theorem

Relates the mass moment of inertia about an arbitrary axis to the mass moment of inertia about a parallel axis through the center of mass.

IO=IG+md2I_O = I_G + m d^2

Variables

SymbolDescriptionUnit
IOI_OMass moment of inertia about axis Okg·m²
IGI_GMass moment of inertia about parallel axis through center of mass Gkg·m²
mmTotal mass of the bodykg
ddPerpendicular distance between the two parallel axesm
Key Takeaways
  • Equations of Motion: ΣF=maG\Sigma F = m a_G and ΣMG=IGα\Sigma M_G = I_G \alpha
  • Kinetic Diagrams explicitly show the effective vectors maGm\mathbf{a}_G and IGαI_G\mathbf{\alpha}, establishing equivalence with the FBD.
  • Mass Moment of Inertia: I=∫r2dmI = \int r^2 dm
  • Parallel Axis Theorem (IO=IG+md2I_O = I_G + md^2) allows calculating II about any axis parallel to one through the center of mass.
  • For rotation about a fixed axis, summing moments about the axis is often simpler (ΣMO=IOα\Sigma M_O = I_O\alpha).
  • Free Body Diagrams must include reactions at supports (pins, rollers) to correctly determine forces and moments.