Kinetics of Rigid Bodies: Force and Acceleration

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

∑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
Where:
  • GG represents the center of mass.
  • mm is the mass of the rigid body.
  • aGa_G is the acceleration of the center of mass.
  • IGI_G is the mass moment of inertia about the center of mass.
  • α\alpha is the angular acceleration.

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:
∑MP (from FBD)=∑(Mk)P (from KD)\sum M_P \text{ (from FBD)} = \sum (M_k)_P \text{ (from KD)}
Where ∑(Mk)P\sum (M_k)_P is the sum of the moments of the kinetic vectors (maGm\mathbf{a}_G and IGαI_G \mathbf{\alpha}) about point PP.

Important

D'Alembert's Principle: 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 of Inertia (II)

A measure of a body's resistance to angular acceleration.
I=∫r2 dmI = \int r^2 \, dm
Common Shapes:
  • Slender Rod (about end): I=13mL2I = \frac{1}{3} mL^2
  • Slender Rod (about center): IG=112mL2I_G = \frac{1}{12} mL^2
  • Thin Disk (about center): IG=12mr2I_G = \frac{1}{2} mr^2
  • Solid Sphere (about center): IG=25mr2I_G = \frac{2}{5} mr^2
  • Rectangular Plate (about center): IG=112m(a2+b2)I_G = \frac{1}{12} m(a^2 + b^2)
Interact with the simulation below to explore mass moment of inertia for different shapes.

Mass Moment of Inertia Calculator

I_G = 1/2 m r^2
Moment of Inertia (IGI_G)1.250kg·m²

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 With vs. Without Slipping

  • Rolling Without Slipping: The point of contact with the ground (CC) acts as an instantaneous center of zero velocity (vC=0v_C = 0). The acceleration of the center of mass is strictly tied to the angular acceleration: aG=αra_G = \alpha r. The friction force FfF_f is a static friction force. It is determined from the equations of motion and must satisfy the condition: Ff≤μsNF_f \le \mu_s N.
  • Rolling With Slipping: The point of contact is sliding (vC≠0v_C \neq 0), so aGa_G and α\alpha are independent. The kinematic relationship aG=αra_G = \alpha r no longer holds. Instead, the friction force is kinetic and acts opposite to the direction of slip: Ff=μkNF_f = \mu_k N.

Important

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):

Fixed Axis Rotation

∑MO=IOα\sum M_O = I_O \alpha
Where IOI_O is the moment of inertia about the fixed axis OO. The Parallel Axis Theorem relates IOI_O to IGI_G:
IO=IG+md2I_O = I_G + m d^2
Where dd is the distance between the center of mass GG and the axis OO.
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.
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