Kinetics of Rigid Bodies: Force and Acceleration

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.

\sum F_x = m(a_G)_x

\sum F_y = m(a_G)_y

\sum M_G = I_G \alpha

Variables

Symbol	Description	Unit
$\sum F_x, \sum F_y$	Sum of external forces in x and y directions	N
$m$	Mass of the rigid body	kg
$(a_G)_x, (a_G)_y$	Acceleration components of the center of mass G	m/s²
$\sum M_G$	Sum of external moments about the center of mass G	N·m
$I_G$	Mass moment of inertia about the center of mass G	kg·m²
$\alpha$	Angular acceleration	rad/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 $\sum \mathbf{F} = m\mathbf{a}_G$ and $\sum \mathbf{M}_G = I_G \mathbf{\alpha}$ 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 ( $m\mathbf{a}_G$ ) applied at the mass center $G$ , and the "effective" couple moment ( $I_G \mathbf{\alpha}$ ). The vector $m\mathbf{a}_G$ must be oriented in the actual direction of acceleration, and $I_G \mathbf{\alpha}$ 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 $P$ , not just the mass center.

Free-Body and Kinetic Diagrams

Keep applied actions on the free-body diagram and the rigid body's inertial response on the kinetic diagram.

The diagrams correspond through ΣF = m a_G and ΣM_G = I_G α; the kinetic vectors follow the actual acceleration directions.

Free-body diagram

Applied forces, support reactions, weight, and couples
Use external loads only on the isolated body

Kinetic diagram

m a_G at the mass center G
I_G α as the kinetic couple with the chosen angular sign

$Applying the FBD=KD Method$

STEP-BY-STEP

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

Moment Equivalence Equation

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

\sum M_P \text{ (from FBD)} = \sum (M_k)_P \text{ (from KD)}

Variables

Symbol	Description	Unit
$\sum M_P$	Sum of moments of all external forces and couples about point P	N·m
$\sum (M_k)_P$	Sum of the moments of the kinetic vectors (m\mathbf{a}_G and I_G\mathbf{\alpha}) about point P	N·m

$D'Alembert's Principle Usage$

Similar to particles, we can rewrite the equations as $\sum \mathbf{F} - m\mathbf{a}_G = 0$ and $\sum \mathbf{M}_G - I_G \mathbf{\alpha} = 0$ . Here, $-m\mathbf{a}_G$ is the inertia force and $-I_G \mathbf{\alpha}$ 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 ( $I$ )

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 ( $I_m = \int r^2 dm$ , units: $kg \cdot m^2$ ) used in dynamics with the area moment of inertia ( $I_x = \int y^2 dA$ , units: $m^4$ or $in^4$ ) 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 = \int r^2 \, dm

Variables

Symbol	Description	Unit
$I$	Mass moment of inertia	kg·m²
$r$	Perpendicular distance from the axis of rotation to the mass element dm	m
$dm$	Differential mass element	kg

$Mass Moment of Inertia for Common Shapes$

Slender Rod (about end): $I = \frac{1}{3} mL^2$
Slender Rod (about center): $I_G = \frac{1}{12} mL^2$
Thin Disk (about center): $I_G = \frac{1}{2} mr^2$
Solid Sphere (about center): $I_G = \frac{2}{5} mr^2$
Rectangular Plate (about center): $I_G = \frac{1}{12} m(a^2 + b^2)$

$Interactive Simulation$

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

Mass Moment of Inertia & Parallel Axis Theorem

Shape & Forces

Body Shape

Mass (m)8.0 kg

Radius (R)1.00 m

Torque (

\tau

)3.0 N·m

Parallel Axis Theorem

Centroidal Moment of Inertia:

I_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:

I_O = I_G + m d^2

I_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

💡 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.

$Rolling Motion Conditions$

Rolling Without Slipping: For a body rolling on a fixed surface, the material point at contact has zero velocity at that instant. On a horizontal surface, with $+x$ to the right and counterclockwise-positive angular quantities, a disk satisfies $v_{Gx} = -\omega R$ and $a_{Gx} = -\alpha R$ . The signed relation changes with the chosen convention. Static friction is solved from the equations of motion; its direction may be either way, and the solution is admissible only when $|F_s| \le \mu_s N$ while contact is maintained.
Rolling With Slipping: The contact point has nonzero velocity relative to the fixed surface, so the no-slip kinematic constraint is not applied. Under the Coulomb model, kinetic friction has magnitude $\mu_k N$ and acts opposite the relative slip direction.

$Assumption of Rolling Without Slipping$

When solving rolling problems without explicitly knowing if slip occurs, assume rolling without slipping and use signed contact kinematics consistent with the coordinate system. Solve for the required static friction without presuming its direction, then check $|F_s| \le \mu_s N$ and verify that the normal reaction maintains contact. If static-friction capacity is exceeded, remove the no-slip constraint and re-solve with kinetic friction opposing the relative slip.

Rolling Without Slip: Assumption and Friction Check

Solve the no-slip model with static friction, test contact and friction capacity, and revise the model if slip occurs.

Use signed kinematic relations consistent with the chosen positive rotation. For a disk rolling on a fixed horizontal surface, the contact point has zero velocity while it touches the surface; its acceleration is generally not zero. Do not set static friction equal to its limiting value unless impending slip is part of the stated condition. If sliding begins, the Coulomb kinetic-friction model uses magnitude μ_k N opposite the relative slip and the no-slip constraint is removed.

Rolling Contact and Friction States

Compare the admissible rolling constraint with sliding contact and distinguish their friction laws.

For a disk on a fixed horizontal surface, with x positive right and θ positive counterclockwise, v_Gx = −ωR and a_Gx = −αR. Zero contact velocity does not mean the material point's acceleration is zero.

Rolling without slip

Contact-point velocity is zero at the instant of contact
Solve signed static friction; require |F_s| ≤ μ_s N

Sliding contact

Relative slip at contact is nonzero
Use kinetic friction opposite slip; remove the no-slip constraint

Rotation about a Fixed Axis

For rotation about a fixed point $O$ (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.

\sum M_O = I_O \alpha

Variables

Symbol	Description	Unit
$\sum M_O$	Sum of external moments about fixed axis O	N·m
$I_O$	Mass moment of inertia about fixed axis O	kg·m²
$\alpha$	Angular acceleration	rad/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.

I_O = I_G + m d^2

Variables

Symbol	Description	Unit
$I_O$	Mass moment of inertia about axis O	kg·m²
$I_G$	Mass moment of inertia about parallel axis through center of mass G	kg·m²
$m$	Total mass of the body	kg
$d$	Perpendicular distance between the two parallel axes	m

Parallel-Axis Shift for Mass Moment of Inertia

Relate inertia about the mass-center axis to an axis parallel to it through point O.

The parallel-axis theorem applies to parallel axes; d is their perpendicular separation.

Axis through G

Centroidal mass moment: I_G
The reference axis passes through the mass center

Parallel axis through O

Perpendicular axis separation is d
Offset-axis inertia is I_O = I_G + m d²

Key Takeaways

Equations of Motion: $\Sigma F = m a_G$ and $\Sigma M_G = I_G \alpha$
Kinetic Diagrams explicitly show the effective vectors $m\mathbf{a}_G$ and $I_G\mathbf{\alpha}$ , establishing equivalence with the FBD.
Mass Moment of Inertia: $I = \int r^2 dm$
Parallel Axis Theorem ( $I_O = I_G + md^2$ ) allows calculating $I$ 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 ( $\Sigma M_O = I_O\alpha$ ).
Free Body Diagrams must include reactions at supports (pins, rollers) to correctly determine forces and moments.