Back to All Subjects

Dynamics Of Rigid Bodies Simulations

A collection of interactive 3D visualizations and simulations to help you master concepts in dynamics of rigid bodies.

Kinematics of Particles - Theory & Concepts - Radial Transverse

Study of the geometry of motion of particles without considering the forces causing the motion, including rectilinear and curvilinear motion.

Radial and Transverse Kinematics Visualizer

Controls

Mathematical Kinematics

Velocity components:
vr=r˙=20.0 m/sv_r = \dot{r} = 20.0\text{ m/s}
vθ=rθ˙=(120)(0.50)=60.0 m/sv_\theta = r\dot{\theta} = (120)(0.50) = 60.0\text{ m/s}
v=vr2+vθ2=63.2 m/sv = \sqrt{v_r^2 + v_\theta^2} = 63.2\text{ m/s}
Acceleration (r¨=0,ddottheta=0\ddot{r}=0, \\ddot{\\theta}=0):
ar=−rθ˙2=−(120)(0.50)2=−30.0 m/s2a_r = -r\dot{\theta}^2 = -(120)(0.50)^2 = -30.0\text{ m/s}^2
aθ=2r˙θ˙=2(20)(0.50)=20.0 m/s2a_\theta = 2\dot{r}\dot{\theta} = 2(20)(0.50) = 20.0\text{ m/s}^2
a=ar2+aθ2=36.1 m/s2a = \sqrt{a_r^2 + a_\theta^2} = 36.1\text{ m/s}^2
Polar Grid Kinematics Space50m100m150m200m0°90°180°270°Ov_rv_θva_ra_θ
vrv_r (Radial Velocity)
vθv_\theta (Transverse Velocity)
vv (Total Velocity)
ara_r (Radial Accel.)
aθa_\theta (Transverse Accel.)

Kinetics of Particles: Force and Acceleration - Theory & Concepts - Orbital Mechanics

Application of Newton's Second Law to determine the motion of a particle subjected to unbalanced forces.

Kinetics of Particles: Work and Energy - Theory & Concepts - Work Energy

Application of the Principle of Work and Energy to solve kinetics problems involving displacement and speed, including conservative and non-conservative forces.

Work-Energy & Incline Simulator

Incline: 15° | μ_k: 0.10

Control Parameters

Incline Angle (θ\theta)15°
Mass (m)2.0 kg
Spring stiffness (k)500 N/m
Initial Compression (c)0.20 m
Friction (μk\mu_k)0.10

Work-Energy Conservation

T1+V1+U1→2=T2+V2T_1 + V_1 + U_{1 \to 2} = T_2 + V_2
T1=0 JT_1 = 0 \text{ J}
V1=PEsp,1=12(500)(0.20)2=10.0 JV_1 = PE_{sp, 1} = \frac{1}{2}(500)(0.20)^2 = 10.0 \text{ J}
U1→2=−Wf=−0.0 JU_{1 \to 2} = -W_f = -0.0 \text{ J}
Etotal,1=10.0−0.0=10.0 JE_{total, 1} = 10.0 - 0.0 = 10.0 \text{ J}
T2=KE=12(2)(0.00)2=0.0 JT_2 = KE = \frac{1}{2}(2)(0.00)^2 = 0.0 \text{ J}
V2=PEsp+PEg=10.0+0.0=10.0 JV_2 = PE_{sp} + PE_g = 10.0 + 0.0 = 10.0 \text{ J}
Etotal,2=0.0+10.0=10.0 JE_{total, 2} = 0.0 + 10.0 = 10.0 \text{ J}
Work-Energy Incline Systemx = 02.0kgFsfNmg sinθmg cosθmg-0.2m0.0m0.5m1.0m1.5m
Position (x):-0.200 m
Velocity (v):0.00 m/s
Height (h):0.00 m

Dynamic Energy Allocation (Joules)

PE Spring10.0J
PE Gravity0.0J
KE (Motion)0.0J
Friction Loss0.0J
Total E10.0J

Kinetics of Particles: Impulse and Momentum - Theory & Concepts - Collision

Application of the principle of linear impulse and momentum to solve problems involving force, mass, velocity, and time, including systems of particles.

Collision Impulse & Momentum Simulator

State: approaching

Collision Settings

Object A (Blue)
Mass (m1m_1)3.0 kg
Initial Velocity (v1v_1)3.0 m/s
Object B (Red)
Mass (m2m_2)2.0 kg
Initial Velocity (v2v_2)-2.0 m/s
Restitution (ee)0.70

Step-by-Step Solver

1. Momentum Conservation
m1v1+m2v2=m1v1′+m2v2′m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'
3 × 3.0 + 2 × -2.0 = 5.00 N·s
2. Restitution Relation
e(v1−v2)=v2′−v1′e(v_1 - v_2) = v_2' - v_1'
0.70 × (3.0 - (-2.0)) = 3.50 m/s
3. Solved Final Velocities
v1′=m1v1+m2v2−m2e(v1−v2)m1+m2v_1' = \frac{m_1 v_1 + m_2 v_2 - m_2 e(v_1 - v_2)}{m_1 + m_2}
v_1' = -0.40 m/s
v2′=v1′+e(v1−v2)v_2' = v_1' + e(v_1 - v_2)
v_2' = 3.10 m/s
Collision Space and PlotsCoMA3.0kgB2.0kg3.0 m/s-2.0 m/sx (pos)t (time)
Momentum Balance (Ns)
Initial Momentum:5.00
Final Momentum:5.00
Conservation verified: 100% constant
Energy Balance (Joules)
Initial KE:17.50 J
Final KE:9.85 J
Thermal Loss:7.65 J

Kinetics of Particles: Impulse and Momentum - Theory & Concepts - Coefficient Of Restitution

Application of the principle of linear impulse and momentum to solve problems involving force, mass, velocity, and time, including systems of particles.

Coefficient of Restitution Simulator

e: 0.75 | h_0: 4.0 m

Simulation Settings

Restitution Coeff (ee)0.75
Release Height (h0)4.0 m

Mathematical Derivation

Coefficient of Restitution
e=−v+v−e = -\frac{v^+}{v^-}
Waiting for first bounce...
Peak Height Decay
hnext=e2hprevh_{next} = e^2 h_{prev}
Initial Height: h0=4.00 mh_0 = 4.00 \text{ m}
Current Bounces: 0
Next Max Height: h1=4.00 mh_{1} = 4.00 \text{ m}
Coefficient of Restitution System0m1m2m3m4m5mRelease (h0 = 4.0m)4.00mHeight (m)Time (s)0s2s4s6s8s
Current Time0.00 s
Height (y)4.00 m
Velocity (v)0.00 m/s

Kinematics of Rigid Bodies - Theory & Concepts - Rigid Body Translation

Analysis of the motion of solid bodies where distances between any two points remain constant, including translation, rotation, and general plane motion.

Rigid Body Translation & Rotation

Motion Control

Kinematics Equation Solver

Governing Principle:

In translation, all lines in the body remain parallel. Thus:

vA=vBandaA=aB\mathbf{v}_A = \mathbf{v}_B \quad \text{and} \quad \mathbf{a}_A = \mathbf{a}_B
Live Values:
Point A
v: 96.0 m/s
a: 0.0 m/s²
Point B
v: 96.0 m/s
a: 0.0 m/s²
✓ Verified: Point A and Point B have identical motion vectors.
Rigid Body Motion Trajectory WorkspaceABv_Av_B
Velocity at A (v_A)
Velocity at B (v_B)
Acceleration (a_A)
Acceleration (a_B)

Kinematics of Rigid Bodies - Theory & Concepts - I C

Analysis of the motion of solid bodies where distances between any two points remain constant, including translation, rotation, and general plane motion.

Instantaneous Center of Zero Velocity (IC) Visualizer

Mechanism Setup

s < 0 (Braking)s = 0 (Pure)s > 0 (Spin/Skid)

Mathematical Equations

Wheel IC Location & Slip Mechanics:

IC offset from center O is given by slip ratio:

yIC−yO=R(1+s)=2.0(1+0.00)=2.00 my_{IC} - y_O = R(1 + s) = 2.0(1 + 0.00) = 2.00\text{ m}

Pure rolling: IC is exactly at the point of contact.

Dragged Point P Velocity:
vP=ω×rP/ICv_P = \omega \times r_{P/IC}
d from IC: 3.27 m|v_P: 6.55 m/s
Instantaneous Center Geometric WorkspaceCenter OContactv_Ov_AICP (Drag)v_P
💡 Drag point P to analyze local velocity.
IC (Instant Center, v=0)
Velocity vector v_P
Distance ray r_P/IC

Kinetics of Rigid Bodies: Force and Acceleration - Theory & Concepts - Mass Moment Of Inertia

Application of Newton's Second Law and Euler's Equations to determine the general plane motion of rigid bodies.

Mass Moment of Inertia & Parallel Axis Theorem

Shape & Forces

Parallel Axis Theorem

Centroidal Moment of Inertia:
IG=12mR2=0.5×8×12=4.000 kg⋅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 kg⋅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

Kinetics of Rigid Bodies: Work and Energy - Theory & Concepts - Rolling Sphere

Energy methods applied to rigid bodies, including rotational kinetic energy, potential energy of rigid bodies, and conservation of energy.

Rigid Body Incline Rolling (Inertia Race)

Ramp & Incline

Dynamics Equations

Acceleration Formula:
a=gsin⁡θ1+βa = \frac{g \sin\theta}{1 + \beta}
a₁: 3.50 m/s² (t_end: 2.39s)
Speed at bottom:
vmax=2gLsin⁡θ1+βv_{max} = \sqrt{\frac{2 g L \sin\theta}{1 + \beta}}
Shape 1 v_max:8.37 m/s
Ramp Roll Incline Workspaceθmgmg sinθNf_s
Energy Conservation BalanceTotal: 98.1 J
PE: 100%

Kinetics of Rigid Bodies: Impulse and Momentum - Theory & Concepts - Bullet Rod

Analysis of impulsive forces and moments acting on rigid bodies, introducing angular momentum and the principle of impulse and momentum.

Angular Momentum: Bullet Striking a Rod

System Parameters

Angular Momentum Solver

Conservation of HOH_O:
mv0d=Itotal ωm v_0 d = I_{\text{total}}\,\omega
Itotal=13ML2+md2I_{\text{total}} = \tfrac{1}{3}ML^2 + md^2
=1.333+0.0025=1.3358 kg⋅m2= 1.333 + 0.0025 = 1.3358\,\text{kg·m}^2
H0=2.0000 kg⋅m2/sH_0 = 2.0000\,\text{kg·m}^2/\text{s}
ω=1.50 rad/s\omega = 1.50\,\text{rad/s}
KE before:800.0 J
KE after:1.5 J
Energy lost:798.5 J (99.8%)
Bullet–Rod Angular Momentum WorkspaceO (Pivot)Ld = 0.50mReady — click Fire!
Bullet (m = 0.010 kg)
Rod (M = 4 kg)
Impact point d = 0.50 m

Three-Dimensional Kinematics of Rigid Bodies - Theory & Concepts - Coriolis

Analysis of the motion of rigid bodies in three dimensions, including translation, rotation about a fixed axis, general motion, and Euler's theorem.

Coriolis Effect — Rotating Reference Frame

Controls

CCW (−)CW (+)
Coriolis Deflection Preview
Trajectory previewFixedRotating

Coriolis Equations

Rotating Frame Acceleration:
aabs=arel+Ω˙×r+Ω×(Ω×r)+2Ω×vrel\mathbf{a}_{\text{abs}} = \mathbf{a}_{\text{rel}} + \dot{\boldsymbol{\Omega}} \times \mathbf{r} + \boldsymbol{\Omega} \times (\boldsymbol{\Omega} \times \mathbf{r}) + 2\boldsymbol{\Omega} \times \mathbf{v}_{\text{rel}}
Coriolis Term:
aCor=2Ω×vrel\mathbf{a}_{\text{Cor}} = 2\boldsymbol{\Omega} \times \mathbf{v}_{\text{rel}}
∣aCor∣=2×1.5×2.0=6.00 m/s2|\mathbf{a}_{\text{Cor}}| = 2 \times 1.5 \times 2.0 = 6.00\,\text{m/s}^2
Fixed observer: particle moves straight
Rotating observer: particle curves
🌍 Fixed (Inertial) Observer

Fixed: Particle travels in straight line; disc rotates under it.

Rotating: Particle appears to deflect — this is the Coriolis pseudo-force.

Drag to orbit view.

Three-Dimensional Kinetics of Rigid Bodies - Theory & Concepts - Gyroscopic

Study of the relationships between the forces and moments acting on rigid bodies and the resulting 3D motion, including Eulerian angles, gyroscopic motion, and torque-free motion.

Gyroscopic Precession & Torque Dynamics

Controls

Equations

Rotor Moment of Inertia:
Is=12mR2=0.5×1.5×0.42=0.120 kg⋅m2I_s = \frac{1}{2} m R^2 = 0.5 \times 1.5 \times 0.4^2 = 0.120\,\text{kg}\cdot\text{m}^2
Gravity Torque Magnitude:
τ=mgdsin⁡θ=1.5×9.81×0.35×sin⁡(60∘)=4.46 N⋅m\tau = m g d \sin\theta = 1.5 \times 9.81 \times 0.35 \times \sin(60^\circ) = 4.46\,\text{N}\cdot\text{m}
Spin Angular Momentum:
Hs=Isωs=0.120×20.0=2.40 kg⋅m2/sH_s = I_s \omega_s = 0.120 \times 20.0 = 2.40\,\text{kg}\cdot\text{m}^2/\text{s}
Precession Rate (steady):
Ω=τHssin⁡θ=mgdIsωs=5.1502.40=2.15 rad/s\Omega = \frac{\tau}{H_s \sin\theta} = \frac{m g d}{I_s \omega_s} = \frac{5.150}{2.40} = 2.15\,\text{rad/s}
Precession Vector (Ω)
Angular Momentum (H)
Gravity Torque Vector (τ)

Concept: The gravitational force creates a torque τ\tau perpendicular to the spin axis.

Instead of falling down, the gyroscope precesses horizontally because the torque changes the direction of the spin angular momentum vector H\mathbf{H}.

Drag to rotate the view camera.

Mechanical Vibrations - Theory & Concepts

Introduction to free and forced vibrations of single-degree-of-freedom systems, including torsional vibrations and damping.

Mechanical Vibrations — Mass-Spring-Damper SystemUnderdamped (ζ = 0.06)

System Parameters

Vibration Equations

Equation of Motion:
mx¨+cx˙+kx=0m\ddot{x} + c\dot{x} + kx = 0
ωn=km=3.16 rad/s\omega_n = \sqrt{\frac{k}{m}} = 3.16\,\text{rad/s}
ζ=ccc=c2km=0.063\zeta = \frac{c}{c_c} = \frac{c}{2\sqrt{km}} = 0.063
ωd=ωn1−ζ2=3.16 rad/s\omega_d = \omega_n\sqrt{1-\zeta^2} = 3.16\,\text{rad/s}
x(t):1.0000 m
ẋ(t):-0.0250 m/s
ẍ(t):-4.993 m/s²
t:0.00 s
Mechanical Vibration Model and Responsekceq.m+1.00mt(s)x(m)036912-1.00.01.0UnderdampedModelResponse x(t)

Mechanical Vibrations - Theory & Concepts - Pendulum

Introduction to free and forced vibrations of single-degree-of-freedom systems, including torsional vibrations and damping.

Simple Pendulum — Exact Numerical Integration

Pendulum Setup

Dynamics Equations

Equation of Motion:
θ¨+gLsin⁡θ=0\ddot{\theta} + \frac{g}{L}\sin\theta = 0
Small-Angle Period:
T=2πLg=2.457 sT = 2\pi\sqrt{\frac{L}{g}} = 2.457\,\text{s}
θ (current):30.0°
ω\omega:0.000 rad/s
TT (current):0.00 s
Normal (NN):16.99 N
Pendulum Motion Workspacem30.0°Method: RK4 integration (exact for large θ)
Energy Balance
PE3.94 J
KE0.00 J
Total E₀3.94 J
Phase Portrait (θ, ω)Phase portraitθω