Dynamics

Robot dynamics relates joint torques to motion. Learn gravity, inertia, Coriolis and centrifugal terms, the manipulator equation, Lagrange and Newton–Euler methods, payload effects, and how to size joint torques for a duty cycle.

Kinematics tells us where the robot goes; dynamics tells us what torques the motors must provide to get there — against gravity, inertia and the coupling between moving links.

Dynamic models are used for actuator sizing, model-based control (feedforward), simulation and collision detection.

An analogy

Swinging a heavy bag on a long arm is much harder than lifting it close to your body: distance from the joint multiplies both the gravity torque and the inertia.

What joints must overcome

Gravity (holding the arm and payload up), inertia (accelerating masses), friction (gearbox, seals) and process forces (cutting, pressing). Gravity torque is largest when the arm is horizontal and stretched out.

Coupled dynamics

Moving one joint creates torques on others: centrifugal forces from rotation, Coriolis effects from combined joint speeds, and changing inertia as the arm reconfigures. The mass matrix M(q) depends on configuration — a stretched arm is harder to rotate.

Lagrange and Newton–Euler

Lagrange's method derives equations from kinetic and potential energy — clean for analysis. The recursive Newton–Euler algorithm propagates velocities outward and forces inward — efficient (O(n)) for real-time control.

Using the model

Computed-torque control cancels nonlinear dynamics; feedforward torque reduces tracking error; collision detection compares measured and predicted torques. Parameters are identified from excitation trajectories. Payload changes (mass, CG, inertia) must be entered correctly on industrial robots for good performance and safety.

Formulas

Worked example: Holding torque

Given: Link 5 kg, CG at 0.2 m; Payload 2 kg at 0.4 m; Arm horizontal

Find: Gravity torque at the shoulder.

Answer: ≈ 17.7 N·m

Worked example: Adding acceleration

Given: Link as rod L = 0.4 m, 5 kg; payload 2 kg at 0.4 m; α = 5 rad/s²

Find: Total torque.

Answer: ≈ 20.6 N·m

Common mistakes

Applications

Key takeaways

Going further