Rigid body dynamics

Orientation or attitude descriptions

Several methods to describe orientations of a rigid body in three dimensions have been developed. They are summarized in the following sections.

Tait–Bryan angles

Main article: Euler angles § Tait–Bryan angles

These are three angles, also known as yaw, pitch and roll, Navigation angles and Cardan angles. Mathematically they constitute a set of six possibilities inside the twelve possible sets of Euler angles, the ordering being the one best used for describing the orientation of a vehicle such as an airplane. In aerospace engineering they are usually referred to as Euler angles.

Newton's second law in three dimensions

To consider rigid body dynamics in three-dimensional space, Newton's second law must be extended to define the relationship between the movement of a rigid body and the system of forces and torques that act on it.

Newton formulated his second law for a particle as, "The change of motion of an object is proportional to the force impressed and is made in the direction of the straight line in which the force is impressed."^[3] Because Newton generally referred to mass times velocity as the "motion" of a particle, the phrase "change of motion" refers to the mass times acceleration of the particle, and so this law is usually written as $${\displaystyle \mathbf {F} =m\mathbf {a} ,}$$ where F is understood to be the only external force acting on the particle, m is the mass of the particle, and a is its acceleration vector. The extension of Newton's second law to rigid bodies is achieved by considering a rigid system of particles.

Rigid system of particles

If a system of N particles, P_i, i=1,...,N, are assembled into a rigid body, then Newton's second law can be applied to each of the particles in the body. If F_i is the external force applied to particle P_i with mass m_i, then $${\displaystyle \mathbf {F} _{i}+\sum _{j=1}^{N}\mathbf {F} _{ij}=m_{i}\mathbf {a} _{i},\quad i=1,\ldots ,N,}$$ where F_ij is the internal force of particle P_j acting on particle P_i that maintains the constant distance between these particles.

An important simplification to these force equations is obtained by introducing the resultant force and torque that acts on the rigid system. This resultant force and torque is obtained by choosing one of the particles in the system as a reference point, R, where each of the external forces are applied with the addition of an associated torque. The resultant force F and torque T are given by the formulas, $${\displaystyle \mathbf {F} =\sum _{i=1}^{N}\mathbf {F} _{i},\quad \mathbf {T} =\sum _{i=1}^{N}(\mathbf {R} _{i}-\mathbf {R} )\times \mathbf {F} _{i},}$$ where R_i is the vector that defines the position of particle P_i.

Newton's second law for a particle combines with these formulas for the resultant force and torque to yield, $${\displaystyle \mathbf {F} =\sum _{i=1}^{N}m_{i}\mathbf {a} _{i},\quad \mathbf {T} =\sum _{i=1}^{N}(\mathbf {R} _{i}-\mathbf {R} )\times (m_{i}\mathbf {a} _{i}),}$$ where the internal forces F_ij cancel in pairs. The kinematics of a rigid body yields the formula for the acceleration of the particle P_i in terms of the position R and acceleration a of the reference particle as well as the angular velocity vector ω and angular acceleration vector α of the rigid system of particles as, $${\displaystyle \mathbf {a} _{i}=\alpha \times (\mathbf {R} _{i}-\mathbf {R} )+\omega \times (\omega \times (\mathbf {R} _{i}-\mathbf {R} ))+\mathbf {a} .}$$

Mass properties

The mass properties of the rigid body are represented by its center of mass and inertia matrix. Choose the reference point R so that it satisfies the condition $${\displaystyle \sum _{i=1}^{N}m_{i}(\mathbf {R} _{i}-\mathbf {R} )=0,}$$

then it is known as the center of mass of the system.

The inertia matrix [I_R] of the system relative to the reference point R is defined by $${\displaystyle [I_{R}]=\sum _{i=1}^{N}m_{i}\left(\mathbf {I} \left(\mathbf {S} _{i}^{\textsf {T}}\mathbf {S} _{i}\right)-\mathbf {S} _{i}\mathbf {S} _{i}^{\textsf {T}}\right),}$$

where ${\displaystyle \mathbf {S} _{i}}$ is the column vector R_i − R; ${\displaystyle \mathbf {S} _{i}^{\textsf {T}}}$ is its transpose, and ${\displaystyle \mathbf {I} }$ is the 3 by 3 identity matrix.

${\displaystyle \mathbf {S} _{i}^{\textsf {T}}\mathbf {S} _{i}}$ is the scalar product of ${\displaystyle \mathbf {S} _{i}}$ with itself, while ${\displaystyle \mathbf {S} _{i}\mathbf {S} _{i}^{\textsf {T}}}$ is the tensor product of ${\displaystyle \mathbf {S} _{i}}$ with itself.

Force-torque equations

Using the center of mass and inertia matrix, the force and torque equations for a single rigid body take the form $${\displaystyle \mathbf {F} =m\mathbf {a} ,\quad \mathbf {T} =[I_{R}]\alpha +\omega \times [I_{R}]\omega ,}$$ and are known as Newton's second law of motion for a rigid body.

The dynamics of an interconnected system of rigid bodies, B_i, j = 1, ..., M, is formulated by isolating each rigid body and introducing the interaction forces. The resultant of the external and interaction forces on each body, yields the force-torque equations $${\displaystyle \mathbf {F} _{j}=m_{j}\mathbf {a} _{j},\quad \mathbf {T} _{j}=[I_{R}]_{j}\alpha _{j}+\omega _{j}\times [I_{R}]_{j}\omega _{j},\quad j=1,\ldots ,M.}$$

Newton's formulation yields 6M equations that define the dynamics of a system of M rigid bodies.^[4]

Rotation in three dimensions

A rotating object, whether under the influence of torques or not, may exhibit the behaviours of precession and nutation. The fundamental equation describing the behavior of a rotating solid body is Euler's equation of motion: $${\displaystyle {\boldsymbol {\tau }}={\frac {D\mathbf {L} }{Dt}}={\frac {d\mathbf {L} }{dt}}+{\boldsymbol {\omega }}\times \mathbf {L} ={\frac {d(I{\boldsymbol {\omega }})}{dt}}+{\boldsymbol {\omega }}\times {I{\boldsymbol {\omega }}}=I{\boldsymbol {\alpha }}+{\boldsymbol {\omega }}\times {I{\boldsymbol {\omega }}}}$$ where the pseudovectors τ and L are, respectively, the torques on the body and its angular momentum, the scalar I is its moment of inertia, the vector ω is its angular velocity, the vector α is its angular acceleration, D is the differential in an inertial reference frame and d is the differential in a relative reference frame fixed with the body.

The solution to this equation when there is no applied torque is discussed in the articles Euler's equation of motion and Poinsot's ellipsoid.

It follows from Euler's equation that a torque τ applied perpendicular to the axis of rotation, and therefore perpendicular to L, results in a rotation about an axis perpendicular to both τ and L. This motion is called precession. The angular velocity of precession Ω_P is given by the cross product:^{[citation needed]} $${\displaystyle {\boldsymbol {\tau }}={\boldsymbol {\Omega }}_{\mathrm {P} }\times \mathbf {L} .}$$

Precession can be demonstrated by placing a spinning top with its axis horizontal and supported loosely (frictionless toward precession) at one end. Instead of falling, as might be expected, the top appears to defy gravity by remaining with its axis horizontal, when the other end of the axis is left unsupported and the free end of the axis slowly describes a circle in a horizontal plane, the resulting precession turning. This effect is explained by the above equations. The torque on the top is supplied by a couple of forces: gravity acting downward on the device's centre of mass, and an equal force acting upward to support one end of the device. The rotation resulting from this torque is not downward, as might be intuitively expected, causing the device to fall, but perpendicular to both the gravitational torque (horizontal and perpendicular to the axis of rotation) and the axis of rotation (horizontal and outwards from the point of support), i.e., about a vertical axis, causing the device to rotate slowly about the supporting point.

Under a constant torque of magnitude τ, the speed of precession Ω_P is inversely proportional to L, the magnitude of its angular momentum: $${\displaystyle \tau ={\mathit {\Omega }}_{\mathrm {P} }L\sin \theta ,}$$ where θ is the angle between the vectors Ω_P and L. Thus, if the top's spin slows down (for example, due to friction), its angular momentum decreases and so the rate of precession increases. This continues until the device is unable to rotate fast enough to support its own weight, when it stops precessing and falls off its support, mostly because friction against precession cause another precession that goes to cause the fall.

By convention, these three vectors - torque, spin, and precession - are all oriented with respect to each other according to the right-hand rule.

Virtual work

Work is computed from the dot product of each force with the displacement of its point of contact $${\displaystyle \delta W=\sum _{i=1}^{n}\mathbf {F} _{i}\cdot \delta \mathbf {r} _{i}.}$$ If the trajectory of a rigid body is defined by a set of generalized coordinates q_j, j = 1, ..., m, then the virtual displacements δr_i are given by $${\displaystyle \delta \mathbf {r} _{i}=\sum _{j=1}^{m}{\frac {\partial \mathbf {r} _{i}}{\partial q_{j}}}\delta q_{j}=\sum _{j=1}^{m}{\frac {\partial \mathbf {V} _{i}}{\partial {\dot {q}}_{j}}}\delta q_{j}.}$$ The virtual work of this system of forces acting on the body in terms of the generalized coordinates becomes $${\displaystyle \delta W=\mathbf {F} _{1}\cdot \left(\sum _{j=1}^{m}{\frac {\partial \mathbf {V} _{1}}{\partial {\dot {q}}_{j}}}\delta q_{j}\right)+\dots +\mathbf {F} _{n}\cdot \left(\sum _{j=1}^{m}{\frac {\partial \mathbf {V} _{n}}{\partial {\dot {q}}_{j}}}\delta q_{j}\right)}$$

or collecting the coefficients of δq_j $${\displaystyle \delta W=\left(\sum _{i=1}^{n}\mathbf {F} _{i}\cdot {\frac {\partial \mathbf {V} _{i}}{\partial {\dot {q}}_{1}}}\right)\delta q_{1}+\dots +\left(\sum _{1=1}^{n}\mathbf {F} _{i}\cdot {\frac {\partial \mathbf {V} _{i}}{\partial {\dot {q}}_{m}}}\right)\delta q_{m}.}$$

Generalized forces

For simplicity consider a trajectory of a rigid body that is specified by a single generalized coordinate q, such as a rotation angle, then the formula becomes $${\displaystyle \delta W=\left(\sum _{i=1}^{n}\mathbf {F} _{i}\cdot {\frac {\partial \mathbf {V} _{i}}{\partial {\dot {q}}}}\right)\delta q=\left(\sum _{i=1}^{n}\mathbf {F} _{i}\cdot {\frac {\partial ({\boldsymbol {\omega }}\times (\mathbf {R} _{i}-\mathbf {R} )+\mathbf {V} )}{\partial {\dot {q}}}}\right)\delta q.}$$

Introduce the resultant force F and torque T so this equation takes the form $${\displaystyle \delta W=\left(\mathbf {F} \cdot {\frac {\partial \mathbf {V} }{\partial {\dot {q}}}}+\mathbf {T} \cdot {\frac {\partial {\boldsymbol {\omega }}}{\partial {\dot {q}}}}\right)\delta q.}$$

The quantity Q defined by $${\displaystyle Q=\mathbf {F} \cdot {\frac {\partial \mathbf {V} }{\partial {\dot {q}}}}+\mathbf {T} \cdot {\frac {\partial {\boldsymbol {\omega }}}{\partial {\dot {q}}}},}$$

is known as the generalized force associated with the virtual displacement δq. This formula generalizes to the movement of a rigid body defined by more than one generalized coordinate, that is $${\displaystyle \delta W=\sum _{j=1}^{m}Q_{j}\delta q_{j},}$$ where $${\displaystyle Q_{j}=\mathbf {F} \cdot {\frac {\partial \mathbf {V} }{\partial {\dot {q}}_{j}}}+\mathbf {T} \cdot {\frac {\partial {\boldsymbol {\omega }}}{\partial {\dot {q}}_{j}}},\quad j=1,\ldots ,m.}$$

It is useful to note that conservative forces such as gravity and spring forces are derivable from a potential function V(q₁, ..., q_n), known as a potential energy. In this case the generalized forces are given by $${\displaystyle Q_{j}=-{\frac {\partial V}{\partial q_{j}}},\quad j=1,\ldots ,m.}$$

Static equilibrium

The static equilibrium of a mechanical system rigid bodies is defined by the condition that the virtual work of the applied forces is zero for any virtual displacement of the system. This is known as the principle of virtual work.^[5] This is equivalent to the requirement that the generalized forces for any virtual displacement are zero, that is Q_i=0.

Let a mechanical system be constructed from n rigid bodies, B_i, i = 1, ..., n, and let the resultant of the applied forces on each body be the force-torque pairs, F_i and T_i, i = 1, ..., n. Notice that these applied forces do not include the reaction forces where the bodies are connected. Finally, assume that the velocity V_i and angular velocities ω_i, i = 1, ..., n, for each rigid body, are defined by a single generalized coordinate q. Such a system of rigid bodies is said to have one degree of freedom.

The virtual work of the forces and torques, F_i and T_i, applied to this one degree of freedom system is given by $${\displaystyle \delta W=\sum _{i=1}^{n}\left(\mathbf {F} _{i}\cdot {\frac {\partial \mathbf {V} _{i}}{\partial {\dot {q}}}}+\mathbf {T} _{i}\cdot {\frac {\partial {\boldsymbol {\omega }}_{i}}{\partial {\dot {q}}}}\right)\delta q=Q\delta q,}$$ where $${\displaystyle Q=\sum _{i=1}^{n}\left(\mathbf {F} _{i}\cdot {\frac {\partial \mathbf {V} _{i}}{\partial {\dot {q}}}}+\mathbf {T} _{i}\cdot {\frac {\partial {\boldsymbol {\omega }}_{i}}{\partial {\dot {q}}}}\right),}$$ is the generalized force acting on this one degree of freedom system.

If the mechanical system is defined by m generalized coordinates, q_j, j = 1, ..., m, then the system has m degrees of freedom and the virtual work is given by, $${\displaystyle \delta W=\sum _{j=1}^{m}Q_{j}\delta q_{j},}$$ where $${\displaystyle Q_{j}=\sum _{i=1}^{n}\left(\mathbf {F} _{i}\cdot {\frac {\partial \mathbf {V} _{i}}{\partial {\dot {q}}_{j}}}+\mathbf {T} _{i}\cdot {\frac {\partial {\boldsymbol {\omega }}_{i}}{\partial {\dot {q}}_{j}}}\right),\quad j=1,\ldots ,m.}$$ is the generalized force associated with the generalized coordinate q_j. The principle of virtual work states that static equilibrium occurs when these generalized forces acting on the system are zero, that is $${\displaystyle Q_{j}=0,\quad j=1,\ldots ,m.}$$

These m equations define the static equilibrium of the system of rigid bodies.

Generalized inertia forces

Consider a single rigid body which moves under the action of a resultant force F and torque T, with one degree of freedom defined by the generalized coordinate q. Assume the reference point for the resultant force and torque is the center of mass of the body, then the generalized inertia force Q* associated with the generalized coordinate q is given by $${\displaystyle Q^{*}=-(M\mathbf {A} )\cdot {\frac {\partial \mathbf {V} }{\partial {\dot {q}}}}-\left([I_{R}]{\boldsymbol {\alpha }}+{\boldsymbol {\omega }}\times [I_{R}]{\boldsymbol {\omega }}\right)\cdot {\frac {\partial {\boldsymbol {\omega }}}{\partial {\dot {q}}}}.}$$

This inertia force can be computed from the kinetic energy of the rigid body, $${\displaystyle T={\tfrac {1}{2}}M\mathbf {V} \cdot \mathbf {V} +{\tfrac {1}{2}}{\boldsymbol {\omega }}\cdot [I_{R}]{\boldsymbol {\omega }},}$$ by using the formula $${\displaystyle Q^{*}=-\left({\frac {d}{dt}}{\frac {\partial T}{\partial {\dot {q}}}}-{\frac {\partial T}{\partial q}}\right).}$$

A system of n rigid bodies with m generalized coordinates has the kinetic energy $${\displaystyle T=\sum _{i=1}^{n}\left({\tfrac {1}{2}}M\mathbf {V} _{i}\cdot \mathbf {V} _{i}+{\tfrac {1}{2}}{\boldsymbol {\omega }}_{i}\cdot [I_{R}]{\boldsymbol {\omega }}_{i}\right),}$$ which can be used to calculate the m generalized inertia forces^[6] $${\displaystyle Q_{j}^{*}=-\left({\frac {d}{dt}}{\frac {\partial T}{\partial {\dot {q}}_{j}}}-{\frac {\partial T}{\partial q_{j}}}\right),\quad j=1,\ldots ,m.}$$

Dynamic equilibrium

D'Alembert's form of the principle of virtual work states that a system of rigid bodies is in dynamic equilibrium when the virtual work of the sum of the applied forces and the inertial forces is zero for any virtual displacement of the system. Thus, dynamic equilibrium of a system of n rigid bodies with m generalized coordinates requires that $${\displaystyle \delta W=\left(Q_{1}+Q_{1}^{*}\right)\delta q_{1}+\dots +\left(Q_{m}+Q_{m}^{*}\right)\delta q_{m}=0,}$$ for any set of virtual displacements δq_j. This condition yields m equations, $${\displaystyle Q_{j}+Q_{j}^{*}=0,\quad j=1,\ldots ,m,}$$ which can also be written as $${\displaystyle {\frac {d}{dt}}{\frac {\partial T}{\partial {\dot {q}}_{j}}}-{\frac {\partial T}{\partial q_{j}}}=Q_{j},\quad j=1,\ldots ,m.}$$ The result is a set of m equations of motion that define the dynamics of the rigid body system.

Lagrange's equations

If the generalized forces Q_j are derivable from a potential energy V(q₁, ..., q_m), then these equations of motion take the form $${\displaystyle {\frac {d}{dt}}{\frac {\partial T}{\partial {\dot {q}}_{j}}}-{\frac {\partial T}{\partial q_{j}}}=-{\frac {\partial V}{\partial q_{j}}},\quad j=1,\ldots ,m.}$$

In this case, introduce the Lagrangian, L = T − V, so these equations of motion become $${\displaystyle {\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {q}}_{j}}}-{\frac {\partial L}{\partial q_{j}}}=0\quad j=1,\ldots ,m.}$$ These are known as Lagrange's equations of motion.

System of particles

The linear and angular momentum of a rigid system of particles is formulated by measuring the position and velocity of the particles relative to the center of mass. Let the system of particles P_i, i = 1, ..., n be located at the coordinates r_i and velocities v_i. Select a reference point R and compute the relative position and velocity vectors, $${\displaystyle \mathbf {r} _{i}=\left(\mathbf {r} _{i}-\mathbf {R} \right)+\mathbf {R} ,\quad \mathbf {v} _{i}={\frac {d}{dt}}(\mathbf {r} _{i}-\mathbf {R} )+\mathbf {V} .}$$

The total linear and angular momentum vectors relative to the reference point R are $${\displaystyle \mathbf {p} ={\frac {d}{dt}}\left(\sum _{i=1}^{n}m_{i}\left(\mathbf {r} _{i}-\mathbf {R} \right)\right)+\left(\sum _{i=1}^{n}m_{i}\right)\mathbf {V} ,}$$ and $${\displaystyle \mathbf {L} =\sum _{i=1}^{n}m_{i}\left(\mathbf {r} _{i}-\mathbf {R} \right)\times {\frac {d}{dt}}\left(\mathbf {r} _{i}-\mathbf {R} \right)+\left(\sum _{i=1}^{n}m_{i}\left(\mathbf {r} _{i}-\mathbf {R} \right)\right)\times \mathbf {V} .}$$

If R is chosen as the center of mass these equations simplify to $${\displaystyle \mathbf {p} =M\mathbf {V} ,\quad \mathbf {L} =\sum _{i=1}^{n}m_{i}\left(\mathbf {r} _{i}-\mathbf {R} \right)\times {\frac {d}{dt}}\left(\mathbf {r} _{i}-\mathbf {R} \right).}$$

Rigid system of particles

To specialize these formulas to a rigid body, assume the particles are rigidly connected to each other so P_i, i=1,...,n are located by the coordinates r_i and velocities v_i. Select a reference point R and compute the relative position and velocity vectors, $${\displaystyle \mathbf {r} _{i}=(\mathbf {r} _{i}-\mathbf {R} )+\mathbf {R} ,\quad \mathbf {v} _{i}=\omega \times (\mathbf {r} _{i}-\mathbf {R} )+\mathbf {V} ,}$$ where ω is the angular velocity of the system.^[7][8][9]

The linear momentum and angular momentum of this rigid system measured relative to the center of mass R is $${\displaystyle \mathbf {p} =\left(\sum _{i=1}^{n}m_{i}\right)\mathbf {V} ,\quad \mathbf {L} =\sum _{i=1}^{n}m_{i}(\mathbf {r} _{i}-\mathbf {R} )\times \mathbf {v} _{i}=\sum _{i=1}^{n}m_{i}(\mathbf {r} _{i}-\mathbf {R} )\times (\omega \times (\mathbf {r} _{i}-\mathbf {R} )).}$$

These equations simplify to become, $${\displaystyle \mathbf {p} =M\mathbf {V} ,\quad \mathbf {L} =[I_{R}]\omega ,}$$ where M is the total mass of the system and [I_R] is the moment of inertia matrix defined by $${\displaystyle [I_{R}]=-\sum _{i=1}^{n}m_{i}[r_{i}-R][r_{i}-R],}$$ where [r_i − R] is the skew-symmetric matrix constructed from the vector r_i − R.