User:Prof McCarthy/rigid body dynamics

Rigid-body dynamics studies the movement of systems of interconnected bodies under the action of external forces. The assumption that the bodies are rigid, which means that they do not deflect under the action of applied forces, simplifies the analysis by reducing the parameters that describe the configuration of the system to the translation and rotation of reference frames attached to each body.

The dynamics of a rigid body system is defined by its equations of motion, which are derived using either Newtons laws of motion or Lagrangian mechanics. The solution of these equations of motion defines how the configuration of the system of rigid bodies changes as a function of time. The formulation and solution of rigid body dynamics is an important tool in the computer simulation of mechanical systems.

Newton's second law for a system of rigid bodies
In order to consider rigid body dynamics, 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's 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." 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
 * $$\mathbf{F} = m\mathbf{A},$$

where F is the 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, Pi, 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 Fi is the external force applied to particle Pi with mass mi, then
 * $$ \mathbf{F}_i + \sum_{j=1}^N \mathbf{F}_{ij} = m_i\mathbf{A}_i,\quad i=1, \ldots, N,$$

where Fij is the internal force of particle Pj acting on particle Pi 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,
 * $$ \mathbf{F} = \sum_{i=1}^N \mathbf{F}_i,\quad \mathbf{T} = \sum_{i=1}^N (\mathbf{R}_i-\mathbf{R})\times \mathbf{F}_i, $$

where Ri is the vector that defines the position of particle to Pi.

Newton's second law for a particle combines with these formulas for the resultant force and torque to yield,
 * $$ \mathbf{F} = \sum_{i=1}^N m_i\mathbf{A}_i,\quad \mathbf{T} = \sum_{i=1}^N (\mathbf{R}_i-\mathbf{R})\times (m_i\mathbf{A}_i), $$

where the internal forces Fij cancel in pairs. The kinematics of a rigid body yields the formula for the acceleration of the particle Pi 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,
 * $$ \mathbf{A}_i = \alpha\times(\mathbf{R}_i-\mathbf{R}) + \omega\times\omega\times(\mathbf{R}_i-\mathbf{R})  + \mathbf{A}.$$

Mass properties
The mass properties of the rigid system are represented by its center of mass and inertia matrix. Choose the reference point R so that it satisfies the condition
 * $$ \sum_{i=1}^N m_i(\mathbf{R}_i-\mathbf{R})=0,$$

then it is known as the center of mass of the system. The inertia matrix [IR] of the system relative to the reference point R is defined by
 * $$[I_R] =-\sum_{i=1}^N m_i[R_i-R][R_i-R], $$

where the matrix [Ri-R] is the skew symmetric matrix constructed from the relative position vector Ri-R.

Force-torque equations
Using the center of mass and inertia matrix, the force and torque equations for a single rigid body take the form
 * $$ \mathbf{F} = M\mathbf{A},\quad \mathbf{T}=[I_R]\alpha + \omega\times[I_R]\omega,$$

and are known as Newtons second law of motion for a rigid body.

The dynamics of an interconnected system of rigid bodies, Bi, 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
 * $$ \mathbf{F}_j = M_j \mathbf{A}_j, \quad \mathbf{T}_j =[I_R]_j\alpha_j + \omega_j\times[I_R]_j\omega_j,\quad j=1,\ldots,M.$$

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

Virtual work of forces acting on a rigid body
The virtual work of forces acting at various points on a single rigid body can be calculated using the velocities of their point of application and the resultant force and torque of the forces. To see this, let the forces F1, F2 ... Fn act on the points R1, R2 ... Rn in a rigid body.

The trajectories of Ri, i=1,...,n  are defined by the movement of the rigid body. The velocity of the points Ri along their trajectories are
 * $$\mathbf{V}_i = \vec{\omega}\times(\mathbf{R}_i-\mathbf{R}) + \mathbf{V},$$

where ω is the angular velocity vector of the body.

The virtual work the dot product of each force and the virtual displacement of its point of application
 * $$ \delta W = \sum_{i=1}^n \mathbf{F}_i \cdot \delta\mathbf{r}_i.$$

If the trajectory of a rigid body is defined by a set of generalized coordinates qj, j=1, ..., m, then the virtual displacements δri are given by
 * $$ \delta\mathbf{r}_i=\sum_{j=1}^m \frac{\partial \mathbf{r}_i}{\partial q_j}\delta q_j =\sum_{j=1}^m \frac{\partial \mathbf{V}_i}{\partial \dot{q}_j}\delta q_j.$$

The virtual work of this system of forces acting on the body in terms of the generalized coordinates becomes
 * $$ \delta W = \mathbf{F}_1\cdot(\sum_{j=1}^m \frac{\partial \mathbf{V}_1}{\partial \dot{q}_j}\delta q_j) + \ldots + \mathbf{F}_n\cdot(\sum_{j=1}^m \frac{\partial \mathbf{V}_n}{\partial \dot{q}_j}\delta q_j)$$

or collecting the coefficients of δqj
 * $$ \delta W = (\sum_{i=1}^n \mathbf{F}_i\cdot \frac{\partial \mathbf{V}_i}{\partial \dot{q}_1})\delta q_1 + \ldots + (\sum_{1=1}^n \mathbf{F}_i\cdot \frac{\partial \mathbf{V}_i}{\partial \dot{q}_m})\delta q_m. $$

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
 * $$ \delta W = (\sum_{i=1}^n \mathbf{F}_i\cdot \frac{\partial \mathbf{V}_i}{\partial \dot{q}})\delta q =  (\sum_{i=1}^n \mathbf{F}_i\cdot \frac{\partial  (\vec{\omega}\times(\mathbf{R}_i-\mathbf{R}) + \mathbf{V})}{\partial \dot{q}})\delta q. $$

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

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

is known as the generalized force associated with the virtual displacement δq. This formula generalizes easily to the movement of a rigid body defined by more than one generalized coordinate, that is
 * $$ \delta W = \sum_{j=1}^m Q_j \delta q_j, $$

where
 * $$ Q_j = \mathbf{F}\cdot \frac{\partial \mathbf{V}}{\partial \dot{q}_j} + \mathbf{T}\cdot\frac{\partial \vec{\omega}}{\partial \dot{q}_j}.$$

D'Alembert's form of the principle of virtual work
The equations of motion for a mechanical system of rigid bodies can be determined using D'Alembert's form of the principle of virtual work. The principle of virtual work is used to study the static equilibrium of a system of rigid bodies, however by introducing acceleration terms in Newton's laws this approach is generalized to define dynamic equilibrium.

Static equilibrium
The static equilibrium of a mechanical system rigid bodies is defined by the that 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. This is equivalent to the requirement that the generalized forces for any virtual displacement are zero, that is Qi=0.

Let a mechanical system be constructed from n rigid bodies, Bi, i=1,...,n, and let the resultant of the applied forces on each body be the force-torque pairs, Fi and Ti, i=1,...,n. Notice that these applied forces do not include the reaction forces where the bodies are connected. Finally, assume that the velocity Vi and angular velocities ωi, 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, Fi and Ti, applied to this one degree of freedom system is given by
 * $$ \delta W = \sum_{i=1}^n (\mathbf{F}_i\cdot \frac{\partial \mathbf{V}_i}{\partial \dot{q}} + \mathbf{T}_i\cdot\frac{\partial \vec{\omega}_i}{\partial \dot{q}})\delta q = Q\delta q,$$

where
 * $$ Q = \sum_{i=1}^n (\mathbf{F}_i\cdot \frac{\partial \mathbf{V}_i}{\partial \dot{q}} + \mathbf{T}_i\cdot\frac{\partial \vec{\omega}_i}{\partial \dot{q}}),$$

is the generalized force associated with q acting on the mechanical system.

If the the mechanical system is defined by m generalized coordinates, qj, j=1,...,m, then the system has m degrees of freedom and the virtual work is given by,
 * $$ \delta W = \sum_{j=1}^m Q_j\delta q_j,$$

where
 * $$ Q_j = \sum_{i=1}^n (\mathbf{F}_i\cdot \frac{\partial \mathbf{V}_i}{\partial \dot{q}_j} + \mathbf{T}_i\cdot\frac{\partial \vec{\omega}_i}{\partial \dot{q}_j}),$$

is the generalized force associated with the generalized coordinate qj. The principle of virtual work defines the static equilibrium of a system of rigid bodies as occurring when these generalized forces acting on the system are zero.

Generalized inertia forces
Consider a single rigid body which moves under the action of a resultant for 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
 * $$ Q^* = -(M\mathbf{A})\cdot \frac{\partial \mathbf{V}}{\partial \dot{q}}   -   ([I_R]\alpha+ \omega\times[I_R]\omega)\cdot \frac{\partial \vec{\omega}}{\partial \dot{q}}.$$

This inertia force can be computed from the kinetic energy of the rigid body given by
 * $$ T = \frac{1}{2}M\mathbf{V}\cdot\mathbf{V} + \frac{1}{2}\vec{\omega}\cdot [I_R]\vec{\omega},$$

by the formula
 * $$ Q^* = -\left(\frac{d}{dt} \frac{\partial T}{\partial \dot{q}} -\frac{\partial T}{\partial q}\right).$$

A system of n rigid bodies with m generalized coordinates has the kinetic energy
 * $$T = \sum_{i=1}^n(\frac{1}{2}M\mathbf{V}_i\cdot\mathbf{V}_i + \frac{1}{2}\vec{\omega}_i\cdot [I_R]\vec{\omega}_i),$$

which can be used to calculate the m generalized inertia forces
 * $$ Q^*_j = -\left(\frac{d}{dt} \frac{\partial T}{\partial \dot{q}_j} -\frac{\partial T}{\partial q_j}\right),\quad j=1, \ldots, m.$$

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 the virtual work
 * $$ \delta W = (Q_1 + Q^*_1)\delta q_1 + \ldots + (Q_m + Q^*_m)\delta q_m = 0,$$

be zero for any set of virtual displacements δqj. This condition yields m equations,
 * $$ Q_j + Q^*_j = 0, j=1, \ldots, m,$$

which can also be written as
 * $$ \frac{d}{dt} \frac{\partial T}{\partial \dot{q}_j} -\frac{\partial T}{\partial q_j} = Q_j, \quad j=1,\ldots,m.$$

The result is a set of m equation of motion that define the dynamics of the rigid body system.

Rigid-body linear momentum
Newton's Second Law states that the rate of change of the linear momentum of a particle with constant mass is equal to the sum of all external forces acting on the particle:


 * $$\frac{\mathrm{d}(m \mathbf{v})}{\mathrm{d}t}=\sum_{i=1}^N \mathbf{f}_i $$

where m is the particle's mass, v is the particle's velocity, their product mv is the linear momentum, and fi is one of the N number of forces acting on the particle.

Because the mass is constant, this is equivalent to


 * $$m \frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}=\sum_{i=1}^N \mathbf{f}_i.$$

To generalize, assume a body of finite mass and size is composed of such particles, each with infinitesimal mass dm. Each particle has a position vector r. There exist internal forces, acting between any two particles, and external forces, acting only on the outside of the mass. Since velocity v is the derivative of position r with respect to time, the derivative of velocity dv/dt is the second derivative of position d2r/dt2, and the linear momentum equation of any given particle is


 * $$\mathrm{d}m \frac{\mathrm{d}^2\mathbf{r}}{\mathrm{d}t^2}= \sum_{i=1}^M \mathbf{f}_{i,\text{internal}} + \sum_{j=1}^N \mathbf{f}_{j,\mathrm{external}}.$$

When the linear momentum equations for all particles are added together, the internal forces sum to zero according to Newton's third law, which states that any such force has opposite magnitudes on the two particles. By accounting for all particles, the left side becomes an integral over the entire body, and the second derivative operator can be moved out of the integral, so


 * $$ \frac{\mathrm{d}^2}{\mathrm{d}t^2} \int \mathbf{r}\, \mathrm{d}m = \sum_{j=1}^N \mathbf{f}_{j,\mathrm{external}}$$.

Let M be the total mass, which is constant, so the left side can be multiplied and divided by M, so


 * $$ M \frac{\mathrm{d}^2}{\mathrm{d}t^2}\!\left(\frac{\int \mathbf{r}\, \mathrm{d}m}{M}\right) = \sum_{j=1}^N \mathbf{f}_{j,\mathrm{external}}$$.

The expression $$\frac{\int \mathbf{r}\, \mathrm{d}m}{M}$$ is the formula for the position of the center of mass. Denoting this by rcm, the equation reduces to


 * $$ M \frac{\mathrm{d}^2 \mathbf{r}_\mathrm{cm}}{\mathrm{d}t^2} = \sum_{j=1}^N \mathbf{f}_{j,\mathrm{external}}.$$

Thus, linear momentum equations can be extended to rigid bodies by denoting that they describe the motion of the center of mass of the body. This is known as Euler's first law.

Rigid-body angular momentum
The most general equation for rotation of a rigid body in three dimensions about an arbitrary origin O with axes x, y, z is


 * $$M b_{G/O} \times \frac{\mathrm{d}^2 R_O}{\mathrm{d}t^2} + \frac{\mathrm{d}(\mathbf{I}\boldsymbol{\omega})}{\mathrm{d}t} = \sum_{j=1}^N \tau_{O,j} $$

where the moment of inertia tensor, $$\mathbf{I}$$, is given by
 * $$ \mathbf{I} = \begin{pmatrix}

I_{xx} & I_{xy} & I_{xz} \\ I_{yx} & I_{yy} & I_{yz} \\ I_{zx} & I_{zy} & I_{zz} \end{pmatrix}$$


 * $$ \mathbf{I} =

\begin{pmatrix} \int (y^2+z^2)\, \mathrm{d}m & -\int xy\, \mathrm{d}m & -\int xz\, \mathrm{d}m\\ -\int xy\, \mathrm{d}m & \int (x^2+z^2)\, \mathrm{d}m & -\int yz\, \mathrm{d}m \\ -\int xz\, \mathrm{d}m & -\int yz\, \mathrm{d}m & \int (x^2+y^2)\, \mathrm{d}m \end{pmatrix}$$

Given that Euler's rotation theorem states that there is always an instantaneous axis of rotation, the angular velocity, $$\boldsymbol{\omega}$$, can be given by a vector over this axis
 * $$\quad \boldsymbol{\omega} = \omega_x \mathbf{\hat{i}} + \omega_y \mathbf{\hat{j}} + \omega_z \mathbf{\hat{k}} $$

where $$\scriptstyle{(\mathbf{\hat{i}},\ \mathbf{\hat{j}},\  \mathbf{\hat{k}})}$$ is a set of mutually perpendicular unit vectors fixed in a reference frame.

Rotating a rigid body is equivalent to rotating a Poinsot ellipsoid.

Angular momentum and torque
Similarly, the angular momentum $$\mathbf{L}$$ for a system of particles with linear momenta $$p_{i}$$ and distances $$r_{i}$$ from the rotation axis is defined



\mathbf{L} = \sum_{i=1}^{N} \mathbf{r}_{i} \times \mathbf{p}_{i} = \sum_{i=1}^{N} m_{i} \mathbf{r}_{i} \times \mathbf{v}_{i} $$

For a rigid body rotating with angular velocity $$\omega$$ about the rotation axis $$\mathbf{\hat{n}}$$ (a unit vector), the velocity vector $$\mathbf{v}_{i}$$ may be written as a vector cross product



\mathbf{v}_{i} = \omega \mathbf{\hat{n}} \times \mathbf{r}_{i} \ \stackrel{\mathrm{def}}{=}\ \boldsymbol\omega \times \mathbf{r}_{i} $$

where
 * angular velocity vector $$\boldsymbol\omega \ \stackrel{\mathrm{def}}{=}\ \omega \mathbf{\hat{n}}$$
 * $$\mathbf{r}_{i}$$ is the shortest vector from the rotation axis to the point mass.

Substituting the formula for $$\mathbf{v}_{i}$$ into the definition of $$\mathbf{L}$$ yields



\mathbf{L} = \sum_{i=1}^{N} m_{i} \mathbf{r}_{i} \times (\boldsymbol\omega \times \mathbf{r}_{i}) = \boldsymbol\omega \sum_{i=1}^{N} m_{i} r_{i}^{2} = I \omega \mathbf{\hat{n}} $$

where we have introduced the special case that the position vectors of all particles are perpendicular to the rotation axis (e.g., a flywheel): $$\boldsymbol\omega \cdot \mathbf{r}_{i} = 0$$.

The torque $$\mathbf{N}$$ is defined as the rate of change of the angular momentum $$\mathbf{L}$$



\mathbf{N} \ \stackrel{\mathrm{def}}{=}\ \frac{d\mathbf{L}}{dt} $$

If I is constant (because the inertia tensor is the identity, because we work in the intrinsecal frame, or because the torque is driving the rotation around the same axis $$\mathbf{\hat{n}}$$ so that $$\mathrm{I}$$ is not changing) then we may write



\mathbf{N} \ \stackrel{\mathrm{def}}{=}\ I \frac{d\omega}{dt}\mathbf{\hat{n}} = I \alpha \mathbf{\hat{n}} $$

where
 * $$\alpha$$ is called the angular acceleration (or rotational acceleration) about the rotation axis $$\mathbf{\hat{n}}$$.

Notice that if I is not constant in the external reference frame (i.e. the three main axes of the body are different) then we cannot take the I outside the derivate. In this cases we can have torque-free precession.