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Rigid body dynamics 

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In physics, rigid body dynamics is the study of the motion of rigid bodies. Unlike point like particles, rigid bodies occupy space and as such have geometrical properties, such as a center of mass, natural rotation axis, moments of inertia, etc. Rigid bodies are also characterized as being non-deformable, whereas deformable bodies are characterized by being deformable.

Rigid-body dynamics are commonly used in video games.

Contents

Rigid body linear momentum

The equation for particle linear momentum is

\frac{\mathrm{d}(m v)}{\mathrm{d}t}=\sum_{i=1}^N f_i

where:

  • m is the particle's mass.
  • v is the particle's velocity.
  • fi is one of the N forces acting on the particle.

Assuming constant mass, this reduces to

m \frac{\mathrm{d}v}{\mathrm{d}t}=\sum_{i=1}^N f_i.

To generalize assume a body of finite mass and size is composed of such particles. There exist internal forces, acting between any two particles, and external forces, acting only on the outside of the mass. Each particle has:

  • a mass dm.
  • a position vector r.

Thus, the linear momentum equation of any given particle would look like this:

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

If the equation for each particle were added together, the internal forces would cancel out, since by Newton's third law, any such force would have opposite magnitudes on the two particles. Also, the left side would become an integral over the entire body, and the second derivative operator could come out of the integral, leaving

\frac{\mathrm{d}^2}{\mathrm{d}t^2} \int r\, \mathrm{d}m = \sum_{j=1}^N f_{j,\mathrm{external}}.

Letting M be the total mass, the left side can be multiplied and divided by M without changing the validity:

M \frac{\mathrm{d}^2 \frac{\int r\, \mathrm{d}m}{M}}{\mathrm{d}t^2} = \sum_{j=1}^N f_{j,\mathrm{external}}

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

M \frac{\mathrm{d}^2 r_{cm}}{\mathrm{d}t^2} = \sum_{j=1}^N 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.

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}

and the angular velocity, \boldsymbol{\omega}, is given by

\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.

Moving any rigid body is equivalent to moving a Poinsot's ellipsoid.

Angular momentum and torque

Similarly, the angular momentum \mathbf{L} for a system of particles with linear momenta pi and distances ri 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 ω 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 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

α 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 (ie. 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.

Applications

Computer physics engines use rigid body dynamics to increase interactivity and realism in video games.

See also

Theory

Simulators

External links
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