Rigid body dynamics

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In physics, Rigid body dynamics differs from particle dynamics in that the body takes up space and can rotate, which introduces other considerations.

Note: This article has much overlap with the rigid rotor and rigid body articles. Articles should eventually be merged.

Equations from particle dynamics can be generalized to rigid body dynamics as follows:

1 Rigid body linear momentum
2 Rigid body angular momentum
3 Angular momentum and torque
4 Applications
5 See also
6 External links

[edit] 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.
f_i 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 $d m$ .
a position vector r.

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

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

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 r_cm, 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.

[edit] 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:

$\mathbf{I} = \begin{pmatrix} \int y^2+z^2\, \mathrm{d}m & -\int xy\, \mathrm{d}m & -\int xz\, \mathrm{d}m\\ -\int xy\, \mathrm{d}m & \int x^2+z^2\, \mathrm{d}m & -\int yz\, \mathrm{d}m \\ -\int xz\, \mathrm{d}m & -\int yz\, \mathrm{d}m & \int x^2+y^2\, \mathrm{d}m \end{pmatrix} \quad \hbox{and} \quad \boldsymbol{\omega} = \begin{pmatrix} \omega_x \\ \omega_y \\ \omega_z \end{pmatrix}.$

Here $\mathbf{I}$ is the moment of inertia tensor and $\boldsymbol{\omega}$ is the angular velocity (a vector). Based on this, a theorem states that any rigid body is equivalent when moving to a Poinsot's ellipsoid.

Further

ω_q is the angular velocity about axis q.
M is the total mass.
b_G/O is the vector from O to the body's center of mass.
R_O is the position of O.
t is time.
$\int \, \mathrm{d}m$ is an integral over the mass of the body.

$τ O, j$ is one of the N moments about O.

This equation follows from equation for linear momentum of a particle and kinematics; no additional observations of nature are necessary to arrive at it.

There are many special cases that simplify this equation. The first term goes to zero if any of three conditions are met:

O is a fixed point (since its second derivative would be zero).
A set of axes is chosen with its origin attached to the body's center of mass (since this would reduce the vector b to zero).
The vector b always points in the direction of the acceleration of O (since the cross product of parallel vectors is zero).

Also, if the axes are chosen are the principle axes (i.e., the moments about the xy, xz, and yz planes is zero), the off-diagonal terms of the matrix are zero. This case is further discussed by Euler's equations.

When learning about angular motion, students are generally first exposed to the case of rotation only in the x-y plane and a fixed axis or axis at the center of mass with constant rotational inertia. That equation is

$\int \left(x^2+y^2\right)\, \mathrm{d}m \frac{\mathrm{d}\omega_z}{\mathrm{d}t} = \sum_{j=1}^N \tau_{O,j,z}.$

[edit] 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 $ω$ 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.