Wednesday, 10 August 2011

Chapter 4 : Dynamics Of Particles.


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 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 (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.

6-DOF Dynamic Model

The model can be considered in 2 independent parts. For this model, the orthogonal matrix representation of rotation is used.


Read more: http://www.answers.com/topic/rigid-body-dynamics#ixzz1UcQtYctA
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