Rigid body dynamics

From Academic Kids

Rigid body dynamics differs from particle dynamics in that the body takes up space and can rotate, which introduces other considerations. Equations from particle dynamics can be generalized to rigid body dynamics as follows:

Rigid body linear momentum

The equation for particle linear momentum is

<math>\frac{d(m v)}{dt}=\sum_{i=1}^N f_i <math>

where m is the particle's mass, v is its velocity, and fi is one of the N forces acting on it. Assuming constant mass, this reduces to

<math>m \frac{dv}{dt}=\sum_{i=1}^N f_i. <math>

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 and a position vector r. Thus, the linear momentum equation of any given particle would look like this:

<math>dm \frac{d^2r}{dt^2}= \sum_{i=1}^M f_{i,internal} + \sum_{j=1}^N f_{j,\mathrm{external}}.<math>

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

<math> \frac{d^2}{dt^2} \int r\, dm = \sum_{j=1}^N f_{j,\mathrm{external}}.<math>

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

<math> M \frac{d^2 \frac{\int r\, dm}{M}}{dt^2} = \sum_{j=1}^N f_{j,\mathrm{external}}<math>

However, <math>\frac{\int r\, dm}{M}<math> is the formula for the position of center of mass. Denoting this by rcm, the equation reduces to

<math> M \frac{d^2 r_{cm}}{dt^2} = \sum_{j=1}^N f_{j,\mathrm{external}}.<math>

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

<math>M b_{G/O} \times \frac{d^2 R_O}{dt^2} + \frac{d\begin{pmatrix} \int y^2+z^2\, dm & -\int xy\, dm & -\int xz\, dm\\ -\int xy\, dm & \int x^2+z^2\, dm & -\int yz\, dm \\ -\int xz\, dm & -\int yz\, dm & \int x^2+y^2\, dm \end{pmatrix}\begin{pmatrix} \omega_x \\ \omega_y \\ \omega_z \end{pmatrix}}{dt} = \sum_{j=1}^N \tau_{O,j} <math>

where M is the total mass, bG/O is the vector from O to the body's center of mass, RO is the position of O, t is time, <math> \int \, dm<math> is an integral over the mass of the body, ωq is the angular velocity about axis q, and τO,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

<math>\int x^2+y^2\, dm \frac{d\omega_z}{dt} = \sum_{j=1}^N \tau_{O,j,z}.<math>

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