|
T=
| 1 | |
| 2 |
| sf{T}M | |||
|
|
where
| sf{T} | |||
|
|
T=
| 1 | |
| 2 |
m|v|2=
| 1 | |
| 2 |
v ⋅ mv
and can be derived from it, by expressing the position of each particle of the system in terms of .
In general, the mass matrix depends on the state, and therefore varies with time.
Lagrangian mechanics yields an ordinary differential equation (actually, a system of coupled differential equations) that describes the evolution of a system in terms of an arbitrary vector of generalized coordinates that completely defines the position of every particle in the system. The kinetic energy formula above is one term of that equation, that represents the total kinetic energy of all the particles.
For example, consider a system consisting of two point-like masses confined to a straight track. The state of that systems can be described by a vector of two generalized coordinates, namely the positions of the two particles along the track.
q=\begin{bmatrix}x1&x2\end{bmatrix}sf{T}
Supposing the particles have masses, the kinetic energy of the system is
T=
| 2 | |
| \sum | |
| i=1 |
| 1 | |
| 2 |
mi
| xi |
2
This formula can also be written as
T=
| 1 | |
| 2 |
| q |
| |||
where
M=\begin{bmatrix}m1&0\ 0&m2\end{bmatrix}
More generally, consider a system of particles labelled by an index, where the position of particle number is defined by free Cartesian coordinates (where). Let be the column vector comprising all those coordinates. The mass matrix is the diagonal block matrix where in each block the diagonal elements are the mass of the corresponding particle:
M=\operatorname{diag}\left[m1
| I | |
| n1 |
,m2
| I | |
| n2 |
,\ldots,mN
| I | |
| nN |
\right]
where is the identity matrix, or more fully:
M=\begin{bmatrix} m1& … &0&0& … &0& … &0& … &0\\ \vdots&\ddots&\vdots&\vdots&\ddots&\vdots&\ddots&\vdots&\ddots&\vdots\\ 0& … &m1&0& … &0& … &0& … &0\\ 0& … &0&m2& … &0& … &0& … &0\\ \vdots&\ddots&\vdots&\vdots&\ddots&\vdots&\ddots&\vdots&\ddots&\vdots\\ 0& … &0&0& … &m2& … &0& … &0\\ \vdots&\ddots&\vdots&\vdots&\ddots&\vdots&\ddots&\vdots&\ddots&\vdots\\ 0& … &0&0& … &0& … &mN& … &0\\ \vdots&\ddots&\vdots&\vdots&\ddots&\vdots&\ddots&\vdots&\ddots&\vdots\\ 0& … &0&0& … &0& … &0& … &mN\\ \end{bmatrix}
For a less trivial example, consider two point-like objects with masses, attached to the ends of a rigid massless bar with length, the assembly being free to rotate and slide over a fixed plane. The state of the system can be described by the generalized coordinate vector
q=\begin{bmatrix}x&y&\alpha\end{bmatrix}
where are the Cartesian coordinates of the bar's midpoint and is the angle of the bar from some arbitrary reference direction. The positions and velocities of the two particles are
\begin{align} x1&=(x,y)+R(\cos\alpha,\sin\alpha)&v1&=\left(
| x, |
| y\right) |
+R
| \alpha(-\sin\alpha, |
\cos\alpha)\\ x2&=(x,y)-R(\cos\alpha,\sin\alpha)&v2&=\left(
| x, |
| y\right) |
-R
| \alpha(-\sin\alpha, |
\cos\alpha) \end{align}
and their total kinetic energy is
2T=m
| x |
2+m
| y |
2+
| |||
| mR |
2-2Rd\sin(\alpha)
| x |
| \alpha |
+2Rd\cos(\alpha)
| y |
| \alpha |
where
m=m1+m2
d=m1-m2
T=
| 1 | |
| 2 |
| q |
| |||
where
M=\begin{bmatrix} m&0&-Rd\sin\alpha\\ 0&m&Rd\cos\alpha\\ -Rd\sin\alpha&Rd\cos\alpha&R2m \end{bmatrix}
Note that the matrix depends on the current angle of the bar.
For discrete approximations of continuum mechanics as in the finite element method, there may be more than one way to construct the mass matrix, depending on desired computational accuracy and performance. For example, a lumped-mass method, in which the deformation of each element is ignored, creates a diagonal mass matrix and negates the need to integrate mass across the deformed element.