Question

Fortran 90: Calculate the center of mass of water, shift origin to the center of mass, construct moment of inertia tensor, determine the principal moments of inertia, print the eigenvalues of the inertia tensor.

Answer 1

Center of Mass is defined by the as the average position of the mass of the system.

• Mathematically, it is defined by the following integral:

                    xCM=1/M£(integral)xdm

• For defined points of mass in a system, this integral can also

be written as the following sum:

      xCM=1/M∑i=1 to n mixi

• In other words, the center of mass is sum of the mass fraction of each point in the system multiplied by its position. In this case M is the total mass of the system.

• The previous equations describe the position of the center of mass in the x direction, but the same equations apply for the y and z directions as well.

Motion of the Center of Mass

• The velocity and acceleration of the center of mass of a system is found the same way as the center of mass:

      vCM=(m1v1+m2v2+........mnvn)/M.

      aCM=(m1a1+m2a2+.......mnan)/M.

• The advantage to using the center of mass to evaluate the

motion of a system is that the center of mass acts the same as a single particle:

       P=MvCM,F=MaCM

Angular Momentum

Expression of the angular momentum of a system of particles about the center of mass,

HG=∑i=1 to n (ri' * mi(w*ri')=∑i=1 to n miri'2w

Here, r' is the position vector relative to the center of mass, v' is the velocity relative to the center of mass.

We note that, in the above expression, an integral is used instead of a summation, since we are now dealing with a continuum distribution of mass.

For a 3D rigid body, the distance between any particle and the center of mass will remain constant, and the particle velocity, relative to the center of mass, will be given by

v'= ω × r'.Here, we have used the vector identity A × (B × C)=(AC)B − (AB)C. We note that, for planar bodies undergoing a 2D motion in its own plane, r' is perpendicular to ω, and the term (r'· ω) is zero. In this case, the vectors ω and HG are always parallel. In the three-dimensional case however, this simplification does not occur, and as a consequence, the angular velocity vector, ω, and the angular momentum vector, HG,are in general, not parallel.

In cartesian coordinates, we have, r' = x'i + y'j + z'k and ω = ωxi + ωyj + ωzk, and the above expression can be expanded to yield,

HG = ωx£m (x'2+ y'2+ z'2) dm − £m(ωxx'+ ωyy'+ ωzz')x'dm )i + ωy£m (x'2+ y'2+ z'2) dm − £m(ωxx'+ ωyy'+ ωzz')x'dm)j

+ ωz£m (x'2+ y'2+ z'2) dm − £m(ωxx'+ ωyy'+ ωzz')x'dm)k

= ( Ixxωx − Ixyωy − Ixz ωz ) i

+ (−Iyxωx + Iyyωy − Iyzωz) j

+ (−Izxωx − Izy ωy + Izzωz) k .

The quantities Ixx, Iyy, and Izz are called moments of inertia with respect to the x, y and z axis, respectively,

and are given by

Ixx = (y'2+ z'2) dm, Iyy = (x'2+ z'2) dm, Izz = (x'2+ y'2) dm.

We observe that the quantity in the integrand is precisely the square of the distance to the x, y and z axis,respectively. They are analogous to the moment of inertia used in the two dimensional case. It is also clear, from their expressions, that the moments of inertia are always positive. The quantities Ixy, Ixz, Iyx, Iyz, Izx and Izy are called products of inertia.

They are a measure of the imbalance in the mass distribution. If we are interested in calculating the angular

momentum with respect to a fixed point O then, the resulting expression would be,

HO = ( (Ixx)O ωx − (Ixy)O ωy − (Ixz )O ωz ) i

+ (−(Iyx)O ωx +(Iyy)O ωy − (Iyz)O ωz) j

+ (−(Izx)O ωx − (Izy )O ωy +(Izz )O ωz) k

Analogously, we can define the tensor of inertia about point O, by writing equatio in matrix form. Thus,

we have HO =[IO] ω ,

where the components of [IO] are the moments and products of inertia about point O given above.

It follows from the definition of the products of inertia, that the tensors of inertia are always symmetric. The

implications of equation are that in many situations of importance, even for bodes of some symmetry, the

angular momentum vector Hand the angular velocity vector ω are not parallel. This introduces considerable complexity into the analysis of the dynamics of rotating bodies in three dimensions.

Principal Axes of Inertia

For a general three-dimensional body, it is always possible to find 3 mutually orthogonal axis (an x,y,z coordinate system) for which the products of inertia are zero, and the inertia matrix takes a diagonal form.

In most problems, this would be the preferred system in which to formulate a problem. For a rotation about only one of these axis, the angular momentum vector is parallel to the angular velocity vector. For symmetric bodies, it may be obvious which axis are principle axis. However, for an irregular-shaped body this coordinate system may be difficult to determine by inspection; we will present a general method to determine these axes in the next section.

But, if the body has symmetries with respect to some of the axis, then some of the products of inertia become zero and we can identify the principal axes. For instance, if the body is symmetric with respect to the plane x'= 0 then, we will have Ix'y' = Iy'x' = Ix'z' = Iz'x' = 0 and x' will be a principal axis. This can be shown by looking at the definition of the products of inertia.

The integral for, say, Ix' y' can be decomposed into two integrals for the two halves of the body at either side

of the plane x' = 0. The integrand on one half, x'y'

, will be equal in magnitude and opposite in sign to the

integrand on the other half (because x' will change sign).

Also, if the body is symmetric with respect to two planes passing through the center of mass which are orthogonal to the coordinate axis, then the tensor of inertia is diagonal, with Ix'y' = Ix'z' = Iy'z'= 0.

Another case of practical importance is when we consider axis symmetric bodies of revolution. In this case, if one of the axis coincides with the axis of symmetry, the tensor of inertia has a simple diagonal form. For an axis symmetric body, the moments of inertia about the two axis in the plane will be equal. Therefore, the moment about any axis in this plane is equal to one of these. And therefore, any axis in the plane isa principal axis. One can extend this to show that if the moment of inertia is equal about two axis in the plane (IPP = Ixx), whether or not they are orthogonal, then all axes in the plane are principal axes and the moment of inertia is the same about all of them. In its inertial properties, the body behaves like a circular cylinder.

The tensor of inertia will take different forms when expressed in different axes. When the axes are such that

the tensor of inertia is diagonal, then these axes are called the principal axes of inertia.

The Search for Principal Axes and Moments of Inertia as an Eigenvalue Problem:

Three orthogonal principal axes of inertia always exist even though in bodies without symmetries their directions may not be obvious. To find the principle axis of a general body consider the body in that rotates about an unknown principal axis. The the total angular momentum vector is I' ω in the direction of the principle axis. For rotation about the principal axis, the angular momentum and the angular velocity are in the same direction. We seek a coordinate axes x, y and z, about which a rotation ωx, ωy and ωz, which is aligned with this coordinate direction, will be parallel to the angular momentum vector and related by the equation

HGx

HGy     Iωx   Iωy   I 0 0 0 I 0 ωx ωy Iωz 0 0 I ωz HGz

The structure of the solution for finding the principal axes of inertia and their magnitudes is a characteristic-value problem. The three eigenvalues give the directions of the three principal axis, and the three eigen­ vectors give the moments of inertia with respect to each of these axis.

where we will write Ix = Ixx, Iy = Iyy and Iz = Izz. Also, in principal axes we will then have

HG = Ixωxi + Iyωyj + Iz ωz k .

Here, we have use the fact that y'and z' are the coordinates relative to the center of mass and therefore their integrals over the body are equal to zero.

Rotation of Axes

In some situations, we will know the tensor of inertia with respect to some axes xyz and, we will be interested

in calculating the tensor of inertia with respect to another set of axis x'y'z'. We denote by i, j and k the unit vectors along the direction of xyz axes, and by i', j'and k' the unit vectors along the direction of x'y'z' axes. The transformation of the inertia tensor can be accomplished by considering the transformation of the angular momentum vector H' and the angular velocity vector ω'. We begin with the expression of H' and ω' in the x1,x2,x3 system.

There are several options to find a set of principal axis for this problem. One is to use the symmetry to identify a coordinate system in which the products of inertia vanish. The transformation sketched in the figure is a good candidate to obtain the principal axes; that is a rotation of θ = π/4 about the x3 axis, followed by a rotation of ψ = sin−1 axis.