Generalization of parallel axis theorem for rotational inertia
A. R. Abdulghany∗
Physics Department, Faculty of Science, Cairo University, Giza, Egypt.
(Received April 13, 2017; Accepted June 8, 2017)
Abstract
This paper discusses two levels of generalization of the parallel axis theorem for rotational inertia. The first relates the moments of inertia about any two parallel axes, whether or not they are passing through the center of mass. The second relates the inertia tensors about any two points.
I. INTRODUCTION
Rotational inertia appears in a wide variety of research fields. For example, it is related to the structure of atomic nuclei,1 molecules2 and the neutron stars.3 Moreover, the rotational inertia plays a crucial role in the design of wind turbines,4 artificial limbs,5 robots,6 objects
for 3D printing7 and in countless other applications. The rotational inertia (I) of an object is
described mathematically by a 3×3 symmetric matrix,8the components of the inertia tensor. The diagonal terms (Ixx, Iyy, Izz) are the moments of inertia about the three orthogonal axes
x, y and z. The off-diagonal terms (Ixy, Iyx, Ixz, Izx, Iyz, Izy) are the products of inertia in
directions perpendicular to the axial rotations,
I = Ixx Ixy Ixz Iyx Iyy Iyz Izx Izy Izz . (1)
The elements of the inertia tensor are given by,
I = −Xmi[ri][ri], (2)
where [ri] is a skew symmetric matrix associated with the position ri ≡ (xi, yi, zi) given by,
[ri] = 0 −zi yi zi 0 −xi −yi xi 0 . (3)
The product in the inertia tensor is
−[ri][ri] = (y2i + zi2) −xiyi −xizi −yixi (x2i + z2i) −yizi −zixi −ziyi (x2i + yi2) . (4)
From which the moments of inertia are
Ixx = X (y2i + zi2)mi, Iyy = X (x2i + zi2)mi, (5) Izz = X (x2i + yi2)mi,
and the products of inertia are Ixy = Iyx = − X xiyimi, Ixz = Izx= − X xizimi, (6) Iyz = Izy= − X yizimi.
The elements of inertia tensor, in general, depend on the origin of coordinates and on the directions of the three axes relative to the object. The calculations of inertia tensor elements about some points are easier than about other points, due to the geometry of the object. It is convenient, therefore, to find simple relations among the inertia tensors about different points. If such relations exist, the problem of finding the inertia tensor of the object about any point is easier.
There are some theorems to relate the moments of inertia about different axes, such as the parallel axis theorem and the perpendicular axis theorem.9 The usefulness of those
theorems, however, has limitations. For example, in the parallel axis theorem, the reference axis should be passing through the center of mass of the object. The perpendicular axis theorem is only applicable for the objects that lie entirely within a plane. (The generalization of the perpendicular axis theorem is discussed in Ref. 10 for the diagonal elements of the inertia tensor.)
II. THEORY
The parallel axis theorem relates the moment of inertia (I) about any axis to its value about a parallel axis passing through the center of mass by the following expression,
I = IC+ M d2. (7)
where IC and d are the moment of inertia about the center of mass and the distance between
the two axes. If the reference axis does not pass through the center of mass, the above relation is no longer valid. In the present work, we discuss two levels of generalization of the parallel axis theorem. The first level relates the moments of inertia about any two parallel whether or not they pass through the center of mass. The higher level of generalization, the most generalized version of the parallel axis theorem, will relate the inertia tensor about any point to the inertia tensor about any other point and about a set of axes parallel to the set of axes of the first point.
The first generalization is simple. It is easy to use the parallel axis theorem to relate the moments of inertia about any two parallel axes by relating each of them to a third one through the center of mass,
I2 = I1+ M (d22− d 2
1), (8)
where di is the distance between the center of mass and the rotation axis i, associated with
the moment of inertia Ii. The case of the higher level of generalization is harder, because
not only the moments of inertia change but also the products of inertia.
In formal notation, the moment of inertia about an axis (z0) parallel to z-axis is given by,
Izz0 =X[(xi− X)2+ (yi− Y )2]mi, (9)
where X and Y are the x-component and y-component of the z0-axis relative to the z-axis. From this we get,
Izz0 =X(x2i + yi2)mi+ (X2+ Y2) X mi− 2X X (xi)mi− 2Y X (yi)mi.
From the definition of the center of mass, we can find the following relations,
X (xi)mi = M xc, X (yi)mi = M yc, X mi = M,
where M is the total mass of the object, and where xc and yc are the x-component and
y-component of the center of mass position. The moment of inertia about the z0-axis then becomes,
Izz0 = Izz+ M (X2+ Y2) − 2M Xxc− 2M Y yc. (10)
This expression is a general relation between the moments of inertia about any two parallel axes perpendicular to the xy-plane. If the reference axis is along the center of mass, i.e., xc= yc= 0, this expression is simplified to the parallel axis theorem. If the relationship in
Eq. (10) is applied to a new parallel axis, the z00 axis, and the difference Izz0 − I00
zz is taken,
the result is the parallel axis theorem of Eq. (8) in a different notation.
The other moments of inertia and products of inertia can be found using similar anal-ysis. Figure 1 shows the change of moment of inertia, given by Eq. (10), as the axis of rotation moves keeping its direction perpendicular to the xy-plane. The figure shows that Izz0 describes a “paraboloid” with its minimum at the center of mass. The paraboloid of the moment of inertia is axially symmetric about the axis passing through the center of mass.
The product of inertia (Ixy0 ) about a point with x-component and y-component as X and Y , respectively, is given by,
Ixy0 = −X(xi− X)(yi− Y )mi (11)
From this we get,
Ixy0 = −Xxiyimi− XY X mi + Y X ximi+ X X yimi, or simply, Ixy0 = Ixy − M XY + M Y xc+ M Xyc. (12)
Figure 2 shows the change of the product of inertia, given by Eq. (12), as the point of consideration changes. The figure shows that Ixy0 describes a “hyperbolic paraboloid.” The change of Ixy0 with Y at constant X, or with X at constant Y , is linear. If X or Y is equal to its value at the center of mass, the value of Ixy0 remains constant. The surface of Ixy0 has reflection symmetry about the plane X + Y = xc+ yc and the plane X − Y = xc− yc. The
intersection with the Ixy0 surface of any plane perpendicular to the xy-plane, and passing through the center of mass, is a curve with an extremum at the center of mass.
To get the generalized tensor form of parallel axis theorem, we define the symmetric matrix [(A, B)] by [(A, B)] = −1 2([rA][rB] + [rB][rA]) = (yAyB+ zAzB) −12(xAyB+ yAxB) −12(xAzB+ zAxB) −1 2(xAyB+ yAxB) (xAxB+ zAzB) − 1 2(yAzB+ zAyB) −1 2(xAzB+ zAxB) − 1 2(yAzB+ zAyB) (xAxB+ yAyB) , (13)
where A and B denote two points with positions (xA, yA, zA) and (xB, yB, zB) respectively.
If A and B have the same coordinates, the matrix [(A, B)] is reduced to Eq. (4).
In terms of the symmetric matrix [(A, B)], the most general form of the parallel axis theorem is given by Eq. (14). This general relation gives the inertia tensor ( I0 ) about a set of orthogonal axes x0, y0 and z0, parallel to the reference set of axes x, y and z, associated with the inertia tensor ( I ),
I0 = I + M [(R, R)] − 2M [(R, C)], (14)
III. SAMPLE CALCULATION
In order to test the usefulness of the generalized parallel axis theorem discussed in this paper, we consider the problem of finding the inertia tensor, about its center of mass, of the compound homogeneous object shown in Fig. 3. The object consists of two parts; the first is a cylinder of radius r and length 8r; the second part is a cuboid of sides 2r, 2r, and 8r. The calculation can be simplified by considering the inertia tensor of each part separately about its center, and Eq. (14) can then be used to get the inertia tensor of each part about the center of mass of the whole object. The inertia tensor of the whole object is finally found by addition of the two tensors associated with the two parts.
A. Calculation of the inertia tensor of the cuboid
The mass of the cuboid is given by
M1 = 32ρr3,
and the mass of the cylinder is given by
M2 = 8πρr3,
where ρ is the mass density. To simplify the equations, we define the unit of mass µ = 8ρr3, then in terms of µ,
M1 = 4µ, M2 = πµ. (15)
For the cuboid, Eq. (14) becomes
I01 = I1+ 4µ[(R1, R1)] − 8µ[(R1, C1)], (16)
where C1 denotes the center of mass of the cuboid, which is the same as the reference point
in this part, then
C1 ≡ (xc, yc, zc) ≡ (0, 0, 0),
The position of the center of mass of the whole object relative to C1 is given by
R1 ≡ (X, Y, Z) ≡
M2r
M1+ M2
(0, 3, 5) = πr
The moment of inertia of the cuboid about its center is, I1 = 1 12M1(4r 2+ 64r2) 0 0 0 121M1(4r2+ 4r2) 0 0 0 121M1(4r2+ 64r2) , or simply I1 = 4µr2 68 12 0 0 0 128 0 0 0 6812 . (17)
The other terms in Eq. (16) are,
[(R1, C1)] = [0], [(R1, R1)] = πr 4 + π 2 34 0 0 0 25 −15 0 −15 9 .
Then the inertia tensor of the cuboid about the center of mass of the whole object is,
I01 = 4µr2 68 12+ 34 π 4+π 2 0 0 0 128 + 25 4+ππ 2 −15 π 4+π 2 0 −15 π 4+π 2 68 12+ 9 π 4+π 2 . (18)
B. Calculation of the inertia tensor of the cylinder
For the cylinder, Eq. (14) becomes
I02 = I2+ πµ[(R2, R2)] − 2πµ[(R2, C2)], (19)
where C2 denotes the center of mass of the cylinder, which is the same as the reference point
in this part, then
C2 ≡ (xc, yc, zc) ≡ (0, 0, 0),
The position of the center of mass of the whole object relative to C2 is given by,
R2 ≡ (X, Y, Z) ≡
M1r
M1+ M2
(0, −3, −5) = 4r
The moment of inertia of the cylinder about its center is, I2 = 1 12M2(3r 2+ 64r2) 0 0 0 121 M2(3r2+ 4r2) 0 0 0 12M2r2 , or simply, I2 = πµr2 67 12 0 0 0 6712 0 0 0 12 . (20)
The other terms in Eq. s(19) are,
[(R2, C2)] = [0], [(R2, R2)] = 4r 4 + π 2 34 0 0 0 25 −15 0 −15 9 .
The inertia tensor of the cylinder about the center of mass of the whole object is then,
I02 = πµr2 67 12+ 34 4 4+π 2 0 0 0 6712 + 25 4+π4 2 −15 4 4+π 2 0 −15 4 4+π 2 1 2 + 9 4 4+π 2 . (21)
C. Calculation of the inertia tensor of compound object
The inertia tensor of the compound object is the sum of the two tensors associated with the two parts,
I0 = I01+ I02, (22)
from which we have,
I0 = µr2 100.034 0 0 0 64.197 −26.394 0 −26.394 40.074 . (23)
This example is chosen to be simple for manual calculation, but the procedure used here could be used for any complicated object. It is obvious that the generalized parallel axis theorem introduced in this paper provides a simple and powerful method to calculate the
inertia tensor about any point in the space. The strength of this generalization is the aggre-gation of all relations between inertia tensor elements about different points in one simple equation, without dealing with the center of mass of the aggregation as a reference. This generalization could be very powerful in the calculation of rotational inertia of compound objects that can be divided into simple units, such that the inertia tensor of each unit is known at least about one point. This type of analysis is useful for studying physical ob-jects through 3D imaging11 and designing objects for 3D printing,7,12 for which the object is considered as a set of slices or voxels.
1 Y. Alhassid, G.F. Bertsch, L. Fang, and S. Liu, “Nuclear moment of inertia and spin distribution
of nuclear levels,” Physical Review C 72 (6), 064326 (2005).
2 C.H. Townes, and A.L. Schawlow, Microwave spectroscopy (Courier Corporation, 2013). 3 M. Bejger, and P. Haensel, “Moments of inertia for neutron and strange stars: Limits derived
for the Crab pulsar,” Astronomy & Astrophysics 396 (3), 917–921 (2002).
4 L. Ran, J.R. Bumby, and P.J. Tavner, “Use of turbine inertia for power smoothing of wind
turbines with a DFIG,” Harmonics and Quality of Power, 2004. 11th International Conference on. IEEE, 106–111 (2004).
5 Y.S. Narang, V.M. Arelekatti, and A.G. Winter, “The effects of prosthesis inertial properties on
prosthetic knee Moment and hip energetics required to achieve able-bodied kinematics,” IEEE Transactions on Neural Systems and Rehabilitation Engineering 24 (7), 754–763 (2016).
6 T. Fedor, J. Vittek, and P. Sindler, “Influence of variable moment of inertia in robot servo
motor control,” ELEKTRO 2014. IEEE, 165–169 (2014).
7 M. B¨acher, E. Whiting, B. Bickel, and O. Sorkine-Hornung, “Spin-it: optimizing moment of
inertia for spinnable objects,” ACM Transactions on Graphics (TOG) 33 (4), 96 (2014).
8 H. Goldstein, Classical mechanics (Pearson Education India, 1965).
9 G.R. Fowles, and G.L. Cassiday, Analytical mechanics (Saunders college, 1999).
10 J.P. McKelvey, “A generalization of the perpendicular axis theorem for the rotational inertia of
rigid bodies,” American Journal of Physics 51 (7), 658–660 (1983).
optical methods,” Optical Engineering 39 (1), 10–22 (2000).
12 R. Prvost, E. Whiting, S. Lefebvre, and O. Sorkine-Hornung, “Make it stand: balancing shapes
FIGURE CAPTIONS
FIG. 1. Change of moment of inertia (Izz0 ) about an axis parallel to the z-axis, and intercepting the xy-plane at point (X, Y ). The center of mass has x-component and y-component as xcand yc,
respectively. Izz is the moment of inertia about the z-axis , i.e., X = Y = 0.
FIG. 2. Change of the product of inertia (Ixy0 ) about a point with x-component and y-component as X and Y , respectively. Ixy is the product of inertia about a point along the z-axis , i.e. X = Y = 0.
FIG. 3. Compound object of uniform density ρ, that consists of two parts. The first part is a cuboid of square base, of sides 2r, and of length 8r. The second part is a cylinder of a circular base, of radius r, and of length 8r. The object is aligned such that the length of the cuboid is parallel to the y-axis , and the length of the cylinder is parallel to the z-axis.
