16.07 Dynamics Fall 2004 ecture D23-3D Rigid Body Kinematics: The Inertia Tensor In this lecture, we will derive expressions for the angular momentum and kinetic energy of a 3D rigid body. We shall see that this introduces the concept of the Inertia Tensor Angular momentum We start form the expression of the angular momentum of a system of particles about the center of mass HG, derived in lecture D17 H=/ Here, T'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 am 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 Thus, we have e=/ r3×(×r)dhm=/I(r We note that, for planar bodies undergoing a 2D motion in its own plane, r' is perpendicular to w, and the term(r. w)is zero. In this case, the vectors w and HG are always parallel. In the three dimensional case however, this simplification does not occur, and as a consequence, the angular velocity vector, w, and the angular momentum vector, HG, are in general, not parallel In cartesian coordinates, we have, r'sr'i+yj+ak and w=wi+w,+w, k, and the above expression can be expanded to yield
J. Peraire 16.07 Dynamics Fall 2004 Version 1.0 Lecture D23 - 3D Rigid Body Kinematics: The Inertia Tensor In this lecture, we will derive expressions for the angular momentum and kinetic energy of a 3D rigid body. We shall see that this introduces the concept of the Inertia Tensor. Angular Momentum We start form the expression of the angular momentum of a system of particles about the center of mass, HG, derived in lecture D17, HG = Z m r ′ × v ′ dm . 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 ′ . Thus, we have, HG = Z m r ′ × (ω × r ′ ) dm = Z m [(r ′ · r ′ )ω − (r ′ · ω)r ′ ] dm . 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, 1
-4++y (/a +(-Izrwr-Iaywy+I22w2)k The quantities lxr, Iyy, and liz are called moments of inertia with respect to the a, y and z axis, respectively and are given by y2+2)dm,m=/(x2+2)dm,:=/(x2+y2)dm We observe that the quantity in the integrand is precisely the square of the distance to the a, y and z axis, espectively. 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 Izy, Irz, lyz, ly2, I2z and l2y are called products of inertia. They can be positive, negative, or zero, and are given by, =4=/ a'y'dm x'z′dm y'z dm If we are interested in calculating the angular momentum with respect to a fixed point O then, the resulting Ho =((Irro wr-(I vo Wy-(I)o w2)i +((lyro wr +(Iyy)o wy-(lyo w2)3 ((I)owr-(Ixy)o wy+(I2)o w2)k Here, the moments of products of inertia have expressions which are analogous to those given above but with a', y and z replaced by z, y and z. Thus, we have that (Ix)o=/(y2+2)dm,() x2+2)dm,(L2)o=/(x2+y2)dm (Lry)o=(Iyr)o=/ xy dm, (xz)o=(ar)o=/xz dm, (Iy:)o=(ay)o=/yzdm
HG = ωx Z m (x ′2 + y ′2 + z ′2 ) dm − Z m (ωxx ′ + ωyy ′ + ωzz ′ )x ′ dm i + ωy Z m (x ′2 + y ′2 + z ′2 ) dm − Z m (ωxx ′ + ωyy ′ + ωzz ′ )y ′ dm j + ωz Z m (x ′2 + y ′2 + z ′2 ) dm − Z m (ωxx ′ + ωyy ′ + ωzz ′ )z ′ dm k = ( Ixxωx − Ixyωy − Ixzωz) i + (−Iyxωx + Iyyωy − Iyzωz) j + (−Izxωx − Izyωy + Izzωz) k . (1) 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 = Z m (y ′2 + z ′2 ) dm , Iyy = Z m (x ′2 + z ′2 ) dm , Izz = Z m (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 can be positive, negative, or zero, and are given by, Ixy = Iyx = Z m x ′ y ′ dm , Ixz = Izx = Z m x ′ z ′ dm , Iyz = Izy = Z m y ′ z ′ dm . 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 . (2) Here, the moments of products of inertia have expressions which are analogous to those given above but with x ′ , y ′ and z ′ replaced by x, y and z. Thus, we have that (Ixx)O = Z m (y 2 + z 2 ) dm , (Iyy)O = Z m (x 2 + z 2 ) dm , (Izz )O = Z m (x 2 + y 2 ) dm , and, (Ixy)O = (Iyx)O = Z m xy dm , (Ixz)O = (Izx)O = Z m xz dm , (Iyz)O = (Izy)O = Z m yz dm . 2
The Tensor of inertia The expression for angular momentum given by equation 1, can be written in matrix form as, HGy H where [IG] is the tensor of inertia about the center of mass G and with respect to the ayz axes. The tensor of inertia gives us an idea about how the mass is distributed in a rigid body. It follows from the definition of the products of inertia, that the tensor of inertia is always symmetric. In addition, if the body has symmetries with respect to some of the axis, then some of the products of inertia become zero. For instance, if the bod is symmetric with respect to the plane I=0 then, we will have Iry= lyr =Irz=I2r =0. This can be shown by looking at the definition of the products of inertia. The integral for, say, lry can be decompe into two integrals for the two halves of the body at either side of the plane r =0. The integrand on half, r'y, will be equal in magnitude and opposite in sign to the integrand on the other half(because a' will change sign). Therefore, the integrals over the two halves will cancel each other and the product of inertia Iu will be zero Another case of practical importance is when we consider axisymmetric 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. Analogously, we can define the tensor of inertia about point O, by writing expression 2 in matrix form. Thus. where the components of [ Io] are the moments and products of inertia about point O given above Parallel axis Theorem It will often be easier to obtain the tensor of inertia with respect to axis passing through the center of mass. In some problems however, we will need to calculate the tensor of inertia about different axes. The parallel xis theorem introduced in lecture d18 for the two dimensional moments of inertia can be extended and applied to each of the components of the tensor of inertia
The Tensor of Inertia The expression for angular momentum given by equation 1, can be written in matrix form as, HGx HGy HGz = Ixx −Ixy −Ixz −Iyx Iyy −Iyz −Izx −Izy Izz ωx ωy ωz , or, HG = [IG] ω , where [IG] is the tensor of inertia about the center of mass G and with respect to the xyz axes. The tensor of inertia gives us an idea about how the mass is distributed in a rigid body. It follows from the definition of the products of inertia, that the tensor of inertia is always symmetric. In addition, if the body has symmetries with respect to some of the axis, then some of the products of inertia become zero. For instance, if the body is symmetric with respect to the plane x = 0 then, we will have Ixy = Iyx = Ixz = Izx = 0. This can be shown by looking at the definition of the products of inertia. The integral for, say, Ixy 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). Therefore, the integrals over the two halves will cancel each other and the product of inertia Ixy will be zero. Another case of practical importance is when we consider axisymmetric 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. Analogously, we can define the tensor of inertia about point O, by writing expression 2 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. Parallel Axis Theorem It will often be easier to obtain the tensor of inertia with respect to axis passing through the center of mass. In some problems however, we will need to calculate the tensor of inertia about different axes. The parallel axis theorem introduced in lecture D18 for the two dimensional moments of inertia can be extended and applied to each of the components of the tensor of inertia. 3
y In particular we can write, (Ix)o=/(x2+2)dm=/(v+y)2+(ag+2)2)dm (y2+2)+2/ydm+2c/2dm+(+a)/dm Ix+m(+2) Here, we have use the fact that y and 2 are the coordinates relative to the center of mass and therefore heir integrals over the body are equal to zero. Similarly, we can write (lv)O=l+m(x+2),(12)o=l2+m(x+v) (Ixy)o=(lyx)o=Ixg+mcv,(2)o=(I2x)o=Ixz+mxc,(ly=)o=(1xy)o=lyz+myec Rotation of axes In some situations, we will know the tensor of inertia with respect to some axes zyz and, we will be intereste in calculating the tensor of inertia with respect to another set of axis a'y z. We denote by i, i and k the unit vectors along the direction of zyz axes, and by i', j and k the unit vectors along the direction of r'y'z' y
In particular we can write, (Ixx)O = Z m (y 2 + z 2 ) dm = Z m ((yG + y ′ ) 2 + (zG + z ′ ) 2 ) dm = Z m (y ′2 + z ′2 ) + 2yG Z m y ′ dm + 2zG Z m z ′ dm + (y 2 G + z 2 G) Z m dm = Ixx + m(y 2 G + z 2 G) . 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. Similarly, we can write, (Iyy)O = Iyy + m(x 2 G + z 2 G), (Izz)O = Izz + m(x 2 G + y 2 G), and, (Ixy)O = (Iyx)O = Ixy + mxGyG, (Ixz)O = (Izx)O = Ixz + mxGzG, (Iyz)O = (Izy)O = Iyz + myGzG . 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. 4
If [n is the tensor of inertia with respect to the ayz axes(passing through either G or O), then [i= Irzi-Ixyi-Ixzk, and i [Ii=Ixr. More generally, if n is a unit vector, n. [In= Inn, where Inn is the moment of inertia about the axis defined by n and passing through the point to which [I refers. Similarly, ve can dot [/i with j to obtain j [i=-Ixy, and, more generally, m[/n=-Imn, where m and n are two orthogonal unit vectors Therefore, if we want to calculate the tensor of inertia with respect to axis r'y'a, we will have 1r k i'k ik ik 12 ki k ADDITIONAL READING L Meriam and L G. Kraige, Engineering Mechanics, DYNAMICS, 5th Edition
If [I] is the tensor of inertia with respect to the xyz axes (passing through either G or O), then [I]i = Ixxi − Ixyj − Ixzk, and i · [I]i = Ixx. More generally, if n is a unit vector, n · [I]n = Inn, where Inn is the moment of inertia about the axis defined by n and passing through the point to which [I] refers. Similarly, we can dot [I]i with j to obtain j · [I]i = −Ixy, and, more generally, m · [I]n = −Imn, where m and n are two orthogonal unit vectors. Therefore, if we want to calculate the tensor of inertia with respect to axis x ′y ′ z ′ , we will have Ix′x′ −Ix′y′ −Ix′z ′ −Iy′x′ Iy′y′ −Iy′z ′ −Iz ′x′ −Iz ′y′ Iz ′z ′ = i ′ · i j′ · i k′ · i i ′ · j j′ · i k′ · j i ′ · k i′ · k i′ · k Ixx −Ixy −Ixz −Iyx Iyy −Iyz −Izx −Izy Izz i ′ · i i′ · j i′ · k j ′ · i j′ · j j′ · k k ′ · i k′ · j k′ · k . ADDITIONAL READING J.L. Meriam and L.G. Kraige, Engineering Mechanics, DYNAMICS, 5th Edition 7/7, 7/8, Appendix B 5
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