Angular Momentum: H = I w
The concept of momentum is familiar to all of us that have listened to sports broadcasters. Teams or individuals frequently have "the momentum". What does this mean? It means they are going some place and it is going to be difficult to turn them aside from their intended destination. Newton, a great sports fan, quantified this concept. Newton introduced linear momentum, typically denoted G, and defined it as the mass of the system multiplied by the velocity of the mass center of the system. To be meaningful this velocity must be evaluated relative to a fixed or inertial coordinate system. Thus in terms of the notation developed in some of the earlier lessons:
G = M VG/A
G = Linear momentum of the system with respect to the fixed
coordinate system A.
M = Mass of the system.
G = Mass center of the system
A = Fixed or inertial coordinate system
VG/A= Velocity of the system mass center with respect
to the fixed coordinate system A.
Newton captured the difficulty of turning momentum aside with the observation that an external force is required to change the momentum of a system. In particular the time derivative with respect to a fixed or inertial coordinate system of the linear momentum is equal to the net sum of the external forces acting on the system.
S F = dG/dt|A
S F = Sum of the external forces
acting on the system.
Thus if the sum of the external forces is zero, the time derivative of the linear momentum is zero, and that linear momentum must have an unchanging constant value. To change or redirect the momentum, some external action or force must be applied - just as we have learned listening to sportscasters. The postulated relationship between force and changing momentum has been invaluable in the understanding of the motion of mechanical systems. More about this later!
But wait, this can't be the whole story! In watching systems move one observes not only a motion of the mass center, but frequently a spin or rotation about the mass center. Certainly a frisbee's mass center does not move very easily unless that mass center movement is accompanied by a significant rotational movement. And that baby just keeps on spinning until something grabs it. Sounds familiar. Could it be? If there is linear momentum, must there also be angular momentum?
Absolutely! If linear momentum is mass multiplied by velocity, then angular momentum must be angular mass multiplied by angular velocity. What could be sweeter? In fact the angular momentum of a rigid body can be expressed in exactly this form in two situations. The angular momentum of a rigid body about its mass center is equal to the angular mass matrix of the rigid body about its mass center multiplied by its angular velocity relative to a fixed or inertial coordinate system (what a mouthful). The symbol H is typically used for angular momentum. Thus the above statement can be formulated as the following equation:
HG = IG wB/A
B = Rigid body whose angular momentum is to be evaluated.
G = Mass center of the rigid body.
A = Fixed or inertial coordinate system.
wB/A = Angular velocity
of the rigid body B relative to the fixed coordinate system A.
IG = Angular mass matrix or inertia
tensor of the body B about its mass center G.
HG = Angular momentum of the rigid body B about its
mass center G relative to the fixed coordinate system A.
The formula for angular momentum is deceptive in its simplicity. While it looks like an ordinary multiplication, don't forget that the inertia tensor is a three by three matrix. Thus the three components of the angular momentum vector are given by the three results obtained by multiplying the three by three angular mass matrix with the three components of the angular velocity vector. Whoa! This has some very significant ramifications. Note that while linear momentum was always parallel to the mass center velocity (mass is a scalar), angular momentum is hardly ever parallel to angular velocity. If the angular mass matrix has any off-diagonal components or if any one of the diagonal components of the mass matrix is not equal to any of the other diagonal components, then the angular momentum may very well be in a different direction then the angular velocity. This makes angular momentum much more difficult to intuitively understand than linear momentum (hence sportscasters hardly ever talk about angular momentum). We will see the impact of this more complex nature of angular momentum when we consider some examples.
A second critical issue arises from the definition of angular momentum. Note that as we have seen earlier, the components of the angular velocity vector can be expressed in different coordinate systems. Likewise the components of the angular mass matrix can be expressed in different coordinate systems. For the definition of the angular momentum to be valid, the components of the angular mass matrix and the angular velocity vector must be expressed in the same coordinate system. Under these circumstances the three components of the angular momentum produced by the matrix multiplication are relative to that same coordinate system. It is crucial to note that it is often most convenient to express the angular mass matrix in terms of axes fixed or attached to the body in question. Thus we must express the angular velocity vector in that same body fixed coordinate system. The angular momentum vector that results is thus expressed in that body fixed (attached to the body) coordinate system. This observation becomes especially important if for some reason one has to take a time derivative of the angular momentum vector.
It is great to be able to calculate the angular momentum of a rigid body about its mass center, but not all bodies spin about their mass center. Do we have any other options? That brings us to the second situation for which the evaluation of the angular momentum of a rigid body is particularly straightforward. Consider a point O that is part of your rigid body but is also fixed in space (pivot point for the rigid body). Then the angular momentum of the rigid body about the point is given by the angular mass matrix (inertia tensor) about the point O of the rigid body multiplied by the angular velocity of the rigid body:
HO = IO wB/A
B = Rigid body whose angular momentum is to be evaluated.
O = Point attached to the body B that is also fixed in the inertial
coordinate system A.
A = Fixed or inertial coordinate system.
wB/A = Angular velocity
of the rigid body B relative to the fixed coordinate system A.
IO = Angular mass matrix or inertia
tensor of the body B about point O.
HO = Angular momentum of the rigid body B about point
O relative to the fixed coordinate system A.
For pivoted bodies the above expression for angular momentum is frequently useful. Note that the use of this formula normally requires the use of the inertia tensor transfer theorems developed in an earlier lesson to evaluate the angular mass matrix about the pivot point O. The observations about the complex nature of the angular momentum of a rigid body about its mass center are equally appropriate to the angular momentum of a rigid body about its pivot point. Care must be exercised in the use of these formulas or you will find that it is your head that is spinning rather than the rigid body!
Wow! Spinning is crazy! Have we hit all the highlights? No! Your thermodynamics teacher would rue the day you forgot to bring energy considerations to the table. If a translating object has kinetic energy, then it seems reasonable that a rotating object has kinetic energy. Recall that the kinetic energy of a translating rigid body is given by:
Kinetic Energy = 1/2 M VG2
Kinetic Energy = Translational kinetic energy of the rigid body.
M = Mass of the rigid body.
VG = Speed of the mass center of the rigid body relative
to a fixed or inertial coordinate system.
In order to relate this to momentum, note that the speed squared is exactly equal to the dot product of the velocity vector with itself. Thus:
Kinetic Energy. = 1/2 vG/A .
M vG/A
vG/A = Velocity of the mass center of the rigid body
relative to a fixed or inertial coordinate system A
A = Fixed or inertial coordinate system.
But wait, one sees a linear momentum in that formula:
Kinetic Energy = 1/2 vG/A .
G
If that works for translation, one can count on the following to account
for both the translational and rotational kinetic energy of a rigid body:
Kinetic Energy = 1/2 vG/A .
G + 1/2 wB/A .
HG
Kinetic Energy = Total kinetic energy of the rigid body relative to
the fixed coordinate system A.
A = Fixed or inertial coordinate system
B = Body
G = Mass center of the body B.
vG/A = Velocity of the mass center G of the body
B relative to the fixed coordinate system A.
wB/A = Angular velocity
of the body B relative to the fixed coordinate system A.
G = Linear momentum of the rigid body B relative to the fixed
coordinate system A.
HG = Angular momentum of the rigid body B about its
mass center G relative to the fixed coordinate system A.
Thus we can handle rotational kinetic energy as well as translational. Note also that our analogy between translational equations and rotational equations continue - only the rotational terms involve more algebra.
Hey wait, how about our other special case, a rigid body with a fixed
pivot point. Is there a simplified expression for the kinetic energy
of such a rigid body? You bet!
Kinetic Energy = 1/2 wB/A
. HO
Kinetic Energy = Total kinetic energy of the rigid body relative to
the fixed coordinate system A.
A = Fixed or inertial coordinate system
B = Body
O = Point attached to the rigid body B that is also fixed in the coordinate
system A (fixed pivot point of the body B).
wB/A = Angular velocity
of the body B relative to the fixed coordinate system A.
HO = Angular momentum of the rigid body B about its
fixed pivot point O relative to the fixed coordinate system A.
For pivoted bodies the above formula is very convenient. Remember though, in any analysis of angular motion, angular mass is a matrix and matrix manipulations including multiplication will be required! Note also that the importance of expressing the components of all of the vectors and matrices involved in a single common coordinate system prior to any multiplication! Note though that the single common coordinate system chosen for the expression of all vectors and matrices need not necessarily be a fixed or inertial coordinate system. Thus even though we are talking about momentums relative to a fixed coordinate system, those momentums can still be expressed in terms of unit vectors that may themselves be changing. This is an important observation because in many cases the expressions for the vectors of interest are much more conveniently obtained using components relative to a set of moving axes. Good luck on your journey into the world of three dimensional rotations!
O.K., so we have some nice formulas for angular momentum and kinetic
energy. How do we use them? Click here
to answer exactly that question!
Copyright (1998) by Nels Madsen. Last Updated:
February 28, 1998