In Taylor, read sections 10.1 to 10.3 for today, and 10.3 to 10.5 for Monday.
Since we will often deal with the CM, it is useful to express the angular momentum in terms of the CM of the system. Recall how we used CM coordinates in the example from last lecture. Let's do that again here, r'i = ri - RCM, and likewise we can express the velocity in terms of the CM velocity and the part relative to the CM.
This can be expanded to give four terms:
This can be simplified by noting that the second and third terms on the right are zero, and the last term can be expressed in terms of individual momenta to get
To reach this solution I've used ∑i mi v'i = 0 and ∑i mir'i = 0. Can you see why these are true? Hint, try calculating RCM by expressing the coordinates in terms of the CM coordinates and the coordinate of the CM.
The above result says that the total angular momentum of a system of particles is composed of two parts: the angular momentum of the CM and the angular momentum about the CM. The second part is what we normally think of as the rotation or spinning of something. The first part we might consider as the angular momentum due to the orbit, or path of our system. This is quite a convenient occurance; the center of mass is a special point indeed.
where Lorbit = RCM × PCM and Lrot = ∑i r'i × p'i.
This is beautifully illustrated by the motion of a planet. The planet (or its CM) orbits the sun, and therefore has orbital angular momentum. It can also spin on an axis through the CM, giving it rotational angular momentum. Note also, that the directions of the orbital and rotational angular momenta don't have to line up. For the planets in our solar system, they generally don't (the earth's axis of rotation is tilted by 23.5° to the direction of its orbital angular momentum).
For a system of n particles the kinetic energy is the sum of the individual kinetic energies:
Now if we express vi in terms of the CM velocity and the velocity relative to the CM, we can break the kinetic energy into a piece due to the motion of the CM and a piece due to the relative motions of the particles.
Inserting this into the kinetic energy expression, and noting that the remaining dot product (third term) sums to zero, we get:
In words, the total kinetic energy of a system is equal to the kinetic energy due to the motion of the CM plus the kinetic energy due to motion of particles of the system relative to their CM. This latter part can come from rotation of the system (a collective motion of all the particles of the system), or the motion of individual particles of the system. If the system of particles are part of a rigid body, then all the relative velocities, v'i, are due to the object's rotation about the CM.
An alternative way to think of the kinetic energy comes from realizing that the derivation, up to the simplification of the expansion for vi², doesn't depend on using the CM. That is, we can replace VCM by the velocity of any point on the object, V, such that vi = V + v'i, and write the kinetic energy as
Now if we choose the reference point to be one with zero (instantaneous) velocity such that V=0, the kinetic energy is just the second term
Written this way, we find that the kinetic energy of the object is the rotational energy about any fixed point.
We can write the potential energy for the conservative forces on a system. (Any non-conservative forces must be handled separately.) Write the potential for due to internal and external forces separately, U = Uext + Uint. Since the external force can be treated as acting on a point mass located at the CM, with mass equal to the total mass of the system, we can show that the external potential is dependent on the position of the CM only, Uext = Uext(RCM).
The internal potential energy depends on the relative separation of masses in the object, Uint = ∑i<jUij(rij) where rij is the separation between two particles. In a rigid body, the separations are fixed, therefore the internal potential energy is a constant. The overall potential energy then reduces to U = Uext(RCM).
For the remainder of this chapter we will concentrate on the rotation of a body about its CM (projectile or satellite for example) or about a fixed point (a spinning top). We'll begin with the special, but common, case of an object rotating on a fixed axis. We can imagine this to be a rotating wheel, disc, or fan. For simplicity, we'll take the z axis as the axis of rotation, and we can write the angular velocity vector ω = ωzhat = (0, 0, ω)
The angular momentum of the body is the sum of the angular momenta of its parts:
In chapter 9 we learned that for a rotating object, va = ω×ra. For the choice of angular velocity and an arbitrary position ra = (xa, ya, za), this yields va = (-ωya, ωxa, 0). The angular momenta terms become la = ra×mava = maω(-zaxa, -zaya, xa² + ya²). Notice that there are three components to the angular momentum despite the fixed axis of rotation.
Summing the angular momenta terms yields the total angular momentum. The total angular momentum has three non-zero components, L = (Lx, Ly, Lz), and I'll treat them each below.
The quantity xa² + ya² is just the perpendicular distance from the axis of rotation, squared, ra⊥². So we can write
where
is the usual expression for the moment of inertia about the z axis. This is the standard, elementary physics equations for angular momentum.
If the system of particles are part of a rigid body, then all the relative velocities, v'a, are due to the object's rotation about the CM and can be related to the angular velocity and distance from the axis of rotation (passing through the CM), v'i = ωra⊥:
It may be a bit surprising that there are angular momentum components in the x and y directions, considering that the rotation is fixed about the z axis. These terms exist and are generally non-zero:
The basic point that the angular velocity vector and angular momentum vectors can, and often will, point in different directions. The basic rule that L = Iω is a special case, not a generally valid rule. A common, everyday, situation where this occurs is when a car has an unbalanced wheel. An unbalanced wheel is one where Lx or Ly is non-zero, where the z axis is in the direction of the axle. Let's assume Lx is the piece that is non-zero. But, the x-axis is fixed to the wheel, so as the wheel rotates, the direction of Lx changes, meaning that Ldot ≠ 0. A torque is required to change L; with your car, the forces needed to create the torque ultimately arise from friction between the tires and the road. To create a torque in the direction of the axle, this friction is along the direction of travel. To create a torque perpendicular to the axle, in the direction of the rotating x-axis, a side-to-side force is needed, causing the car to shake or jitter.
Balanced wheels are important to smooth handling of a car, especially at high speed. Two types of wheel balancing are generally performed: static balancing ensures that mass is distributed evenly around the wheel when laid horizontally; dynamic balancing ensures that mass is distributed evenly in all three planes such that when spun, Lx = Ly = 0. Dynamic balancing is preferred. A wheel can be balanced statically, but have non-zero components of angular momentum in the x or y directions.
© 2006 Robert Harr