Section 18–3 Angular momentum

(Angular momentum of a particle / Three formulas / Angular momentum of a planet)

In this section, Feynman discusses the angular momentum of a particle, three formulas of angular momentum, and the angular momentum of a planet. This section could be titled as “angular momentum of a particle,” whereas the next section could be titled as “angular momentum of a system of particles” instead of “conservation of angular momentum.” (Feynman has also discussed the conservation of angular momentum in this section.)

1. Angular momentum of a particle:

“First, of course, we should consider just one particle. In Fig. 18–3 is one particle of mass m, and an axis O; the particle is not necessarily rotating in a circle about O… (Feynman et al., 1963, section 18–3 Angular momentum).”

Feynman starts with a simple example: a particle of mass m may move elliptically like a planet going around the sun (an axis O) or in some other curve. The moving particle is under the influence of forces, and it accelerates in accordance with the formula of F = ma, for example, the x-component of force is the mass times the x-component of acceleration. On the other hand, it may not seem natural to derive the formula of angular momentum using τ = xF_y − yF_x = xm(d²y/dt²) − ym(d²x/dt²). One may explain that the formula of angular momentum, L = r × p, is merely a definition. Alternatively, we can derive this formula of angular momentum by taking a cross product of r with Newton’s second law of motion such that r × F = τ = r × dp/dt and then deduce that L = r × p if there is no torque (τ = 0).

In chapter 20, Feynman explains that the angular momentum vector for a particle is equal to the cross product of a displacement vector and the linear momentum vector: L = r × p. Curiously, the direction of angular momentum does not seem intuitive to students or physicists. In a textbook on mechanics, Kleppner & Kolenkow (1973) write that “probably the strangest aspect of angular momentum is its direction. The vectors r and p determine a plane (sometimes known as the plane of motion), and by the properties of the cross product, L is perpendicular to this plane. There is nothing particularly ‘natural’ about the definition of angular momentum (p. 233).” More important, this formula of angular momentum is useful for physicists in daily life applications.

2. Three formulas:

“So we have three formulas for angular momentum, just as we have three formulas for the torque: L = xp_y−yp_x = r.p_tang = p.lever arm (Feynman et al., 1963, section 18–3 Angular momentum).”

Feynman mentions that the second form of the equation L = xm(dy/dt) − ym(dx/dt) = xp_y−yp_x is relativistically valid. Subsequently, he says that we have three formulas for angular momentum, just as we have three formulas for the torque: L = xp_y − yp_x = r ´ p_tang = p ´ lever arm. Thus, there are four formulas of angular momentum within this page. In general, angular momentum may also be written as L = r mv sin q. In the case of L = r ´ p_tang, it may be expressed as L = r (mv sin q) = rmv_^ where v_^ = v sin q is the tangential component of the motion and mv_^ is the tangential momentum (p_tang). We can also express the angular momentum as L = (r sin q) mv, or L = r_^ mv, where r_^ = r sin q is the effective length of the moment arm, or simply lever arm.

In chapter 19, there are more formulas of angular momentum. Interestingly, Feynman clarifies that the present discussion is nonrelativistic, but the second form for L (xp_y−yp_x) is relativistically correct. Although the first form of angular momentum, L = xm(dy/dt) − ym(dx/dt), is not relativistically correct, he revises it to a possibly correct form without defining the p_x and p_y. To be relativistically correct, we can include the Lorentz factor in the first form of angular momentum such that it becomes L = xmg(dy/dt) − ymg(dx/dt). Furthermore, physics teachers should emphasize that the angular momentum of a particle depends upon the position of the axis about which it is to be calculated.

3. Angular momentum of a planet:

“But there is no tangential force, so there is no torque about an axis at the sun! Therefore, the angular momentum of the planet going around the sun must remain constant (Feynman et al., 1963, section 18–3 Angular momentum).”

Feynman mentions that the angular momentum of a planet depends on where we select the axis of rotation, and there is no torque about an axis at the sun. Therefore, the angular momentum of the planet around the sun must remain constant because the torque is zero for a central force. It also means that the tangential component of velocity, times the mass, times the radius, will be constant. One may prefer Feynman to represent the area swept out by the radius vector in a short period of time as dA = r(r dθ)/2 in which r dθ is the base of the thin triangle. Using the angular momentum formula, L = mr² dθ/dt, we can express dA/dt = (r²/2)(dθ/dt) = L/2m and deduces that angular momentum is constant for a central force.

Feynman concludes that Kepler’s second law (or law of equal areas in equal times) is equivalent to the statement of the law of conservation of angular momentum when there is no torque produced by the central force. He shows that if we consider a short period of time Δt, a planet can move from P to Q (Fig. 18–3) and sweep through the area OPQ (by disregarding the relatively smaller area QQ′P). Mathematically, it is about half the base PQ (= r dθ) times the height, OR (= r), that is, dA = r(r dθ)/2. This means that dA/dt = ½ (r) d(rθ)/dt = ½ (lever arm) (velocity) in which d(rθ)/dt is the velocity of the planet and r is equal to the lever arm. In essence, the rate of change of area as swept by the planet is proportional to its angular momentum that is constant.

Questions for discussion:

1. How would you define the angular momentum of a particle?

2. How would you explain the three formulas of angular momentum?

3. Does the angular momentum of a planet depend on the axis of rotation selected?

The moral of the lesson: the angular momentum of a particle is equal to the magnitude of the tangential momentum times the length of the lever arm and it depends upon the position of the axis about which it is to be calculated.

References:

1. Feynman, R. P., Leighton, R. B., & Sands, M. (1963). The Feynman Lectures on Physics, Vol I: Mainly mechanics, radiation, and heat. Reading, MA: Addison-Wesley.

2. Kleppner, D., & Kolenkow, R. (1973). An Introduction to Mechanics. Singapore: McGraw-Hill.