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.

Section 18–2 Rotation of a rigid body

(Kinematics of rotation / Dynamics of rotation / Formulas of torque)

In this section, Feynman discusses kinematics of rotation, dynamics of rotation, and three formulas of torque.

1. Kinematics of rotation:

“… let us study the kinematics of rotations. The angle will change with time, and just as we talked about position and velocity in one dimension, we may talk about angular position and angular velocity in plane rotation (Feynman et al., 1963, section 18–2 Rotation of a rigid body).”

According to Feynman, we may simplify the motion of an object that wobbles, shakes, and bends to a simple rotation and idealize a rigid body that does not exist. In the kinematics of motion, the angular displacement θ refers to how far the body has rotated; it replaces the distance s, which represents how far the body has moved. Similarly, angular velocity ω = dθ/dt tells us how much the angle changes with respect to time, just as v = ds/dt describes how far the body moves in a second. One may prefer Feynman to add that it is useful to introduce the angular velocity vector, ω, which points along the axis of rotation, with a magnitude equal to the angular speed. The choice of a direction of the angular velocity vector is a matter of convention using a right-hand rule or left-hand rule.

Feynman says that if the angle is measured in radians, then the angular velocity ω will be so and so many radians per second. As a suggestion, one may explain that the advantage of using the radian instead of the degree is that the radian leads to a simple formula in which we can express the length of the arc of a circle in terms of its radius. On the other hand, he adds that if OP is called r, then the length PQ is rΔθ, because of the way angles are defined. Mathematically, the change in x is simply the projection of rΔθ in the x-direction: Δx = −PQsin θ = −(rΔθ)(y/r) = −yΔθ. However, Feynman could have clarified that this involves an approximation: the real distance moved is a small part of the arc of a circle, but it is almost equal to the length PQ that is a straight line for a small angle, Δθ.

2. Dynamics of rotation:

“Let us now move on to consider the dynamics of rotation. Here a new concept, force, must be introduced (Feynman et al., 1963, section 18–2 Rotation of a rigid body).”

Feynman explains that torque is a “twist” and relates it to the work done in turning an object. Additionally, the definition of the torque is arranged such that the theorem of work has an analog: force times distance is translational work and torque times angle is rotational work. Quantitatively, the translational work done is ΔW = F_xΔx+F_yΔy, whereas the rotational work is ΔW = (xF_y−yF_x)Δθ. In chapter 20, Feynman elaborates that torque is a vector because “it is a miracle of good luck that we can associate a single axis with a plane, and therefore that we can associate a vector with the torque; it is a special property of three-dimensional space (Feynman et al., 1963, section 20–1 Torques in three dimensions).” Conversely, in a hypothetical two-dimensional world, the torque is an ordinary scalar.

Although Feynman uses the word equilibrium, there are three kinds of equilibrium in this context: static equilibrium, translational equilibrium, and rotational equilibrium. In summary, the two conditions of static equilibrium refer to translational equilibrium in which the sum of the forces must be zero and rotational equilibrium in which the sum of all the torques must be zero. Importantly, these two conditions imply there are no translational work and no rotational work that mean “no energy transfer” and thus, a stationary body should continue to remain at rest. Furthermore, it can be proved that “if the vector sum of the forces on an object is zero, and if the sum of the torques about any one point is zero, then the sum of the torques about any other point is also zero (French, 1971, p. 125).”

3. Formulas of torque:

“There is still a third formula for the torque which is very interesting (Feynman et al., 1963, section 18–2 Rotation of a rigid body).”

Feynman mentions that there is a third formula for the torque without clearly stating the formula in symbolic forms. For instance, the formula for the torque may also be stated using a cross product such that t = r ´ F = |r| |F| sin α. In addition, the second formula of torque can be expressed as t = (F sin α) r in which α is the smaller angle between F and r. Feynman interprets this formula of torque as a quantity that is equal to the tangential component of force (perpendicular to the radius, OP) times the radius. In other words, this torque is a product of the “effective force” (F sin α) and the radius, r. In the case where we push right on the axis, there will not be any twisting at all because of r = 0.

In contrast to the second formula of torque that is t = (F sin α) r, we may express the third formula of torque as t = F (r sin α). Feynman explains that if we extend the line of action of the force and draw the line OS, the perpendicular distance to the line of action of the force is also known as the lever arm of the force. This lever arm is shorter than the radius r in just the same proportion as the tangential part of the force is lesser than the total force. The third formula of torque can be interpreted as a quantity that is equal to the magnitude of the total force times the effective length (r sin α). In short, we may have either a shorter effective length (r sin α) or smaller effective force (F sin α).

Note: there are other formulas of torque (e.g., t = dL/dt or t = 2mwr dr/dt) that can be found in the next chapter.

Questions for discussion:

1. What are the idealizations or approximations involved in the kinematics of motion?

2. How would you explain the conditions of static equilibrium?

3. How would you interpret the three formulas of torque?

The moral of the lesson: in Fig. 18–2, if we extend the line of action of the force and draw the line OS, the perpendicular distance to the line of action of the force (the lever arm of the force), we should realize that this lever arm is shorter than r in just the same proportion as the tangential part of the force is lesser than the total force; thus, we may write t = (F sin α) r = F (r sin α).

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. French, A. (1971). Newtonian Mechanics. New York: W. W. Norton.