Section 20–3 The gyroscope

(Uniform precession / Oscillatory precession / Damping precession)

In this section, Feynman discusses uniform precession (steady precession), oscillatory precession (nutation), and damping precession. To be precise, the section could be titled as “The precessional motion of gyroscope.” The discussion could be shifted to the last section because the concept of precession is not intuitive and many students may have difficulties understanding it. In Surely You’re Joking, Mr. Feynman!, Feynman made a mistake in saying that a wobbling plate spins twice as fast as it wobbles, but the truth is the plate wobbles twice as fast as it spins (Chao, 1989).

1. Uniform precession:

“… we have not proved (and it is not true) that a uniform precession is the most general motion a spinning body can undergo as the result of a given torque (Feynman et al., 1963, section 20–3 The gyroscope).”

In Fig. 20–3, Feynman shows how a horizontal torque causes a top to precess in the sense that its spin axis rotates in a circular cone about the vertical. Using the equation, τ = dL/dt = Ω × L₀, he explains that the direction of the precessional motion is in the direction of the torque, or at right angles to the forces producing the torque. Alternatively, we can use the phrase “torque axis,” “spin axis,” and “precession axis” to describe the motion of the top. That is, the torque axis (τ) should be perpendicular to the spin axis (L) and precession axis (Ω) in accordance with τ = Ω × L₀. Thus, uniform precession of a spinning top can be defined as rotational motion of the spin axis around the precession axis (e.g., vertical axis) due to a changing torque (constant magnitude) in the horizontal plane.

The magnitude of an angular momentum vector ΔL is expressed as L₀Δθ and the time rate of change of the angular momentum is τ = ΔL/Δt = L₀Δθ/Δt = L₀Ω. We need to idealize a simple top (or gyroscope) as a rigid body that has a symmetry axis and frictionless pivot. As a good approximation, the applied torque should be small relative to the angular momentum of the rapidly spinning top for a simple precession. In a sense, Feynman was rather sloppy by asking students to take the directions of the various quantities into account to see that τ = Ω × L₀. For a detailed explanation, one may prefer the equation dL/dt = dL/dt + w ´ L that can be used to derive the Euler’s equations of motion for a rotating rigid body (Morin, 2003).

2. Oscillatory precession:

“The general motion involves also a “wobbling” about the mean precession. This “wobbling” is called nutation (Feynman et al., 1963, section 20–3 The gyroscope).”

Feynman says that the general motion of a gyroscope involves a wobbling that is about the mean precession. This wobbling is called nutation (Latin word for nodding) and the gyro does fall below the mean precession, but as soon as it falls, it keeps turning and returns again to the initial level. This motion may also be known as oscillatory precession because the spin axis oscillates about the mean precession angle as it rotates about the precession axis. Feynman adds that all the formulas in the world such as the equation (20.15) does not describe the general motion of a gyroscope because it is valid only for uniform precession. However, we can use Euler’s equations of a rigid body to describe the general motion in terms of the Euler angles.

The general motion of a gyroscope is a superposition of torque-induced (uniform) precession and torque-free (oscillatory) precession. In Tips on Physics, Feynman clarifies that: “…if you throw an object into space alone, like a plate or a coin, you see it doesn’t just turn around one axis. What it does is a combination of spinning around its main axis, and spinning around some other cockeyed axis in such a nice balance, that the net result is that the angular momentum is constant (Feynman et al., 2006, p. 124).” The constancy of angular momentum implies that the additional motion due to nutation is torque-free. Specifically, nutational motion depends on the initial angular displacement and angular velocity of the spin axis just like how the pendula motion depends on its initial height and velocity.

3. Damping precession:

“The answer is that the cycloidal motion of the end of the axis damps down to the average, steady motion of the center of the equivalent rolling circle (Feynman et al., 1963, section 20–3 The gyroscope).”

We can use the phrase damping precession to describe how the nutational motion of the spin axis is damped by frictional forces. According to Feynman, the nutational motion is too quick for the eye to follow, and it damps out quickly because of the friction in the gimbal bearings. In Tips on Physics, Feynman elaborates that “when the airplane quiets down and goes in a straight line for a while, you’ll find that the gyro doesn’t point north anymore, because of friction in the gimbals. The airplane has been turning, slowly, and there has been friction, small torques have been generated, the gyro has had precessional motions, and it is no longer pointing in exactly the same direction (Feynman et al, 2006, p. 98).” Thus, pilots need to reset the directional gyro against the compass regularly.

Feynman explains that the axis of the gyro will be eventually a little bit lower than it was at the start. We should clarify that the gravitational potential energy of the gyro is converted into the rotational kinetic energy and thermal energy (due to friction between the bearings and the pivot). In Tips on Physics, Feynman adds that “the earth is not rigid; it’s got liquid goop on the inside, and so, first of all, its period is different from that of a rigid body, and secondly, the motion is damped out so it should stop eventually - that’s why it’s so small. What makes it nutate at all, despite the damping, are various irregular effects which jiggle the earth, such as the sudden motions of winds, and ocean currents (Feynman et al., 2006, p. 125).” In essence, the Earth also precesses like a gyroscope and its precession is affected by frictional forces.

Questions for discussion:

1. How would you explain the torque on a top using Newton’s second law of motion?

2. Why does the downward force of gravity make the top to move sidewise?

3. How would you explain the axis of the gyro is a little bit lower than it was at the start when the precessional motion settles down?

The moral of the lesson: the top does fall a little bit in the sense that its spin axis is lowered a little bit to allow precession about the vertical axis; the lowering of the spin axis causes a conversion of gravitational potential energy to rotational kinetic energy.

References:

1. Feynman, R. P. (1997). Surely You’re Joking, Mr. Feynman! : Adventures of a Curious Character. New York: Norton.

2. Feynman, R. P., Gottlieb, & M. A., Leighton, R. (2006). Feynman’s tips on physics: reflections, advice, insights, practice: a problem-solving supplement to the Feynman lectures on physics. San Francisco: Pearson Addison-Wesley.

3. 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.

4. Morin, D. (2003). Introductory Classical Mechanics. Cambridge: Cambridge University Press.

5. Chao, F. B. (1989). Feynman's Dining Hall Dynamics. Physics Today, 42(2), 15.

Section 20–2 The rotation equations using cross products

(Angular momentum vector / Angular velocity vector / Velocity vector)

In this section, Feynman discusses an angular momentum vector, angular velocity vector, and velocity vector of a rotating object.

1. Angular momentum vector:

“By the same token, the angular momentum vector, if there is only one particle present, r is the distance from the origin multiplied by the vector momentum… (Feynman et al., 1963, section 20–2 The rotation equations using cross products).”

Feynman expresses an angular momentum vector as L = r × p in which r is a displacement vector (instead of distance) and p is a linear momentum vector. In chapter 52, Feynman provides a better explanation on the angular momentum vector: “when we write a formula which says that the angular momentum is L = r × p, that equation is all right, because if we change to a left-hand coordinate system, we change the sign of L, but p and r do not change; the cross-product sign is changed, since we must change from a right-hand rule to a left-hand rule.” We may clarify that angular quantities are axial vectors, whereas linear quantities are polar vectors in the Gibbs vector system. Alternatively, the angular momentum vector is defined as a bivector in the Clifford vector system.

Feynman states a theorem of angular momentum: if the total external torque is zero, then the total vector angular momentum of the system is a constant. This theorem is also called the law of conservation of angular momentum and he rephrases it as “if there is no torque on a given system, its angular momentum cannot change.” In Chapter 52, Feynman says that “[i]nvariance under rotation through a fixed angle in space corresponds to the conservation of angular momentum.” This is also known as the Noether’s theorem that relates rotational symmetry to the law of conservation of angular momentum. One may add that the angular momentum vector of a rigid body is not definitely in the same direction as its angular velocity vector.

2. Angular velocity vector:

“That is to say, simply, angular velocity is a vector, where we draw the magnitudes of the rotations in the three planes as projections at right angles to those planes (Feynman et al., 1963, section 20–2 The rotation equations using cross products).”

Feynman asks whether angular velocity is a vector and discusses the rotation of an object about two axes simultaneously. The net result of the two rotations is that the object simply rotates about a new axis. However, Feynman did not specify that angular velocity vector is a pseudo vector. One may state a theorem of angular velocity addition: “Given a primed coordinate system rotating with angular velocity  w₁ with respect to an unprimed system, and a starred coordinate system rotating with angular velocity w₂ relative to the primed system, the angular velocity of the starred system relative to the unprimed system is w₁ + w₂ (Symon, 1971, p. 452).” Similarly, a gyroscope rotating about a horizontal axis and vertical axis can be related to the commutative law for addition: w_x + w_z = w_z + w_x.

Feynman says that angular velocity is a vector and we can draw the magnitudes of the rotations in the three planes as projections at right angles to those planes. He adds a footnote that suggests a derivation by compounding the angular displacements of the particles of the body during an infinitesimal time Δt. Importantly, angular displacements can be distinguished as infinitesimal rotations and finite rotations. Physics teachers should illustrate how angular displacements of an object (e.g., a book) about two perpendicular axes in two different orders do not commute. In general, a finite angular displacement is neither a polar vector nor an axial vector because it does not obey the commutative law for addition: θ_xi + θ_yj ¹ θ_yj + θ_xi.

3. Velocity vector:

“We shall leave it as a problem for the student to show that the velocity of a particle in a rigid body is given by v = ω × r… (Feynman et al., 1963, section 20–2 The rotation equations using cross products).”

If a rigid body is rotating about an axis with angular velocity ω, Feynman asks, “What is the velocity of a point at a certain radial position r?” He leaves it as a problem for students to show that the velocity of a particle in a rotating rigid body is given by the equation, v = ω × r, where ω is the angular velocity and r is the displacement vector (instead of position). Although the velocity of an object is usually a real vector, this should not be assumed when it is expressed as a cross product. In a sense, the velocity of the point in the rotating body is an axial vector because it is not moving in a straight line, but rotating about a point. However, Feynman uses the term vector because he classifies a vector as a polar vector and axial vector.

As another example, the Coriolis force can be expressed using a cross product: F_c = 2mv×ω. That is, if a particle is moving with velocity v in a coordinate system that is rotating with angular velocity ω, there is a pseudo force F_c from the perspective of a rotating frame. Physics teachers could elaborate that the cross product of an axial vector v and another axial vector ω is also an axial vector. In other words, the Coriolis force in a rotating coordinate system is not only a pseudo force, but it is also expressed as an axial vector (or pseudo vector). One may conclude that the cross product of two vectors (axial vector × axial vector or axial vector × polar vector) is always an axial vector.

Questions for discussion:

1. How would you define an angular momentum vector?

2. How would you define an angular velocity vector?

3. Is the instantaneous velocity of a particle in a rotating rigid body, v = ω × r, a real vector?

The moral of the lesson: angular momentum vectors, angular velocity vectors and the velocity of a particle in a rotating rigid body can be expressed using the cross product that is an axial vector.

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. Symon, K. R. (1971). Mechanics (3rd ed.). Addison-Wesley, MA: Reading.