Section 20–1 Torques in three dimensions

(Torque vector / Transformation of torque / Axial vector)

In this section, Feynman discusses the concept of a torque vector, the transformation of the torque by rotation, and an axial vector.

1. Torque vector:

“…from Newton’s laws we see that we did not have to assume that the motion was in a plane; when we differentiate xp_y − yp_x, we get xF_y − yF_x, so this theorem is still right (Feynman et al., 1963, section 20–1 Torques in three dimensions).”

According to Feynman, if a rigid body is rotating in three dimensions, what we deduced for two dimensions is still applicable. For example, it is still true that xF_y − yF_x is the torque in the xy-plane, or the torque is “around the z-axis.” This torque is also equal to the rate of change of angular momentum, xp_y − yp_x, by using Newton’s second law of motion. Feynman’s explanation is not sufficient because one may prefer further discussion of the difference in sign in xF_y − yF_x. As a suggestion, it is good to let students derive the torque vector by resolving a force into two components, F_x and F_y. They should be able to deduce independently that the moments due to the two components are clockwise and anti-clockwise respectively, or vice versa.

Feynman cautions that one may get the wrong sign for a quantity if the coordinates are not handled in the right way. He clarifies that we may write τ_yz = zF_y – yF_z because a coordinate system may be either “right-handed” or “left-handed.” The handedness of a coordinate system or rotation is merely a convention in which we can assign clockwise to be the positive direction. To be more precise, if a screw is right-handed, it means that the screw is moved forward when it is rotated clockwise. Clockwise and anti-clockwise are also dependent on the perspective of an observer. An operational definition of handedness can be provided using Wu’s experiment on parity violation that is not based on a convention (See Chapter 52).

2. Transformation of torque:

“We wanted to get a rule for finding torques in new axes in terms of torques in old axes, and now we have the rule (Feynman et al., 1963, section 20–1 Torques in three dimensions).”

Feynman shows the transformation of a torque based on the coordinates of two systems that are related by x′ = xcos θ + ysin θ, y′ = ycos θ – xsin θ, z′ = z. Specifically, we can transform the torque in a new co-ordinate system rotated anti-clockwise by a angle θ using the following equations: τ_x′ = τ_xcos θ + τ_ysin θ, τ_y′ = τ_ycos θ − τ_xsin θ, τ_z′ = τ_z. Some may find Feynman’s method of direct substitution to find the torque in new axes unnecessary or lack of insights. A shorter and simpler method is to make use of the fact that the torque is a vector that is invariant under rotation. Although the torque can be expressed as a sum of vectors (e.g., τ_x′y′ = xF_y − yF_x), it is not a real vector that has all vector properties.

Feynman elaborates that a torque is a twist on a plane and it does behave like a vector. The torque vector is perpendicular to the plane of the twist and its length is proportional to the strength of the twist. To a certain extent, the three components of the torque transform like a real vector under rotation. As an alternative, we can simply transform a torque vector using a two dimensional rotation matrix: (i) first column: the point (1, 0) is rotated by an angle of θ anti-clockwise and moved to (cos θ, sin θ). (ii) second column: the point (0, 1) is rotated by an angle of θ anti-clockwise and moved to (−sin θ, cos θ). The rotation is a linear transformation and it can be expressed as R(x, y) = R[x(1, 0)+y(0, 1)] = xR(1, 0)+yR(0, 1).

3. Axial vector:

“Vectors which involve just one cross product in their definition are called axial vectors or pseudo vectors (Feynman et al., 1963, section 20–1 Torques in three dimensions).”

Feynman explains that axial vectors involve just a cross product (or vector product) in the definition and provides examples such as torque, angular momentum, angular velocity, and magnetic field. On the contrary, we have polar vectors such as coordinate, force, momentum, velocity, and electric field. In Chapter 52, Feynman adds that “a ‘vector’ which, on reflection, does not change about as the polar vector does, but is reversed relative to the polar vectors and to the geometry of the space; such a vector is called an axial vector (Feynman et al, 1963, section 52–5 Polar and axial vectors).” In a sense, the term vector product is a misnomer because it actually produces an axial vector. Feynman says that the torque is a vector, but it is not really a vector that it is loosely stated in books or websites.

Feynman elaborates that the cross product is very important for representing the features of rotation. He asks why the torque is a vector and then says that it is a miracle of good luck that we can associate a single axis with a plane, and thus we can associate a vector with the torque. Historically, the concept of torque as an axial vector is based on the Gibbs vector system (Chappell et al., 2016). However, Jackson (1999) identifies a dangerous aspect of vector notation and writes that “[t]he writing of a vector as a does not tell us whether it is a polar or an axial vector (p. 270).” Interestingly, the Clifford vector system includes the concept of wedge product and distinguishes the electric and magnetic field as a vector and bivector respectively (without the need of the axial vector).

Questions for discussion:

1. How would you define a torque vector?

2. How would you deduce the torque vector in another coordinate system that is rotated by an angle?

3. How would you define an axial vector?

The moral of the lesson: a torque vector is perpendicular to the plane of the twist, its length is proportional to the strength of the twist, and it behaves like an axial vector.

References:

1. Chappell, J. M., Iqbal, A., Hartnett, J. G., & Abbott, D. (2016). The vector algebra war: a historical perspective. IEEE Access, 4, 1997-2004.

2. 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.

3. Jackson, J. D. (1999). Classical Electrodynamics (3rd ed.). John Wiley & Sons, New York.

Section 19–4 Rotational kinetic energy

(Centrifugal force / Coriolis force / Centripetal force)

In this section, Feynman discusses the concept of centrifugal force, Coriolis force, and centripetal force from a perspective of work done on an object during rotation. His discussion of rotational kinetic energy is uncommon because he did not include typical examples such as an object rolling down a slope that has both translational kinetic energy and rotational kinetic energy. To be more precise, this section could be titled “Coriolis force” because this is his main focus. Instead of ending the chapter abruptly, there could be a mention of real-life examples relating to the Coriolis force such as long-range artillery trajectories, Foucault pendulum, or typhoons.

1. Centrifugal force:

“When we are rotating, there is a centrifugal force on the weights (Feynman et al., 1963, section 19–4 Rotational kinetic energy).”

According to Feynman, there is a centrifugal force on the weights when they are being rotated. An observer in an inertial frame of reference may elaborate that there is actually a tension (centripetal force) that pulls the weights. Feynman adds that the work we do against the centrifugal force ought to agree with the difference in the rotational kinetic energy of the weights. As a suggestion, one may explain that the rotational work should be done by a real force such as frictional force. In short, the centrifugal force exists as a fictitious (pseudo) force and it is a useful concept for an observer in a rotating frame of reference.

Feynman clarifies that the rotational work cannot be contributed by the centrifugal force because it is a radial force. This means that the centrifugal force is not the entire story in a rotating system. In chapter 12, Feynman says that “[a]nother example of pseudo force is what is often called ‘centrifugal force’ (Feynman et al., 1963, section 12–5 Pseudo forces).” However, physics teachers may derive the centrifugal force using (dr/dt)_fixed = (dr/dt)_rotating + w ´ r to help understanding the nature of forces in a rotating frame of reference in which r is a displacement vector (Thornton & Marion, 2004). In the derivation, we can identify three forces: (1) centrifugal force (proportional to w²), (2) Coriolis force (proportional to w), and (3) Euler force (proportional to dw/dt).

2. Coriolis force:

“Now let us develop a formula to show how this Coriolis force really works (Feynman et al., 1963, section 19–4 Rotational kinetic energy).”

Feynman explains that the Coriolis force has a very strange property: an object moving in a rotating system would appear to be pushed sidewise. For example, a person moving radially inward on a carousel would feel a sidewise force that is parallel to −2mω × v. We should clarify that the Coriolis force does arise from the motion of the object because the force is velocity-dependent and vanishes if it stops. In a sense, we may feel the existence of Coriolis force (or effect) when we walk radially inward or outward on a carousel. Importantly, the Coriolis force is a fictitious force that does not obey Newton’s third law (action = reaction) because it does not arise from an interaction between two objects.

Feynman suggests his students walk along the radius of a carousel whereby one has to lean over and push sidewise. (If we cannot counter this sidewise force with a frictional force of magnitude 2mωv, we would experience a tangential acceleration of 2ωv.) Feynman simply derives a formula of Coriolis force using τ = F_cr = dL/dt = d(mωr²)/dt = 2mωr(dr/dt), in which F_c is the Coriolis force. If we do not assume the angular speed to be constant, we can obtain mr²(dω/dt) that is known as the Euler force. However, this derivation does not help to understand the factor of 2 in 2mωv. One may prefer a physical explanation of the two equal components of Coriolis force using a general derivation of Coriolis force.

3. Centripetal force:

“This is simply the centripetal force that Moe would expect, having nothing to do with rotation (Feynman et al., 1963, section 19–4 Rotational kinetic energy).”

Unlike the previous example, an object is moving tangentially at a constant speed around the circumference of a carousel. In this example, Moe (rotating frame) observes the object moving at a velocity v_M at an instant, whereas Joe (inertial frame) sees it moving at a velocity v_J = v_M + ωr. One may add that the centripetal force is not a new or different force, but it may be a frictional force between the object and the carousel. We can provide an explanation of the direction of the Coriolis force using the cross product, but this is covered in the next chapter. Intuitively, the sidewise force is radially outward because of an inertial effect in which the object tends to be thrown out of the carousel.

Feynman explains the centripetal force on an object using the equation F_r = −mv_J²/r = −mω²r −mv_M²/r −2mv_Mω. Alternatively, from the perspective of an inertial observer, a person moving tangentially at a constant speed is essentially maintained by a frictional force (Morin, 2003). We can split the frictional force acting on the person walking on the carousel into three parts: F_friction = m(v_M + ωr)²/r = mv_M²/r + 2mv_Mωr + mω²r. The first term (mv_M²/r) is the additional frictional force on the person’s feet if he walks in a circle of radius r. The second term (2mv_Mωr) is the frictional force needed to counter the Coriolis force. The third term (mω²r) is the minimal frictional force needed that is equal to the centrifugal force due to the rotation of the carousel.

Questions for discussion:

1. How would you explain the nature of the Coriolis force?

2. How would you explain the Coriolis force for an object moving radially at a constant speed in a rotating system?

3. How would you explain the Coriolis force for an object moving tangentially (along a circumference) at a constant speed in a rotating system?

The moral of the lesson: the Coriolis force on an object is tangential when its velocity is radial, and the Coriolis force on the object is radial when its velocity is tangential.

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

3. Thornton, S. T. & Marion, J. B. (2004). Classical Dynamics of Particles and Systems (5th Edition). Belmont, CA: Thomson Learning-Brooks/Cole.

Section 19–3 Finding the moment of inertia

(General formula / Parallel axis theorem / Perpendicular axis theorem)

In this section, Feynman discusses the general formula of moment of inertia, and the use of parallel axis theorem and perpendicular axis theorem to find the moment of inertia of an object.

1. General formula:

“Now we must sum all the masses times the x-distances squared (the y’s being all zero in this case) (Feynman et al., 1963, section 19–3 Finding the moment of inertia).”

Feynman expresses the general formula of moment of inertia of an object about the z-axis as I = Sm_i(x_i²+y_i²). He explains that the distance in the expression is not a three-dimensional distance, but only a two-dimensional distance squared, even for a three-dimensional object. We may use the symbol I_z instead of I to emphasize the rotation is about the z-axis in the three-dimensional space. In a sense, Feynman only provides a mathematical definition of moment of inertia. However, it can be theoretically defined as an object’s tendency to resist angular acceleration that is a sum of the products of the mass of each particle in the body with the square of its distance from the axis of rotation. The moment of inertia of a complex system can be operationally defined by suspending the system from three points to form a trifilar pendulum.

The total moment of inertia of an object is the sum of the moments of inertia of the pieces. If the object is a rod, I = òx² dm and “the sum” means the integral of x² times the little elements of mass. That is, it is possible to divide the rod into small elements of length dx and the corresponding parts of the mass are proportional to dx. Curiously, Feynman did not explicitly relate the additive property of moment of inertia to any empirical evidence. In volume II, Feynman adds that “[t]he moment of inertia, then, is a tensor of the second rank whose terms are a property of the body and relate L to ω by L_i = S_jI_ijω_j (Feynman et al., 1964).” Importantly, the moment of inertia should not be treated as an isolated concept, but its additive property could be related to equations of angular momentum.

2. Parallel axis theorem:

“In applying the parallel-axis theorem, it is of course important to remember that the axis for I_c must be parallel to the axis about which the moment of inertia is wanted (Feynman et al., 1963, section 19–3 Finding the moment of inertia).”

Feynman states the parallel axis theorem as “[t]he moment of inertia about any given axis is equal to the moment of inertia about a parallel axis through the CM plus the total mass times the square of the distance from the axis to the CM.” He mentions the importance of remembering the axis for I_c must be parallel to the axis about which the moment of inertia is wanted. However, Feynman could have emphasized that the parallel axis theorem is valid only with the axis that passes through the center of mass of the object instead of any other point. Furthermore, this theorem holds for arbitrary non-planar objects that have certain thickness. Simply phrased, we may apply the theorem for objects that are two-dimensional, three-dimensional, or one dimensional.

In deriving the parallel axis theorem, Feynman provides an excellent explanation of the second term (2X_CMSm_ix′_i): x′ is measured from the center of mass, and in these axes the average position of all the particles, weighted by the masses, is zero. The theorem is applicable to any object supported on a pivot at the center of mass such that the object may rotate about an origin and spin about the axis through the center of mass. On the other hand, the parallel theorem could be derived from the perspective of an object whose center of mass rotates about the origin and spins about its center of mass. Thus, its angular momentum can be split into two parts: the motion of the center of mass and the motion around the center of mass. For example, a planet rotating about the sun has an orbital angular momentum as well as spin angular momentum.

3. Perpendicular axis theorem:

“… the moment of inertia of this figure about the z-axis is equal to the sum of the moments of inertia about the x- and y-axes (Feynman et al., 1963, section 19–3 Finding the moment of inertia).”

The perpendicular theorem is stated as “[i]f the object is a plane figure, the moment of inertia about an axis perpendicular to the plane is equal to the sum of the moments of inertia about any two mutually perpendicular axes lying in the plane and intersecting at the perpendicular axis.” Feynman mentions earlier that we shall restrict to two-dimensional objects for the most part of this section. In essence, the perpendicular axis theorem is only applicable to idealized objects that lie entirely within a plane. Physics teachers should compare the two theorems by clarifying that the parallel axis theorem is applicable to non-planar objects, whereas the perpendicular theorem is applicable only to planar objects. Better still, one may define planar objects as two-dimensional objects that have no thickness.

In deriving the perpendicular axis theorem, Feynman shows that I_x = Sm_i(y_i²+z_i²) = Sm_iy_i² and I_y = Sm_i(x_i²+z_i²) = Sm_ix_i² because z_i = 0. Thus, I_z = I_x + I_y and we can say that the moment of inertia of a planar object about the z-axis is equal to its moment of inertia about the x-axis plus its moment of inertia about the y-axis. The proof is based on an ideal condition in which the object has no thickness and thus z = 0 for all points within the object. However, we can still apply the theorem in real life for very thin objects that may give approximately useful results. In other words, the theorem is possibly applicable to objects whose surface area (A) is significantly greater than its thickness (A >> t) from the perspective of scaling property of center of mass.

Questions for discussion:

1. How would you explain the general formula of moment of inertia of a composite system?

2. What are the physical conditions of the parallel axis theorem?

3. What are the physical conditions of the perpendicular axis theorem?

The moral of the lesson: the general formula of moment of inertia is I = Sm_i(x_i²+y_i²) and the parallel axis theorem is applicable to non-planar objects (the axis passes through CM), whereas the perpendicular theorem is applicable only to planar objects.

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. Feynman, R. P., Leighton, R. B., & Sands, M. (1964). The Feynman Lectures on Physics, Vol II: Mainly electromagnetism and matter. Reading, MA: Addison-Wesley.

Section 19–2 Locating the center of mass

(Pappus’ first theorem / Pappus’ second theorem / Centroid of a triangle)

In this section, Feynman discusses Pappus’ centroid theorem for volume, Pappus’ theorem for surface area, and how to locate the centroid of a triangle.

1. Pappus’ first theorem:

“One such trick makes use of what is called the theorem of Pappus (Feynman et al., 1963, section 19–2 Locating the center of mass).”

Feynman mentions a mathematical trick that is known as the theorem of Pappus. This theorem of center of mass (centroid) can be stated as “[i]f a plane area rotates about an axis in its own plane which does not intersect it, the volume generated is equal to the area times the length of the path of its centroid (Symon, 1971, p. 224).” That is, if we take any closed area in a plane and generate a solid by rotation such that each point is moved perpendicular to the plane of the area, the resulting solid has a total volume equal to the area of the cross-section times the distance moved by the centroid. However, Feynman did not prove Pappus’ theorem formally. There could be a simple example using a rectangular area (A) to generate a cylindrical volume (V) in which V = (2πr)A, where 2πr is the distance moved by the centroid.

Pappus’ centroid theorem for volume (or Pappus–Guldin theorem) shows a connection between surface area, volume of revolution, and centroid. Feynman elaborates that the theorem is true for a curved path because the outer parts go around farther, but the inner parts go around lesser, and these effects balance out. Importantly, two necessary conditions of Pappus’ theorem are (1) a plane sheet of uniform density, and (2) the generated solid does not intersect itself during rotation. Feynman says that the theorem is true if we move the closed area in a circle or in some other curve. About seven years after Feynman's lecture, Adolph Winkler Goodman and Gary Goodman (1969) developed a generalization of Pappus’ theorem that allows the closed area to move in a natural manner on any sufficiently smooth simple closed curve.

2. Pappus’ second theorem:

“There is another theorem of Pappus which is a special case of the above one, and therefore equally true (Feynman et al., 1963, section 19–2 Locating the center of mass).”

Pappus’ centroid theorem for surface area is sometimes described as a special case of Pappus’ first theorem. The surface area that is swept by a plane curve after rotation is equal to the distance moved by the centroid times the length of the curve. Pappus’ second theorem can be stated as “[i]f a plane curve rotates about an axis in its own plane which does not intersect it, the area of the surface of revolution which it generates is equal to the length of the curve multiplied by the length of the path of its centroid (Symon, 1971, p. 224).” This theorem can be proved using A = ò_C 2py ds = 2p ò_C yds = 2pYs in which Y is the y-coordinate of the centroid and s is the length of the curve. There could be a simple example using a straight line of length (s) to generate a cylindrical area (A) in which A = (2πr)s, where 2πr is the distance moved by the centroid.

Feynman suggests applying Pappus’ centroid theorem to a semicircular piece of wire with uniform mass density to find its centroid. In this case, there is no mass in the center of the semicircular wire, except only within the wire. However, he did not show how to solve this problem during his lecture. We may idealize the wire (length: πR) as having a very narrow thickness, and apply Pappus’ theorem to generate a spherical surface area (4πR²): (2πY)(πR) = 4πR². Solving this equation, we obtain Y = 2R/π where Y is the y-coordinate of the centroid and R is the radius of the semicircular wire. Alternatively, we can locate the centroid of the semicircular wire using the general formula, Y = òydm/M = ò (Rsin q)(rR dq)/(rπR) = 2R/π, and by taking limit from π radians to zero.

3. Centroid of a triangle:

“… if we wish to find the center of mass of a right triangle of base D and height H (Fig. 19–2), we might solve the problem in the following way (Feynman et al., 1963, section 19–2 Locating the center of mass).”

Pappus’ theorem can be easily applied to a right triangle of base D and height H. By rotating the triangle 360 degrees about an axis through H (Fig. 19–2), it generates a cone. The volume of the cone with height H and radius D is πD²H/3, which is exactly one third the volume of the smallest cylinder that the cone can fit inside. Based on Pappus’ theorem, the cone’s volume (πD²H/3) is equal to the area of the triangle (½HD) times the circumference (2πX) moved by the centroid. Thus, (2πX)(½HD) = πD²H/3, and we can obtain the x-component of the centroid as X = D/3. Similarly, by rotating the right triangle about the other axis, we can deduce the y-component of the centroid as Y = H/3.

Feynman has selected a good example because the centroid of any uniform triangular area can also be located at where the three medians (the lines from the vertices through the centers of the opposite sides) intersect. As an alternative, we can locate the centroid of a triangle by constructing the perpendicular bisectors of any two sides. More important, Feynman gives a clue that involves slicing the triangle into a lot of little pieces, each parallel to a base without providing the workings. The centroid of the right triangle can be calculated using the formula R = òr dm/M (or Y = òy dm/M) where M = rV = r(HBt) and dm = rty dx = rt(xH/B) dx in which H is the height and B is the base of the triangle. Using integration and taking limit from x = B to x = 0, we can obtain the y-coordinate of the centroid that is H/3.

Questions for discussion:

1. How would you state Pappus’ centroid theorem for volume?

2. How would you state Pappus’ centroid theorem for surface area?

3. How would you locate the centroid of a right triangle?

The moral of the lesson: Pappus’ centroid theorem is a useful mathematical trick that helps to locate the center of mass of an object.

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. Goodman, A. W., & Goodman, G. (1969). Generalizations of the theorems of Pappus. The American Mathematical Monthly, 76(4), 355-366.

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