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.

Section 19–1 Properties of the center of mass

(Symmetrical properties / Scaling properties / Accelerated reference frame)

In this section, Feynman discusses symmetrical properties and scaling properties of the center of mass, as well as a theorem of motion of the center of mass in an accelerated reference frame.

1. Symmetrical properties:

“… if it is just any symmetrical object, then the center of gravity lies somewhere on the axis of symmetry because in those circumstances there are as many positive as negative x’s (Feynman et al., 1963, section 19–1 Properties of the center of mass).”

Center of mass (CM) is a point “inside” an object where the net external force may produce an acceleration of an imaginary particle at this point as if the whole mass of the object were to be concentrated there. Using circular symmetry, Feynman clarifies that CM does not have to be in the material of a body, for example, the CM of a hoop is in the center of the hoop that is not in the hoop itself. In the case of a rectangle that is symmetrical in two planes, we can easily determine its CM that lies on their line of intersection. Similarly, if a body is symmetrical about an axis, its CM also lies on the same axis. More importantly, we can use possible symmetrical properties of CM to simplify physics problems.

Idealization (or simplification): we idealize a rigid body as a system of discrete particles (or a continuous distribution of matter) and gravitational forces are uniform in a small region of space. Feynman also proves a theorem that simplifies the location of the center of mass by assuming a body is composed of two or more parts whose centers of mass are known. However, one may elaborate that the location of the center of mass of a rigid body is uniquely defined, but the center of mass vector is dependent on the selected coordinate system. Furthermore, the center of mass is defined without reference to the gravity of an object. These properties of the center of mass could have been summarized at the beginning or at the end of the section.

2. Scaling properties:

“…Newton’s law has the peculiar property that if it is right on a certain small scale, then it will be right on a larger scale (Feynman et al., 1963, section 19–1 Properties of the center of mass).”

The concept of scaling is not commonly found in current physics textbooks. Feynman explains that Newton’s laws of dynamics hold for the motion of objects at a higher scale and it becomes more accurate as the scale gets larger. On the contrary, quantum mechanics for the small-scale atoms are quite different from Newtonian mechanics that are applicable to large-scale objects. Historically, Galileo (1638) writes that “if the size of a body be diminished, the strength of that body is not diminished in the same proportion; indeed the smaller the body the greater its relative strength (p. 131).” He investigated scaling properties, for example, the relationship between speed and distance of a moving body as well as the irregular shapes of bones. Feynman (1969) has also contributed to the concept of scaling in his interpretation of experimental results on deep inelastic scattering (electron-proton collisions).

Approximation: according to Feynman, we can approximate the motion of bodies at a larger scale by a certain expression in which it keeps reproducing itself on a larger and larger scale. In addition, Newton’s laws of dynamics are similar to the “tail end” of the atomic laws and they can be extrapolated to a very large scale. Interestingly, the laws of motion of particles on a small scale are very peculiar, but a large number of particles also approximately obey Newton’s laws. Currently, one may prefer Feynman to suggest the need of laws of motion for galaxies, for example, Modified Newtonian dynamics (MOND) is developed for motion of bodies at an even larger scale. It provides an alternative explanation for the motion of galaxies that do not appear to obey Newton’s laws.

3. Accelerating reference frame:

“… the theorem that torque equals the rate of change of angular momentum is true in two general cases: (1) a fixed axis in inertial space, (2) an axis through the center of mass, even though the object may be accelerating (Feynman et al., 1963, section 19–1 Properties of the center of mass).”

Feynman states two validity conditions of the theorem concerning the center of mass in which the torque is equal to the rate of change of angular momentum: (1) a fixed axis in an inertial frame, (2) an axis through the center of mass in an accelerating frame. He elaborates that an observer in an accelerating box would expect the same situation (or experience same magnitude of forces) if an object were in a uniform gravitational field whose g value is equal to the acceleration a. One may add that an inertial force acting on the object is equivalent to an apparent gravitational force based on Einstein’s principle of equivalence. In other words, the theorem involving an external torque acting on an accelerating object is equivalent to the same object that is at rest, but it is now under the influence of apparent gravitational field.

Exception (or limitation): when a small object is supported at its center of mass, there is no torque on it because of a parallel gravitational field. This is not strictly true for a large object because gravitational forces are non-uniform, and thus, the center of gravity of the large object departs slightly from its center of mass. However, Feynman could have included Chasles’ theorem that describes the motion of a body as a sum of two independent motions: a translation of the body plus a rotation about an axis. A special case of this theorem is to choose the axis at the center of mass of the body that allows the angular momentum to be split into two components: the motion of the center of mass and the motion around the center of mass. It helps to connect the discussions of translational kinetic energy and rotational kinetic energy in section 19.4.

Questions for discussion:

1. What are the symmetrical properties of the center of mass?

2. What are the scaling properties of the center of mass?

3. What are the validity conditions of the theorem concerning the center of mass in which an external torque is equal to the rate of change of angular momentum?

The moral of the lesson: the center of mass of a rigid body has symmetrical properties and scaling properties, and there are two validity conditions of the theorem concerning the center of mass in which the torque is equal to the rate of change of angular momentum: a fixed axis in an inertial reference frame and an axis through the center of mass in an accelerating reference frame.

References:

1. Feynman, R. P. (1969). Very high-energy collisions of hadrons. Physical Review Letters, 23(24), 1415-1417.

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. Galilei, G. (1638/1914). Dialogues Concerning Two New Sciences (Trans. by Crew, H. and de Salvio, A.). New York: Dover.

Section 18–4 Conservation of angular momentum

(External & internal torques / Constant angular momentum / Moment of inertia)

In this section, Feynman discusses the change of angular momentum due to external and internal torques, constant angular momentum, and moment of inertia of a system of particles. This section could be titled as “angular momentum of a system of particles” because he has discussed the conservation of angular momentum in the previous section.

1. External and internal torques:

“… internal torques balance out pair by pair, and so we have the remarkable theorem that the rate of change of the total angular momentum about any axis is equal to the external torque about that axis (Feynman et al., 1963, section 18–4 Conservation of angular momentum).”

Feynman explains that the total angular momentum of a system of particles is the sum of the angular momenta of all the parts. Thus, the rate of change of the total angular momentum about an axis of rotation is equal to the external torque about the axis. This theorem of angular momentum is applicable to any system of objects whether they form a rigid body or not. In the next chapter, Feynman discusses why the torque is equal to the rate of change of the angular momentum about an axis through the center of mass (CM) of an object that is accelerating. To be precise, French (1971) writes that “[r]egardless of any acceleration that the center of mass of a system of particles may have as a result of a net external force exerted on the system, the rate of change of internal angular momentum about the CM is equal to the total torque of the external forces about the CM (p. 641).”

According to Feynman, if Newton’s third law means that the action and reaction are equal, and they are directed in opposite directions exactly along the same line, then the two torques on two interacting objects are equal and opposite because the lever arms for any axis are equal. Note that he did not specify whether both action and reaction pass through the two objects. Interestingly, Kleppner and Kolenkow (1973) explain that “there is no way to prove from Newton’s laws that the internal torques add to zero. Nevertheless, it is an experimental fact that internal torques always cancel because the angular momentum of an isolated system has never been observed to change (p. 253).” Importantly, action and reaction are forces at a distance that do not necessarily lie on the straight line that connects the two objects.

Note: There are at least two forms of Newton’s third law: 1. Strong form means that action and reaction must act along the line joining the two particles. 2. Weak form means that action and reaction need not act along the line joining the two particles.

2. Constant angular momentum:

“…the law of conservation of angular momentum: if no external torques act upon a system of particles, the angular momentum remains constant (Feynman et al., 1963, section 18–4 Conservation of angular momentum).”

Feynman states the law of conservation of angular momentum as “if no external torques act upon a system of particles, the angular momentum remains constant.” A special case is that of a rigid body in which it has a definite shape while it is rotating around an axis. One may add that the conservation of angular momentum is dependent on the absence of an external torque, but the kinetic energy of a body may not be constant. For example, a planet orbiting about the Sun may increase its speed when it is moving nearer to the Sun. Feynman adds that we should consider a body that is fixed in its geometrical dimensions and is rotating about a fixed axis. However, a “fixed axis” means that the axis is fixed relative to the body and fixed in direction relative to an inertial frame, but it is not necessarily fixed in space.

One may expect Feynman to discuss Kepler’s second law of planetary motion in this section that is titled conservation of angular momentum. Historically, Kepler formulates the law of constant areal velocity as “a line drawn from the sun to a planet sweeps out equal areas in equal time intervals.” In other words, the areal velocity of a planet, orbiting the sun as a focal point, is always constant. Newton was the first physicist to recognize the physical significance of Kepler’s second law that is related to a radial force. Importantly, the law of conservation of angular momentum is applicable to a radial force that is not necessarily inversely proportional to the square of the distance. Therefore, physics teachers should clarify that “no external torque” does not mean that there is no external force.

3. Moment of inertia:

“Velocity is replaced by angular velocity, and we see that the mass is replaced by a new thing which we call the moment of inertia I, which is analogous to the mass (Feynman et al., 1963, section 18–4 Conservation of angular momentum).”

Feynman says that a body has inertia for turning which depends on the masses of its parts and how far away they are from the axis. He adds that the mass of an object never changes, but its moment of inertia can be changed; this is in contrast to his concept of relativistic mass. In Tips on Physics, Feynman elaborates that “for any rigid body, there is an axis through the body’s center of mass about which the moment of inertia is maximal, there is another axis through the body’s center of mass about which the moment of inertia is minimal, and these are always at right angles (Feynman et al., 2006, p. 122).” In short, the momentum of inertia of a rigid body is also dependent on the axis of rotation.

The law of conservation of angular momentum may be rephrased as “if the external torque is zero, then the angular momentum, the moment of inertia (I) times angular velocity (w), remains constant.” Feynman mentions that an important difference between mass and moment of inertia is very dramatic: if we stand on a frictionless rotatable stand with our arms outstretched, we may change our moment of inertia by drawing our arms in, but our mass does not change. However, one may expect Feynman to describe the spinning of a ice-skater that can be dramatic. A more dramatic example would be how a cat can rotate itself in the air after dropped vertically from an upside-down position and can become upright on its feet.

Questions for discussion:

1. How would you state the law of rate of change of angular momentum that is in terms of an external torque?

2. How would you state the law of conservation of angular momentum that is in the absence of an external torque?

3. How would you state the law of conservation of angular momentum that is in terms of moment of inertia?

The moral of the lesson: the angular momentum of a system of particles is constant if there is no external torque, or equivalently, the moment of inertia of an object times its angular velocity remains constant.

References:

1. 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.

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

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