9.4.  Newton’s Second Law for Rotational Motion About a Fixed Axis

The goal of this section is to put Newton’s second law into a form suitable for describing the rotational motion of a rigid object about a fixed axis. We begin by considering a particle moving on a circular path. Figure 9.14 presents a good approximation of this situation by using a small model plane on a guideline of negligible mass. The plane’s engine produces a net external tangential force F_T that gives the plane a tangential acceleration a_T. In accord with Newton’s second law, it follows that F_T=ma_T. The torque t produced by this force is t=F_Tr, where the radius r of the circular path is also the lever arm. As a result, the torque is t=ma_Tr. But the tangential acceleration is related to the angular acceleration a according to a_T=ra (Equation 8.10), where a must be expressed in rad/s². With this substitution for a_T, the torque becomes

(9.4)

Equation 9.4 is the form of Newton’s second law we have been seeking. It indicates that the net external torque t is directly proportional to the angular acceleration a. The constant of proportionality is I=mr², which is called the moment of inertia of the particle. The SI unit for moment of inertia is kg · m².

Figure 9.14  A model airplane on a guideline has a mass m and is flying on a circle of radius r (top view). A net tangential force FT acts on the plane.

If all objects were single particles, it would be just as convenient to use the second law in the form F_T=ma_T as in the form t=Ia. The advantage in using t=Ia is that it can be applied to any rigid body rotating about a fixed axis, and not just to a particle. To illustrate how this advantage arises, Figure 9.15a shows a flat sheet of material that rotates about an axis perpendicular to the sheet. The sheet is composed of a number of mass particles, m₁, m₂, ..., m_N, where N is very large. Only four particles are shown for the sake of clarity. Each particle behaves in the same way as the model airplane in Figure 9.14 and obeys the relation t=(mr²)a:

In these equations each particle has the same angular acceleration a, since the rotating object is assumed to be rigid. Adding together the N equations and factoring out the common value of a, we find that

(9.5)

where is the sum of the external torques, and represents the sum of the individual moments of inertia. The latter quantity is the moment of inertia I of the body:

(9.6)

In this equation, r is the perpendicular radial distance of each particle from the axis of rotation. Combining Equation 9.6 with Equation 9.5 gives the following result:

ROTATIONAL ANALOG OF NEWTON’S SECOND LAW FOR A RIGID BODY ROTATING ABOUT A FIXED AXIS

(9.7)  Requirement: a must be expressed in rad/s2.

Figure 9.15  (a) A rigid body consists of a large number of particles, four of which are shown. (b) The internal forces that particles 3 and 4 exert on each other obey Newton’s law of action and reaction.

The form of the second law for rotational motion, , is similar to that for translational (linear) motion, , and is valid only in an inertial frame. The moment of inertia I plays the same role for rotational motion that the mass m does for translational motion. Thus, I is a measure of the rotational inertia of a body. When using Equation 9.7, a must be expressed in rad/s², because the relation a_T=ra (which requires radian measure) was used in the derivation.

When calculating the sum of torques in Equation 9.7, it is necessary to include only the external torques, those applied by agents outside the body. The torques produced by internal forces need not be considered, because they always combine to produce a net torque of zero. Internal forces are those that one particle within the body exerts on another particle. They always occur in pairs of oppositely directed forces of equal magnitude, in accord with Newton’s third law (see m₃ and m₄ in Figure 9.15b). The forces in such a pair have the same line of action, so they have identical lever arms and produce torques of equal magnitudes. One torque is counterclockwise, while the other is clockwise, the net torque from the pair being zero.

Concept Simulation 9.3

One or two forces can be applied to a rod that is free to rotate about its center. Each force produces a torque about the axis. The user can change the net torque by altering the magnitude and lever arms of the forces. The direction of one of the forces can also be changed. The simulation calculates the net torque and graphically illustrates the resulting angular displacement q  and velocity w of the rod as a function of time.

It can be seen from Equation 9.6 that the moment of inertia depends on both the mass of each particle and its distance from the axis of rotation. The farther a particle is from the axis, the greater is its contribution to the moment of inertia. Therefore, although a rigid object possesses a unique total mass, it does not have a unique moment of inertia, for the moment of inertia depends on the location and orientation of the axis relative to the particles that make up the object. Example 8 shows how the moment of inertia can change when the axis of rotation changes.

Two particles each have a mass m and are fixed to the ends of a thin rigid rod, whose mass can be ignored. The length of the rod is L. Find the moment of inertia when this object rotates relative to an axis that is perpendicular to the rod at (a) one end and (b) the center. (See Figure 9.16.) Reasoning  When the axis of rotation changes, the distance r between the axis and each particle changes. In determining the moment of inertia using , we must be careful to use the distances that apply for each axis. Figure 9.16  Two particles, masses m 1 and m 2, are attached to the ends of a massless rigid rod. The moment of inertia of this object is different, depending on whether the rod rotates about an axis through (a) the end or (b) the center of the rod. Solution (a) Particle 1 lies on the axis, as part a of the drawing shows, and has a zero radial distance: r1=0. In contrast, particle 2 moves on a circle whose radius is r2=L. Noting that m1=m2=m, we find that the moment of inertia is  (9.6)  (b) Part b of the drawing shows that particle 1 no longer lies on the axis but now moves on a circle of radius r1=L/2. Particle 2 moves on a circle with the same radius, r2=L/2. Therefore, This value differs from that in part (a) because the axis of rotation is different.

The procedure illustrated in Example 8 can be extended using integral calculus to evaluate the moment of inertia of a rigid object with a continuous mass distribution, and Table 9.1 gives some typical results. These results depend on the total mass of the object, its shape, and the location and orientation of the axis.

Table 9.1   Moments of inertia for Various Rigid Objects of Mass M

CONCEPTS AT A GLANCE When forces act on a rigid object, they can affect its motion in two ways. They can produce a translational acceleration a (components a_x and a_y). The forces can also produce torques, which can produce an angular acceleration a. In general, we can deal with the resulting combined motion by using Newton’s second law. For the translational motion, we use the law in the form SF=ma. For the rotational motion, we use the law in the form St=Ia. The Concepts-at-a-Glance chart in Figure 9.17 illustrates the essence of this joint usage of Newton’s second law. When a (both components) and a are zero, there is no acceleration of any kind, and the object is in equilibrium. This is the situation already discussed in Section 9.2. If any component of a or a is nonzero, we have accelerated motion, and the object is not in equilibrium. Examples 9, 10, and 11 deal with this type of situation.

Figure 9.17  CONCEPTS AT A GLANCE An object is in equilibrium when its translational acceleration components, ax and ay, and its angular acceleration a are zero. If ax, ay, or a is not zero, the object has an acceleration and is not in equilibrium. As these motorcyclists round the turn, they have translational and angular accelerations, so they are not in equilibrium. (© Pascal Rondeau/Stone/Getty Images)

The motor in an electric saw brings the circular blade from rest up to the rated angular velocity of 80.0 rev/s in 240.0 rev. One type of blade has a moment of inertia of 1.41×10–2 kg · m2. What net torque (assumed constant) must the motor apply to the blade? Reasoning  Newton’s second law for rotational motion (Equation 9.7, St=Ia) can be used to find the net torque St, once the angular acceleration a is determined. The angular acceleration can be calculated from the data in the following table and the appropriate equation of rotational kinematics:  q    a   w   w0   t   1508 rad (240.0 rev)   ?   503 rad/s (80.0 rev/s)   0 rad/s     The data for the angular displacement q  and the angular velocity w have been converted to radian measure because St=Ia requires that a be expressed in radian measure. Solution From Equation 8.8 () it follows that Newton’s second law for rotational motion (Equation 9.7) can now be used to obtain the net torque:

To accelerate a wheelchair, the rider applies a force to a handrail attached to each wheel. The torque generated by this force is the product of the magnitude of the force and the lever arm. As Figure 9.18 illustrates, the lever arm is just the radius of the circular rail, which is designed to be as large as possible. Thus, a relatively large torque can be generated for a given force, allowing the rider to accelerate quickly.

Figure 9.18  A rider applies a force F to the circular handrail. The torque produced by this force is the product of its magnitude and the lever arm about the axis of rotation. (© Dan Coffee/The Image Bank/Getty Images)

Example 9 shows how Newton’s second law for rotational motion is used when design considerations demand an adequately large angular acceleration. There are also situations when it is desirable to have as little angular acceleration as possible, and Conceptual Example 10 deals with one of them.

Archers can shoot with amazing accuracy, especially using modern bows such as the one in Figure 9.19. Notice the bow stabilizer, a long, thin rod that extends from the front of the bow and has a relatively massive cylinder at the tip. Advertisements claim that the stabilizer helps to steady the archer’s aim. Could there be any truth to this claim? Explain. Figure 9.19  The stabilizer helps to steady the archer’s aim, as Conceptual Example 10 discusses. Relative to an axis through the archer’s shoulder, the moment of inertia of the bow is larger with the stabilizer than without it. The counterweights below the archer’s hand help to keep the center of gravity of the bow near the hand. (© Amwell/Stone/Getty Images) Reasoning and Solution To help explain why the stabilizer works, we have added to the photograph an axis for rotation that passes through the archer’s shoulder and is perpendicular to the plane of the paper. Any angular acceleration a about this axis will lead to a rotation of the bow that will degrade the archer’s aim. The acceleration will be created by any unbalanced torques that occur while the archer’s tensed muscles try to hold the drawn bow. Newton’s second law for rotation indicates, however, that the angular acceleration is . The moment of inertia I is in the denominator on the right side of this equation. Therefore, to the extent that I is larger, a given net torque St will create a smaller angular acceleration and less disturbance of the aim. It is to increase the moment of inertia of the bow that the stabilizer has been added. The relatively massive cylinder is particularly effective in increasing the moment of inertia, because it is placed at the tip of the stabilizer, far from the axis of rotation (a large value of r in the equation I=Smr2).

Rotational motion and translational motion sometimes occur together. The next example deals with an interesting situation in which both angular acceleration and translational acceleration must be considered.

Concept Simulation 9.4

A block is hanging from one end of a string. The other end wraps around a pulley that is fastened to the ceiling above the block. When the block is released it accelerates downward. You can adjust the radius and mass of the pulley, as well as the mass of the block. You can also select from three different pulley designs. The simulation determines the linear acceleration of the block, the angular acceleration of the pulley, and the tension in the string. With a calculation analogous to that in Example 11, check to see whether your values agree with those arrived at by the simulation. Related Homework: Problem 42

A crate that weighs 4420 N is being lifted by the mechanism shown in Figure 9.20a. The two cables are wrapped around their respective pulleys, which have radii of 0.600 and 0.200 m. The pulleys are fastened together to form a dual pulley and turn as a single unit about the center axle, relative to which the combined moment of inertia is I=50.0 kg · m2. A tension of magnitude T1=2150 N is maintained in the cable attached to the motor. Find the angular acceleration of the dual pulley and the tension in the cable connected to the crate. Figure 9.20  (a) The crate is lifted upward by the motor and pulley arrangement. The free-body diagram for (b) the dual pulley and (c) the crate. Reasoning  To determine the angular acceleration of the dual pulley and the tension in the cable attached to the crate, we will apply Newton’s second law to the pulley and the crate separately. Three external forces act on the dual pulley, as its free-body diagram shows (Figure 9.20b). These are (1) the tension T1 in the cable connected to the motor, (2) the tension T2 in the cable attached to the crate, and (3) the reaction force P exerted on the dual pulley by the axle. The force P arises because the two cables pull the pulley down and to the left into the axle, and the axle pushes back, thus keeping the pulley in place. The net torque that results from these forces obeys Newton’s second law for rotational motion (Equation 9.7). Two external forces act on the crate, as its free-body diagram indicates (Figure 9.20c). These are (1) the cable tension T'2 and (2) the weight W of the crate. The net force that results from these forces obeys Newton’s second law for translational motion (Equation 4.2b). Solution Using the lever arms 1 and 2 shown in part b of the figure, we can apply the second law to the rotational motion of the pulley. We note that the force P has a zero lever arm, since the line of action of P passes directly through the axle:  (9.7)  This equation contains two unknown quantities, so a second equation is needed and may be obtained by applying Newton’s second law to the upward translational motion of the crate. In doing so, we note that the magnitude of the tension in the cable attached to the crate is T'2=T2 and that the mass of the crate is m=(4420 N)/(9.80 m/s2)=451 kg:  (4.2b)  Because the cable attached to the crate rolls on the pulley without slipping, the linear acceleration ay of the crate is related to the angular acceleration a of the pulley via Equation 8.13: ays=ra=(0.200 m)a. With this substitution for ay, Equation 4.2b becomes This result and Equation 9.7 can be solved simultaneously to yield

We have seen that Newton’s second law for rotational motion, , has the same form as that for translational motion, , so each rotational variable has a translational analog: torque t and force F are analogous quantities, as are moment of inertia I and mass m, and angular acceleration a and linear acceleration a. The other physical concepts developed for studying translational motion, such as kinetic energy and momentum, also have rotational analogs. For future reference, Table 9.2 itemizes these concepts and their rotational analogs.

Table 9.2   Analogies Between Rotational and Translational Concepts

Physical Concept   Rotational   Translational   Displacement   q    s   Velocity   w   v   Acceleration   a   a   The cause of acceleration   Torque t   Force F   Inertia   Moment of inertia I   Mass m   Newton’s second law   St=Ia   SF=ma   Work   tq    Fs   Kinetic energy   ½Iw2   ½mv2   Momentum   L=Iw   p=mv

Three massless rods are free to rotate about an axis at their left end (see the drawing). The same force F is applied to the right end of each rod. Objects with different masses are attached to the rods, but the total mass (3m) of the objects is the same for each rod. Rank the angular acceleration of the rods, largest to smallest. Background: Newton’s second law for rotational motion holds the key here. However, it is also necessary to understand the related concepts of net torque, moment of inertia, and angular acceleration. For similar questions (including calculational counterparts), consult Self-Assessment Test 9.2. This test is described at the end of Section 9.6.

Copyright © 2000-2003 by John Wiley & Sons, Inc. or related companies. All rights reserved.