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 9-15 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 FT that gives the plane a tangential acceleration aT. In accord with Newton’s second law, it follows that FT − maT. The torque produced by this force is , where the radius r of the circular path is also the lever arm. As a result, the torque is . But the tangential acceleration is related to the angular acceleration according to (Equation 8.10), where must be expressed in rad/s2. With this substitution for aT, 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 is directly proportional to the angular acceleration . 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 9-15 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 acts on the plane.

If all objects were single particles, it would be just as convenient to use the second law in the form as in the form . The advantage in using 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 9-16a 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, … , mN, 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 9-15 and obeys the relation :

Figure 9-16 (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.

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

| (9.5)

where the expression 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: must be expressed in rad/s2.

The version of Newton’s second law given in Equation 9.7 applies only for rigid bodies. The word “rigid” means that the distances r1, r2, r3, etc. that locate each particle m1, m2, m3, etc. (see Figure 9-16a) do not change during the rotational motion. In other words, a rigid body is one that does not change its shape while undergoing an angular acceleration in response to an applied net external torque.
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, must be expressed in rad/s2, because the relation (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 9-16b). 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.
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, as indicated by the following Problem solving insight the moment of inertia depends on the location and orientation of the axis relative to the particles that make up the object. Example 9 shows how the moment of inertia can change when the axis of rotation changes. The procedure illustrated in Example 9 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 inartia I for Various Rigig Objects of Mass M

| Example 9 | | | The Moment of Inertia Depends on Where the Axis Is

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 9-17.)

Figure 9-17 Two particles, masses m1 and m2, 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.

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.
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 , 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.

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 ax and ay). The forces can also produce torques, which can produce an angular acceleration . 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 . For the rotational motion, we use the law in the form . The Concepts-at-a-Glance chart in Figure 9-18 illustrates the essence of this joint usage of Newton’s second law. When a (both components) and 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 is nonzero, we have accelerated motion, and the object is not in equilibrium. Examples 10, 11, and 12, Examples 10, 11, and 12 and Examples 10, 11, and 12 deal with this type of situation.

Figure 9-18 CONCEPTS AT A GLANCE An object is in equilibrium when its translational acceleration components, ax and ay, and its angular acceleration are zero. If ax, ay, or is not zero, the object has an acceleration and is not in equilibrium. As this vacationer spirals around and down the water slide, he has translational and angular accelerations and is not in equilibrium. (John Berry/The Image

A N A L Y Z I N G M U L T I P L E - C O N C E P T P R O B L E M S

| Example 10 | | | The Torque of an Electric Saw Motor

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 . What net torque (assumed constant) must the motor apply to the blade?
Reasoning Newton’s second law for rotational motion, (Equation 9.7), can be used to find the net torque . However, when using the second law, we will need a value for the angular acceleration , which can be obtained by using one of the equations of rotational kinematics. In addition, we must remember that the value for must be expressed in rad/s2, not rev/s2, because Equation 9.7 requires radian measure.
Knowns and Unknowns The given data are summarized in the following table:

Description | Symbol | Value | Comment
Explicit Data |  |  |
Final angular velocity |  | 80.0 rev/s | Must be converted to rad/s.
Angular displacement |  | 240.0 rev | Must be converted to rad.
Moment of inertia | I |  |
Implicit Data |  |  |
Initial angular velocity |  | 0 rad/s | Blade starts from rest.
Unknown Variable |  |  |
Net torque applied to blade |  | ? |

Modeling the Problem

| Step 1 Newton’s second law for rotation
| Newton’s second law for rotation (Equation 9.7) specifies the net torque applied to the blade in terms of the blade’s moment of inertia I and angular acceleration . In Step 2, we will obtain the value for that is needed in Equation 9.7.

| (9.7)

| Step 2 Rotational Kinematics
| As the data table indicates, we have data for the angular displacement , the final angular velocity , and the initial angular velocity . With these data, Equation 8.8 from the equations of rotational kinematics can be used to determine the angular acceleration :

| (8.8)

Solving for gives

which can be substituted into Equation 9.7, as shown at the right.

Solution The results of each step can be combined algebraically to show that

In this result for , we must use radian measure for the variables , , and . To convert from revolutions (rev) to radians (rad), we will use the fact that . Thus, the net torque applied by the motor to the blade is

Related Homework: Problems 35, 38, 62 and 76

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 9-19 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 9-19 A rider applies a force 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 10 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 11 deals with one of them.

The physics of wheelchairs.

| Example 11 | | | Archery and Bow Stabilizers

Archers can shoot with amazing accuracy, especially using modern bows such as the one in Figure 9-20. 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 9-20 The stabilizer helps to steady the archer’s aim, as Conceptual Example 11 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. (Rob Griffith/©AP/ Wide World Photos)

The physics of archery and bow stabilizers.

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 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 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 ).

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.

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| Example 12 | | | Hoisting a Crate

A crate of mass 451 kg is being lifted by the mechanism shown in Figure 9-21a. 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 46.0 kg · m2. The cables roll on the dual pulley without slipping. A tension of magnitude 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 9-21 (a) The crate is lifted upward by the motor and pulley arrangement. The free-body diagram is shown 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. Four external forces act on the dual pulley, as its free-body diagram in Figure 9-21b shows. These are (1) the tension in the cable connected to the motor, (2) the tension in the cable attached to the crate, (3) the pulley’s weight , and (4) the reaction force exerted on the dual pulley by the axle. The force arises because the two cables and the pulley’s weight 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 in Figure 9-21c indicates. These are (1) the cable tension and (2) the weight m of the crate. The net force that results from these forces obeys Newton’s second law for translational motion (Equation 4.2b).
Knowns and Unknowns The following table summarizes the given information:

Description | Symbol | Value | Comment
Mass of crate | m | 451 kg |
Radius of outer part of dual pulley |  | 0.600 m | See Figure 9-21b.
Radius of inner part of dual pulley |  | 0.200 m | See Figure 9-21b.
Moment of inertia of dual pulley | I | 46.0 kg · m2 |
Magnitude of tension in cable attached to motor | T1 | 2150 N |
Unknown Variables |  |  |
Angular acceleration of dual pulley |  | ? |
Magnitude of tension in cable attached to crate | T2 | ? |

Modeling the Problem

| Step 1 Newton’s Second Law for Rotation
| Using the lever arms and shown in Figure 9-21b, we can apply the second law to the rotational motion of the dual pulley.

| (9.7)

Note that the forces and have zero lever arms, since their lines of action pass through the axle. Thus, these forces contribute nothing to the net torque. Minor rearrangement of Equation 9.7 gives Equation 1 at the right. In this result, all of the variables are known, except the angular acceleration and the tension T2. Therefore, before Equation 1 can be used to determine , a value for T2 is required. To obtain this value, we turn to Step 2.

| (1)

| Step 2 Newton’s Second Law for Translation
| Applying Newton’s second law to the upward translation motion of the crate gives (see part c of the drawing)

| (4.2b)

Note that the magnitude of the tension in the cable between the crate and the pulley is , so that we can solve Equation 4.2b for the tension and obtain

This result for T2 can be substituted into Equation 1, as shown at the right. The mass m of the crate is known, but the linear acceleration ay of the crate is not, so we proceed to Step 3 to determine its value.

| Step 2 Rolling Motion
| 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 of the pulley via (Equation 8.13), where . Thus, we have

We complete our modeling by substituting this result for ay into Equation 2, as indicated at the right.

Solution Combining the results of each step algebraically, we have

Solving for gives

Equation 1 can be solved for T2 to show that

Related Homework: Problems 40, 42 and 72

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 and force F are analogous quantities, as are moment of inertia I and mass m, and angular acceleration 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 |  | s
Velocity |  | v
Acceleration |  | a
The cause of acceleration | Torque | Force F
Inertia | Moment of inertia I | Mass m
Newton’s second law |  |
Work |  | Fs
Kinetic energy |  |
Momentum |  |

Check Your Understanding 3

Three massless rods are free to rotate about an axis at their left end (see the drawing). The same force 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. (The answer is given at the end of the book.)

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.

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