8.5

Centripetal Acceleration and Tangential Acceleration

When an object picks up speed as it moves around a circle, it has a tangential acceleration, as discussed in the last section. In addition, the object also has a centripetal acceleration, as emphasized in Chapter 5. That chapter deals with uniform circular motion, in which a particle moves at a constant tangential speed on a circular path. The tangential speed vT is the magnitude of the tangential velocity vector. Even when the magnitude of the tangential velocity is constant, an acceleration is present, since the direction of the velocity changes continually. Because the resulting acceleration points toward the center of the circle, it is called the centripetal acceleration. Figure 8-13a shows the centripetal acceleration for a model airplane flying in uniform circular motion on a guide wire. The magnitude of is

| (5.2)

The subscript “T” has now been included in this equation as a reminder that it is the tangential speed that appears in the numerator.

Figure 8-13 (a) If a model airplane flying on a guide wire has a constant tangential speed, the motion is uniform circular motion, and the plane experiences only a centripetal acceleration . (b) Nonuniform circular motion occurs when the tangential speed changes. Then there is a tangential acceleration in addition to the centripetal acceleration.

The centripetal acceleration can be expressed in terms of the angular speed by using (Equation 8.9):

| (8.11)

Only radian measure, such as rad/s, can be used for in this result, since the relation presumes radian measure.
While considering uniform circular motion in Chapter 5, we ignored the details of how the motion is established in the first place. In Figure 8-13b, for instance, the engine of the plane produces a thrust in the tangential direction, and this force leads to a tangential acceleration. In response, the tangential speed of the plane increases from moment to moment, until the situation shown in the drawing results. While the tangential speed is changing, the motion is called nonuniform circular motion.
Figure 8-13b illustrates an important feature of nonuniform circular motion. Since the direction and the magnitude of the tangential velocity are both changing, the airplane experiences two acceleration components simultaneously. The changing direction means that there is a centripetal acceleration . The magnitude of at any moment can be calculated using the value of the instantaneous angular speed and the radius: . The fact that the magnitude of the tangential velocity is changing means that there is also a tangential acceleration . The magnitude of can be determined from the angular acceleration according to , as the previous section explains. If the magnitude FT of the net tangential force and the mass m are known, aT also can be calculated using Newton’s second law, . Figure 8-13b shows the two acceleration components. The total acceleration is given by the vector sum of and . Since and are perpendicular, the magnitude of the total acceleration can be obtained from the Pythagorean theorem as , while the angle in the drawing can be determined from tan . The next example applies these concepts to a discus thrower.

Check Your Understanding 3

The blade of a lawn mower is rotating at an angular speed of 17 rev/s. The tangential speed of the outer edge of the blade is 32 m/s. What is the radius of the blade? (The answer is given at the end of the book.) Background: Angular variables, tangential variables, and the relation between them are the focus of this problem.
For similar questions (including conceptual counterparts), consult Self-Assessment Test 8.2. This test is described at the end of Section 8.6.

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 7 | | | A Discus Thrower

Discus throwers often warm up by standing with both feet flat on the ground and throwing the discus with a twisting motion of their bodies. Figure 8-14a illustrates a top view of such a warm-up throw. Starting from rest, the thrower accelerates the discus to a final angular velocity of +15.0 rad/s in a time of 0.270 s before releasing it. During the acceleration, the discus moves on a circular arc of radius 0.810 m. Find the magnitude a of the total acceleration of the discus just before it is released.

Figure 8-14 (a) A discus thrower and the centripetal acceleration and tangential acceleration of the discus. (b) The total acceleration of the discus just before the discus is released is the vector sum of and .

Reasoning Since the tangential speed of the discus increases as the thrower turns, the discus simultaneously experiences a tangential acceleration and a centripetal acceleration that are oriented at right angles to each other (see the drawing). The magnitude a of the total acceleration is , where ac and aT are the magnitudes of the centripetal and tangential accelerations. The magnitude of the centripetal acceleration can be evaluated from Equation 8.11 (), and the magnitude of the tangential acceleration follows from Equation 8.10 (). The angular acceleration can be found from its definition in Equation 8.4.
Knowns and Unknowns The data for this problem are:

Description | Symbol | Value | Comment
Explicit Data
Final angular velocity |  | +15.0 rad/s | Positive, because discus moves counterclockwise (see drawing).
Time | t | 0.270 s |
Radius of circular arc | r | 0.810 m |
Implicit Data
Initial angular velocity |  | 0 rad/s | Discus starts from rest.
Unknown Variables
Magnitude of total acceleration | a | ? |

Modeling the Problem

| Step 1 Total Acceleration
| Figure 8-14b shows the two perpendicular components of the acceleration of the discus at the moment of its release. The centripetal acceleration arises because the discus is traveling on a circular path; this acceleration always points toward the center of the path (see 5.2). The tangential acceleration arises because the tangential velocity of the discus is increasing; this acceleration is tangent to the path (see Section 8.4). Since and are perpendicular, we can use the Pythagorean theorem to find the magnitude a of the total acceleration, as given by Equation 1 at the right. Values for ac and aT will be obtained in the next two steps.

| (1)

| Step 2 Centripetal Acceleration
| The magnitude ac of the centripetal acceleration can be evaluated from Equation 8.11 as

where r is the radius of the path and is the angular speed of the discus at the moment of its release. Both r and are known, so we can substitute this expression for ac into Equation 1, as indicated at the right. In Step 3, we will evaluate the magnitude aT of the tangential acceleration.

| Step 3 Tangential Acceleration
| According to Equation 8.10, the magnitude aT of the tangential acceleration is , where r is the radius of the path and is the magnitude of the angular acceleration. The angular acceleration is defined (see Equation 8.4) as the change in the angular velocity divided by the time t, or .
Thus, the tangential acceleration can be written as

All the variables on the right side of this equation are known, so we substitute this expression for aT into Equation 1 (see the right column).

Solution Algebraically combining the results of the three steps, we have:
The magnitude of the total acceleration of the discus is

Note that we can also determine the angle between the total acceleration of the discus and its centripetal acceleration (see Figure 8-14b). From trigonometry, we have that tan , so

Related Homework: Problems 41, 60, 62

Check Your Understanding 4

A rotating object starts from rest and has a constant angular acceleration. Three seconds later the centripetal acceleration of a point on the object has a magnitude of 2.0 m/s2. What is the magnitude of the centripetal acceleration of this point six seconds after the motion begins? (The answer is given at the end of the book.)
Background: The concept of centripetal acceleration is the jumping-off point for this question. Furthermore, since the angular acceleration is constant, the equations of kinematics for rotational motion apply (see Table 8-1).
For similar questions (including conceptual counterparts), consult Self-Assessment Test 8.2. This test is described at the end of Section 8.6.

Need more practice?

Interactive LearningWare8.2
A thin rigid rod is rotating with a constant angular acceleration about an axis that passes perpendicularly through one of its ends. At one instant, the total acceleration vector (centripetal plus tangential) at the other end of the rod makes a 60.0° angle with respect to the rod and has a magnitude of 15.0 m/s2. The rod has an angular speed of 2.00 rad/s at this instant. What is the rod’s length?
Related Homework: Problems 60, 62

Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.