5.8

Concepts & Calculations

In uniform circular motion the concepts of acceleration and force play central roles. Concepts & Calculations Example 15 deals with acceleration, which we first discussed in Chapter 2, particularly in connection with the equations of kinematics. This example emphasizes the difference between the acceleration that is used in the equations of kinematics and the acceleration that arises in uniform circular motion.

| Concepts & Calculations Example 15 | | | Acceleration

At time t = 0 s, automobile A is traveling at a speed of 18 m/s along a straight road and is picking up speed with an acceleration that has a magnitude of 3.5 m/s2 (Figure 5-22a). At time t = 0 s, automobile B is traveling at a speed of 18 m/s in uniform circular motion as it negotiates a turn (Figure 5-22b). It has a centripetal acceleration whose magnitude is also 3.5 m/s2. Determine the speed of each automobile when t = 2.0 s.

Figure 5-22 (a) Automobile A accelerates to the right along a straight road. (b) Automobile B travels at a constant speed along a circular turn. It is also accelerating, but the acceleration is a centripetal acceleration.

Concept Questions and Answers Which automobile has a constant acceleration?

Answer Acceleration is a vector, and for it to be constant, both its magnitude and direction must be constant. Automobile A has a constant acceleration, because its acceleration has a constant magnitude of 3.5 m/s2 and its direction always points forward along the straight road. Automobile B has an acceleration with a constant magnitude of 3.5 m/s2, but the acceleration does not have a constant direction. Automobile B is in uniform circular motion, so it has a centripetal acceleration, which points toward the center of the circle at every instant. Therefore, the direction of the acceleration vector continually changes, and the vector is not constant.

For which automobile do the equations of kinematics apply?

Answer The equations of kinematics apply for automobile A, because it has a constant acceleration, which must be the case when you use these equations. They do not apply to automobile B, because it does not have a constant acceleration.

Solution To determine the speed of automobile A at t = 2.0 s, we use Equation 2.4 from the equations of kinematics:

To determine the speed of automobile B we note that this car is in uniform circular motion and, therefore, has a constant speed as it goes around the turn. At a time of t = 2.0 s its speed is the same as it was at t = 0 s—namely, .

The next example deals with force, and it stresses that the centripetal force is the net force along the radial direction. As we will see, this means that all forces along the radial direction must be taken into account when identifying the centripetal force. In addition, the example is a good review of the tension force, which we first encountered in Chapter 4.

| Concept & Calculations Example 16 | | | Centripetal Force

Ball A is attached to one end of a rigid massless rod, while an identical ball B is attached to the center of the rod, as Figure 5-23a illustrates. Each ball has a mass of m = 0.50 kg, and the length of each half of the rod is L = 0.40 m. This arrangement is held by the empty end and is whirled around in a horizontal circle at a constant rate, so each ball is in uniform circular motion. Ball A travels at a constant speed of vA = 5.0 m/s. Find the tension in each half of the rod.

Figure 5-23 Two identical balls attached to a rigid massless rod are whirled around on horizontal circles, as Concept & Calculations Example 16 explains.

Concept Questions and Answers How many tension forces contribute to the centripetal force that acts on ball A?

Answer As Figure 5-23b illustrates, only a single tension force of magnitude TA acts on ball A. It points to the left in the drawing and is due to the tension in the rod between the two balls. This force alone provides the centripetal force keeping ball A on its circular path of radius 2L.

How many tension forces contribute to the centripetal force that acts on ball B?

Answer As Figure 5-23c shows, two tension forces act on ball B. One has a magnitude TB and points to the left in the drawing, which is the positive direction. It is due to the tension in the left half of the rod. The other has a magnitude TA and points to the right. It is due to the tension in the right half of the rod. The centripetal force acting on ball B points toward the center of the circle and is the vector sum of these two forces, or .

Is the speed of ball B the same as the speed of ball A?

Answer No, it is not. The reason is that ball A travels farther than ball B in the same time. Consider a time of one period, which is the same for either ball, since the arrangement is a rigid unit. It is the time for either ball to travel once around its circular path. In this time ball A travels a distance equal to the circumference of its path, which is . In contrast, the circumference for ball B is only . Therefore, the speed of ball B is one-half the speed of ball A, or vB = 2.5 m/s.

Solution Applying Equation 5.3 to each ball, we have

The tension in the right half of the rod follows directly from the first of these equations:

Adding the equations for ball A and ball B gives the following result for the tension in the left half of the rod:

At the end of the problem set for this chapter, you will find homework problems that contain both conceptual and quantitative parts. These problems are grouped under the heading Concepts & Calculations.

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