5.3

Centripetal Force

CONCEPTS AT A GLANCE Newton’s second law indicates that whenever an object accelerates, there must be a net force to create the acceleration. Thus, in uniform circular motion there must be a net force to produce the centripetal acceleration. As the Conceptsat-a-Glance chart in Figure 5-6 indicates, the second law gives this net force as the product of the object’s mass m and its acceleration . This chart is an expanded version of the chart shown previously in Figure 4.9. The net force causing the centripetal acceleration is called the centripetal force and points in the same direction as the acceleration—that is, toward the center of the circle.

Figure 5-6 CONCEPTS AT A GLANCE Uniform circular motion entails a centripetal acceleration . The net force required to produce this acceleration is called the centripetal force and is given by Newton’s second law as . The photograph shows a fairground ride in which a rider spins around on a vertical circle. He experiences a centripetal force that comes in part from gravity and in part from the push of his seat against his back. (Bernd Mellmann/Alamy Images)

| Centripetal Force
| Magnitude: The centripetal force is the name given to the net force required to keep an object of mass m, moving at a speed v, on a circular path of radius r, and it has a magnitude of

| (5.3)

Direction: The centripetal force always points toward the center of the circle and continually changes direction as the object moves.

The phrase “centripetal force” does not denote a new and separate force created by nature. The phrase merely labels the net force pointing toward the center of the circular path, and this net force is the vector sum of all the force components that point along the radial direction. Sometimes the centripetal force consists of a single force such as tension (see Example 5), friction (see Example 7), the normal force or a component thereof (see Examples 8 and 13 and 8 and 13), or the gravitational force (see Examples 9–11 and 9–11). However, there are circumstances when a number of different forces contribute simultaneously to the centripetal force (see Section 5.7).
In some cases, it is easy to identify the source of the centripetal force, as when a model airplane on a guideline flies in a horizontal circle. The only force pulling the plane inward is the tension in the line, so this force alone (or a component of it) is the centripetal force. Example 5 illustrates the fact that higher speeds require greater tensions.

| Example 5 | | | The Effect of Speed on Centripetal Force

The model airplane in Figure 5-7 has a mass of 0.90 kg and moves at a constant speed on a circle that is parallel to the ground. The path of the airplane and its guideline lie in the same horizontal plane, because the weight of the plane is balanced by the lift generated by its wings. Find the tension T in the guideline () for speeds of 19 and 38 m/s.

Figure 5-7 The scale records the tension in the guideline. See Example 5.

Reasoning Since the plane flies on a circular path, it experiences a centripetal acceleration that is directed toward the center of the circle. According to Newton’s second law of motion, this acceleration is produced by a net force that acts on the plane, and this net force is called the centripetal force. The centripetal force is also directed toward the center of the circle. Since the tension T in the guideline is the only force pulling the plane inward, it must be the centripetal force.
Solution Equation 5.3 gives the tension directly: .

Conceptual Example 6 deals with another case where it is easy to identify the source of the centripetal force.

| Conceptual Example 6 | | | A Trapeze Act

In a circus, a man hangs upside down from a trapeze, legs bent over the bar and arms downward, holding his partner (see Figure 5-8). Is it harder for the man to hold his partner when the partner hangs straight down and is stationary or when the partner is swinging through the straight-down position?

Figure 5-8 Does the man hanging upside down from the trapeze have a harder job holding his partner when the team is swinging through the straightdown position or when they are hanging straight down and stationary? (Ludwig’s Flying Trapeze)

Reasoning and Solution When the man and his partner are stationary, the man’s arms must support his partner’s weight. When the two are swinging, however, the man’s arms must do an additional job. Then, the partner is moving on a circular arc and has a centripetal acceleration. The man’s arms must exert an additional pull so that there will be sufficient centripetal force to produce this acceleration. Because of the additional pull, it is harder for the man to hold his partner while swinging than while stationary.
Related Homework: Problem 46

When a car moves at a steady speed around an unbanked curve, the centripetal force keeping the car on the curve comes from the static friction between the road and the tires, as Figure 5-9 indicates. It is static, rather than kinetic friction, because the tires are not slipping with respect to the radial direction. If the static frictional force is insufficient, given the speed and the radius of the turn, the car will skid off the road. Example 7 shows how an icy road can limit safe driving.

Figure 5-9 When the car moves without skidding around a curve, static friction between the road and the tires provides the centripetal force to keep the car on the road.

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 | | | Centripetal Force and Safe Driving

At what maximum speed can a car safely negotiate a horizontal unbanked turn (radius = 51 m) in dry weather (coefficient of static friction = 0.95) and icy weather (coefficient of static friction = 0.10)? Reasoning The speed v at which the car of mass m can negotiate a turn of radius r is related to the centripetal force Fc that is available, according to (Equation 5.3). Static friction provides the centripetal force and can provide a maximum force given by (Equation 4.7), in which is the coefficient of static friction and FN is the normal force. Thus, we will use the fact that to obtain the maximum speed from Equation 5.3. It will also be necessary to evaluate the normal force, which we will do by considering that is must balance the weight of the car. Experience indicates that the maximum speed should be greater for the dry road than for the icy road.
Knowns and Unknowns The data for this problem are as follows:

Description | Symbol | Value | Comment
Radius of turn | r | 51 m |
Coefficient of static friction |  | 0.95 | Dry weather conditions
Coefficient of static friction |  | 0.10 | Icy weather conditions
Unknown Variable |  |  |
Speed of car | v | ?

Modeling the Problem

| Step 1 Speed and Centripetal Force
| According to Equation 5.3, we have

Solving for the speed v gives Equation 1 at the right. To use this result, it is necessary to consider the centripetal force Fc, which we do in Step 2.

| (1)

| Step 2 Static Friction
| The force of static friction is the centripetal force, and the greater its value, the greater is the speed at which the car can negotiate the turn. The maximum force of static friction is given by (Equation 4.7). Thus, the maximum available centripetal force is

which we can substitute into Equation 1, as shown at the right. The next step considers the normal force FN.

| Step 3 The Normal Force
| The fact that the car does not accelerate in the vertical direction tells us that the normal force balances the car’s weight mg, or

This result for the normal force can now be substituted into Equation 2, as indicated at the right.

Solution Algebraically combining the results of each step, we find that
The mass m of the car is eliminated algebraically from this result, and we find that the maximum speeds are

As expected, the dry road allows a greater maximum speed.

Problem solving insight When using an equation to obtain a numerical answer, algebraically solve for the unknown variable in terms of the known variables. Then substitute in the numbers for the known variables, as this example shows.

Related Homework: Problems 14, 16, 23, 59

A passenger in Figure 5-9 must also experience a centripetal force to remain on the circular path. However, if the upholstery is very slippery, there may not be enough static friction to keep him in place as the car makes a tight turn at high speed. Then, when viewed from inside the car, he appears to be thrown toward the outside of the curve. What really happens is that the passenger slides off on a tangent to the circle, until he encounters a source of centripetal force to keep him in place while the car turns. This occurs when the passenger bumps into the side of the car, which pushes on him with the necessary force.
Sometimes the source of the centripetal force is not obvious. A pilot making a turn, for instance, banks or tilts the plane at an angle to create the centripetal force. As a plane flies, the air pushes upward on the wing surfaces with a net lifting force that is perpendicular to the wing surfaces, as Figure 5-10a shows. Part b of the drawing illustrates that when the plane is banked at an angle , a component of the lifting force is directed toward the center of the turn. It is this component that provides the centripetal force. Greater speeds and/or tighter turns require greater centripetal forces. In such situations, the pilot must bank the plane at a larger angle, so that a larger component of the lift points toward the center of the turn. The technique of banking into a turn also has an application in the construction of high-speed roadways, where the road itself is banked to achieve a similar effect, as the next section discusses.

Figure 5-10 (a) The air exerts an upward lifting force on each wing. (b) When a plane executes a circular turn, the plane banks at an angle . The lift component is directed toward the center of the circle and provides the centripetal force.

The physics of flying an airplane in a banked turn.

Need more practice?

Interactive LearningWare5.2
A 0.60-kg and a 1.2-kg model airplane fly in horizontal circles on guidelines that are parallel to the ground. The speeds of the planes are the same, and the same type of cord is used for each guideline. The smallest circle on which the 0.60-kg plane can fly without breaking its guideline has a radius of 3.5 m. What is the radius of the smallest circle on which the 1.2-kg plane can fly?
Related Homework: Problem 58

Check Your Understanding 2

A car is traveling in uniform circular motion on a section of road whose radius is r (see the drawing). The road is slippery, and the car is just on the verge of sliding. (a) If the car’s speed were doubled, what would have to be the smallest radius in order that the car does not slide? Express your answer in terms of r. (b) What would be your answer to part (a) if the car were replaced by one that weighed twice as much, the car’s speed still being doubled? (The answers are given at the end of the book.)
Background: The concepts of centripetal force, the maximum force of static friction, and the normal force are featured in this problem. The relationship between weight and mass also plays a role.
For similar questions (including calculational counterparts), consult Self-Assessment Test 5.1. The test is described next.

Several important ideas were presented in Sections 5.1–5.3:

· | Uniform Circular Motion
· | Centripetal Acceleration
· | Centripetal Force

Use Self-Assessement Test 5.1 to check your understanding of these concepts.

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