Centripetal Force

LEARNING OBJECTIVES

By the end of this section, you will be able to:

• Calculate coefficient of friction on a car tire.
• Calculate ideal speed and angle of a car on a turn.

Any force or combination of forces can cause a centripetal or radial acceleration. Just a few examples are the tension in the rope on a tether ball, the force of Earth’s gravity on the Moon, friction between roller skates and a rink floor, a banked roadway’s force on a car, and forces on the tube of a spinning centrifuge.

Any net force causing uniform circular motion is called a centripetal force. The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration. According to Newton’s second law of motion, net force is mass times acceleration: net F=maF=ma. For uniform circular motion, the acceleration is the centripetal acceleration— a=ac𝑎=𝑎𝑐. Thus, the magnitude of centripetal force FcFc is

Fc=mac.Fc=𝑚ac.

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By using the expressions for centripetal acceleration ac𝑎𝑐 from ac=v2r;ac=rω2𝑎𝑐=𝑣2𝑟;𝑎𝑐=rω2, we get two expressions for the centripetal force FcFc in terms of mass, velocity, angular velocity, and radius of curvature:

Fc=mv2r;Fc=mrω2.𝐹𝑐=𝑚𝑣2𝑟;𝐹𝑐=mr𝜔2.

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You may use whichever expression for centripetal force is more convenient. Centripetal force Fc𝐹c is always perpendicular to the path and pointing to the center of curvature, because ac𝑎𝑐 is perpendicular to the velocity and pointing to the center of curvature.

Note that if you solve the first expression for r𝑟, you get

r=mv2Fc.𝑟=mv2𝐹𝑐.

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This implies that for a given mass and velocity, a large centripetal force causes a small radius of curvature—that is, a tight curve.

Figure 6.9 The frictional force supplies the centripetal force and is numerically equal to it. Centripetal force is perpendicular to velocity and causes uniform circular motion. The larger the FcFc, the smaller the radius of curvature r𝑟 and the sharper the curve. The second curve has the same v𝑣, but a larger FcFc produces a smaller r’𝑟′.

EXAMPLE 6.4

What Coefficient of Friction Do Car Tires Need on a Flat Curve?

(a) Calculate the centripetal force exerted on a 900 kg car that negotiates a 500 m radius curve at 25.0 m/s.

(b) Assuming an unbanked curve, find the minimum static coefficient of friction, between the tires and the road, static friction being the reason that keeps the car from slipping (see Figure 6.10).

Strategy and Solution for (a)

We know that Fc=mv2r𝐹c=mv2𝑟. Thus,

Fc=mv2r=(900 kg)(25.0 m/s)2(500 m)=1125 N.𝐹c=mv2𝑟=(900 kg)(25.0 m/s)2(500 m)=1125 N.

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Strategy for (b)

Figure 6.10 shows the forces acting on the car on an unbanked (level ground) curve. Friction is to the left, keeping the car from slipping, and because it is the only horizontal force acting on the car, the friction is the centripetal force in this case. We know that the maximum static friction (at which the tires roll but do not slip) is μsN𝜇s𝑁, where μs𝜇s is the static coefficient of friction and N is the normal force. The normal force equals the car’s weight on level ground, so that N=mg𝑁=mg. Thus the centripetal force in this situation is

Fc=f=μsN=μsmg.𝐹c=𝑓=𝜇s𝑁=𝜇smg.

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Now we have a relationship between centripetal force and the coefficient of friction. Using the first expression for Fc𝐹c from the equation

Fc=mv2rFc=mrω2},𝐹c=𝑚𝑣2𝑟𝐹c=mr𝜔2},

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mv2r=μsmg.𝑚𝑣2𝑟=𝜇smg.

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We solve this for μs𝜇s, noting that mass cancels, and obtain

μs=v2rg.𝜇s=𝑣2rg.

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Solution for (b)

Substituting the knowns,

μs=(25.0 m/s)2(500 m)(9.80 m/s2)=0.13.𝜇s=(25.0 m/s)2(500 m)(9.80 m/s2)=0.13.

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(Because coefficients of friction are approximate, the answer is given to only two digits.)

Discussion

We could also solve part (a) using the first expression in Fc=mv2rFc=mrω2},𝐹c=𝑚𝑣2𝑟𝐹c=mr𝜔2}, because m,𝑚,v,𝑣, and r𝑟 are given. The coefficient of friction found in part (b) is much smaller than is typically found between tires and roads. The car will still negotiate the curve if the coefficient is greater than 0.13, because static friction is a responsive force, being able to assume a value less than but no more than μsN𝜇s𝑁. A higher coefficient would also allow the car to negotiate the curve at a higher speed, but if the coefficient of friction is less, the safe speed would be less than 25 m/s. Note that mass cancels, implying that in this example, it does not matter how heavily loaded the car is to negotiate the turn. Mass cancels because friction is assumed proportional to the normal force, which in turn is proportional to mass. If the surface of the road were banked, the normal force would be less as will be discussed below.

Figure 6.10 This car on level ground is moving away and turning to the left. The centripetal force causing the car to turn in a circular path is due to friction between the tires and the road. A minimum coefficient of friction is needed, or the car will move in a larger-radius curve and leave the roadway.

Let us now consider banked curves, where the slope of the road helps you negotiate the curve. See Figure 6.11. The greater the angle θ𝜃, the faster you can take the curve. Race tracks for bikes as well as cars, for example, often have steeply banked curves. In an “ideally banked curve,” the angle θ𝜃 is such that you can negotiate the curve at a certain speed without the aid of friction between the tires and the road. We will derive an expression for θ𝜃 for an ideally banked curve and consider an example related to it.

For ideal banking, the net external force equals the horizontal centripetal force in the absence of friction. The components of the normal force N in the horizontal and vertical directions must equal the centripetal force and the weight of the car, respectively. In cases in which forces are not parallel, it is most convenient to consider components along perpendicular axes—in this case, the vertical and horizontal directions.

Figure 6.11 shows a free body diagram for a car on a frictionless banked curve. If the angle θ𝜃 is ideal for the speed and radius, then the net external force will equal the necessary centripetal force. The only two external forces acting on the car are its weight w𝑤 and the normal force of the road N𝑁. (A frictionless surface can only exert a force perpendicular to the surface—that is, a normal force.) These two forces must add to give a net external force that is horizontal toward the center of curvature and has magnitude mv2/rmv2/r. Because this is the crucial force and it is horizontal, we use a coordinate system with vertical and horizontal axes. Only the normal force has a horizontal component, and so this must equal the centripetal force—that is,

Nsinθ=mv2r.𝑁sin𝜃=mv2𝑟.

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Because the car does not leave the surface of the road, the net vertical force must be zero, meaning that the vertical components of the two external forces must be equal in magnitude and opposite in direction. From the figure, we see that the vertical component of the normal force is Ncosθ𝑁cos𝜃, and the only other vertical force is the car’s weight. These must be equal in magnitude; thus,

Ncosθ=mg.𝑁cos𝜃=mg.

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Now we can combine the last two equations to eliminate N𝑁 and get an expression for θ𝜃, as desired. Solving the second equation for N=mg/(cosθ)𝑁=mg/(cos𝜃), and substituting this into the first yields

mgsinθcosθ=mv2rmgsin𝜃cos𝜃=mv2𝑟

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mgtan(θ)tanθ==mv2rv2rg.mgtan(𝜃)=mv2𝑟tan𝜃=𝑣2rg.

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Taking the inverse tangent gives

θ=tan−1(v2rg)(ideally banked curve, no friction).𝜃=tan−1𝑣2rg(ideally banked curve, no friction).

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This expression can be understood by considering how θ𝜃 depends on v𝑣 and r𝑟. A large θ𝜃 will be obtained for a large v𝑣 and a small r𝑟. That is, roads must be steeply banked for high speeds and sharp curves. Friction helps, because it allows you to take the curve at greater or lower speed than if the curve is frictionless. Note that θ𝜃 does not depend on the mass of the vehicle.

Figure 6.11 The car on this banked curve is moving away and turning to the left.

EXAMPLE 6.5

What Is the Ideal Speed to Take a Steeply Banked Tight Curve?

Curves on some test tracks and race courses, such as the Daytona International Speedway in Florida, are very steeply banked. This banking, with the aid of tire friction and very stable car configurations, allows the curves to be taken at very high speed. To illustrate, calculate the speed at which a 100 m radius curve banked at 65.0° should be driven if the road is frictionless.

Strategy

We first note that all terms in the expression for the ideal angle of a banked curve except for speed are known; thus, we need only rearrange it so that speed appears on the left-hand side and then substitute known quantities.

Solution

Starting with

tanθ=v2rgtan𝜃=𝑣2rg

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we get

v=(rgtanθ)1/2.𝑣=(rgtan𝜃)1/2.

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Noting that tan 65.0º = 2.14, we obtain

v==[(100 m)(9.80 m/s2)(2.14)]1/245.8 m/s.𝑣=(100 m)(9.80 m/s2)(2.14)1/2=45.8 m/s.

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Discussion

This is just about 165 km/h, consistent with a very steeply banked and rather sharp curve. Tire friction enables a vehicle to take the curve at significantly higher speeds.

Calculations similar to those in the preceding examples can be performed for a host of interesting situations in which centripetal force is involved—a number of these are presented in this chapter’s Problems and Exercises.