64. Centripetal ForceLearning Objectives
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 By using the expressions for centripetal acceleration ${a}_{c}$ from ${a}_{c}=\frac{{v}^{2}}{r};\phantom{\rule{0.25em}{0ex}}{a}_{c}={\mathrm{r\omega}}^{2}$, we get two expressions for the centripetal force ${\text{F}}_{\text{c}}$ in terms of mass, velocity, angular velocity, and radius of curvature: $${F}_{c}=m\frac{{v}^{2}}{r};\phantom{\rule{0.25em}{0ex}}{F}_{c}=\text{mr}{\omega}^{2}.$$You may use whichever expression for centripetal force is more convenient. Centripetal force ${F}_{\text{c}}$ is always perpendicular to the path and pointing to the center of curvature, because ${\mathbf{a}}_{c}$ is perpendicular to the velocity and pointing to the center of curvature. Note that if you solve the first expression for $r$, you get $$r=\frac{{\mathrm{mv}}^{2}}{{F}_{c}}\text{.}$$This implies that for a given mass and velocity, a large centripetal force causes a small radius of curvature—that is, a tight curve. Example 1: 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 2). Strategy and Solution for (a) We know that ${F}_{\text{c}}=\frac{{\mathrm{mv}}^{\text{2}}}{r}$. Thus, $${F}_{\text{c}}=\frac{{\mathrm{mv}}^{\text{2}}}{r}=\frac{(\text{900 kg})(\text{25.0 m/s}{)}^{\text{2}}}{(\text{500 m})}=\text{1125 N.}$$Strategy for (b) Figure 2 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 ${\mu}_{\text{s}}N$, where ${\mu}_{\text{s}}$ 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 $N=\mathit{mg}$. Thus the centripetal force in this situation is $${F}_{\text{c}}=f={\mu}_{\text{s}}N={\mu}_{\text{s}}\text{mg}\text{.}$$Now we have a relationship between centripetal force and the coefficient of friction. Using the first expression for ${F}_{\text{c}}$ from the equation $$\begin{array}{c}{F}_{\text{c}}=m\frac{{v}^{2}}{r}\\ {F}_{\text{c}}=\text{mr}{\omega}^{2}\end{array}\},$$ $$m\frac{{v}^{2}}{r}={\mu}_{\text{s}}\text{mg}\text{.}$$We solve this for ${\mu}_{\text{s}}$, noting that mass cancels, and obtain $${\mu}_{\text{s}}=\frac{{v}^{2}}{\text{rg}}\text{.}$$Solution for (b) Substituting the knowns, $${\mu}_{\text{s}}=\frac{(\text{25.0 m/s}{)}^{2}}{(\text{500 m})(9\text{.}{\text{80 m/s}}^{2})}=0\text{.}\text{13}\text{.}$$(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 $\begin{array}{c}{F}_{\text{c}}=m\frac{{v}^{2}}{r}\\ {F}_{\text{c}}=\text{mr}{\omega}^{2}\end{array}\},$ because $m,$$v,$ and $r$ are given. The coefficient of friction found in part (b) is much smaller than is typically found between tires and roads. The car will still negotiate the curve if the coefficient is greater than 0.13, because static friction is a responsive force, being able to assume a value less than but no more than ${\mu}_{\text{s}}N$. A higher coefficient would also allow the car to negotiate the curve at a higher speed, but if the coefficient of friction is less, the safe speed would be less than 25 m/s. Note that mass cancels, implying that in this example, it does not matter how heavily loaded the car is to negotiate the turn. Mass cancels because friction is assumed proportional to the normal force, which in turn is proportional to mass. If the surface of the road were banked, the normal force would be less as will be discussed below. Let us now consider For Figure 3 shows a free body diagram for a car on a frictionless banked curve. If the angle $\theta $ is ideal for the speed and radius, then the net external force will equal the necessary centripetal force. The only two external forces acting on the car are its weight $\mathbf{w}$ and the normal force of the road $\mathbf{N}$. (A frictionless surface can only exert a force perpendicular to the surface—that is, a normal force.) These two forces must add to give a net external force that is horizontal toward the center of curvature and has magnitude ${\text{mv}}^{2}\text{/r}$. Because this is the crucial force and it is horizontal, we use a coordinate system with vertical and horizontal axes. Only the normal force has a horizontal component, and so this must equal the centripetal force—that is, $$N\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\frac{{\mathrm{mv}}^{2}}{r}\text{.}$$Because the car does not leave the surface of the road, the net vertical force must be zero, meaning that the vertical components of the two external forces must be equal in magnitude and opposite in direction. From the figure, we see that the vertical component of the normal force is $N\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta $, and the only other vertical force is the car’s weight. These must be equal in magnitude; thus, $$N\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta =\text{mg}\text{.}$$Now we can combine the last two equations to eliminate $N$ and get an expression for $\theta $, as desired. Solving the second equation for $N=\text{mg}/(\text{cos}\phantom{\rule{0.25em}{0ex}}\theta )$, and substituting this into the first yields $$\text{mg}\frac{\text{sin}\phantom{\rule{0.25em}{0ex}}\theta}{\text{cos}\phantom{\rule{0.25em}{0ex}}\theta}=\frac{{\text{mv}}^{2}}{r}$$ $$\begin{array}{rrr}\text{mg}\phantom{\rule{0.25em}{0ex}}\text{tan}(\theta )& =& \frac{{\mathrm{mv}}^{2}}{r}\\ \text{tan}\phantom{\rule{0.25em}{0ex}}\theta & =& \frac{{v}^{2}}{\text{rg.}}\end{array}$$Taking the inverse tangent gives $$\theta ={\text{tan}}^{1}\left(\frac{{v}^{2}}{\text{rg}}\right)\phantom{\rule{0.25em}{0ex}}\text{(ideally banked curve, no friction).}$$This expression can be understood by considering how $\theta $ depends on $v$ and $r$. A large $\theta $ will be obtained for a large $v$ and a small $r$. That is, roads must be steeply banked for high speeds and sharp curves. Friction helps, because it allows you to take the curve at greater or lower speed than if the curve is frictionless. Note that $\theta $ does not depend on the mass of the vehicle. Example 2: What Is the Ideal Speed to Take a Steeply Banked Tight Curve?Curves on some test tracks and race courses, such as the Daytona International Speedway in Florida, are very steeply banked. This banking, with the aid of tire friction and very stable car configurations, allows the curves to be taken at very high speed. To illustrate, calculate the speed at which a 100 m radius curve banked at 65.0° should be driven if the road is frictionless. Strategy We first note that all terms in the expression for the ideal angle of a banked curve except for speed are known; thus, we need only rearrange it so that speed appears on the lefthand side and then substitute known quantities. Solution Starting with $$\text{tan}\phantom{\rule{0.25em}{0ex}}\theta =\frac{{v}^{2}}{\text{rg}}$$we get $$v=(\text{rg}\phantom{\rule{0.25em}{0ex}}\text{tan}\phantom{\rule{0.25em}{0ex}}\theta {)}^{1/2}\text{.}$$Noting that tan 65.0º = 2.14, we obtain $$\begin{array}{lll}v& =& {\left[(\text{100 m})(\mathrm{9.80\; m}{\text{/s}}^{2})(2\text{.}\text{14})\right]}^{1/2}\\ & =& \text{45.8 m/s.}\end{array}$$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. TakeHome Experiment:Ask a friend or relative to swing a golf club or a tennis racquet. Take appropriate measurements to estimate the centripetal acceleration of the end of the club or racquet. You may choose to do this in slow motion. PhET Explorations: Gravity and Orbits:Move the sun, earth, moon and space station to see how it affects their gravitational forces and orbital paths. Visualize the sizes and distances between different heavenly bodies, and turn off gravity to see what would happen without it! Section Summary
Conceptual QuestionsExercise 1If you wish to reduce the stress (which is related to centripetal force) on highspeed tires, would you use large or smalldiameter tires? Explain. Exercise 2Define centripetal force. Can any type of force (for example, tension, gravitational force, friction, and so on) be a centripetal force? Can any combination of forces be a centripetal force? Exercise 3If centripetal force is directed toward the center, why do you feel that you are ‘thrown’ away from the center as a car goes around a curve? Explain. Exercise 4Race car drivers routinely cut corners as shown in Figure 5. Explain how this allows the curve to be taken at the greatest speed. Exercise 5A number of amusement parks have rides that make vertical loops like the one shown in Figure 6. For safety, the cars are attached to the rails in such a way that they cannot fall off. If the car goes over the top at just the right speed, gravity alone will supply the centripetal force. What other force acts and what is its direction if: (a) The car goes over the top at faster than this speed? (b)The car goes over the top at slower than this speed? Exercise 6What is the direction of the force exerted by the car on the passenger as the car goes over the top of the amusement ride pictured in Figure 6 under the following circumstances: (a) The car goes over the top at such a speed that the gravitational force is the only force acting? (b) The car goes over the top faster than this speed? (c) The car goes over the top slower than this speed? Exercise 7As a skater forms a circle, what force is responsible for making her turn? Use a free body diagram in your answer. Exercise 8Suppose a child is riding on a merrygoround at a distance about halfway between its center and edge. She has a lunch box resting on wax paper, so that there is very little friction between it and the merrygoround. Which path shown in Figure 7 will the lunch box take when she lets go? The lunch box leaves a trail in the dust on the merrygoround. Is that trail straight, curved to the left, or curved to the right? Explain your answer. Exercise 9Do you feel yourself thrown to either side when you negotiate a curve that is ideally banked for your car’s speed? What is the direction of the force exerted on you by the car seat? Exercise 10Suppose a mass is moving in a circular path on a frictionless table as shown in figure. In the Earth’s frame of reference, there is no centrifugal force pulling the mass away from the centre of rotation, yet there is a very real force stretching the string attaching the mass to the nail. Using concepts related to centripetal force and Newton’s third law, explain what force stretches the string, identifying its physical origin. Problems ExerciseExercise 1(a) A 22.0 kg child is riding a playground merrygoround that is rotating at 40.0 rev/min. What centripetal force must she exert to stay on if she is 1.25 m from its center? (b) What centripetal force does she need to stay on an amusement park merrygoround that rotates at 3.00 rev/min if she is 8.00 m from its center? (c) Compare each force with her weight. Show/Hide Solution Solutiona) 483 N b) 17.4 N c) 2.24 times her weight, 0.0807 times her weight Exercise 2Calculate the centripetal force on the end of a 100 m (radius) wind turbine blade that is rotating at 0.5 rev/s. Assume the mass is 4 kg. Exercise 3What is the ideal banking angle for a gentle turn of 1.20 km radius on a highway with a 105 km/h speed limit (about 65 mi/h), assuming everyone travels at the limit? Show/Hide Solution Solution$4\text{.}\text{14\xba}$ Exercise 4What is the ideal speed to take a 100 m radius curve banked at a 20.0° angle? Exercise 5(a) What is the radius of a bobsled turn banked at 75.0° and taken at 30.0 m/s, assuming it is ideally banked? (b) Calculate the centripetal acceleration. (c) Does this acceleration seem large to you? Show/Hide Solution Solutiona) 24.6 m b) $\mathrm{36.6\; m}/{\text{s}}^{2}$ c) ${a}_{\text{c}}=3.73\phantom{\rule{0.25em}{0ex}}\mathrm{g.}$ This does not seem too large, but it is clear that bobsledders feel a lot of force on them going through sharply banked turns. Exercise 6Part of riding a bicycle involves leaning at the correct angle when making a turn, as seen in Figure 9. To be stable, the force exerted by the ground must be on a line going through the center of gravity. The force on the bicycle wheel can be resolved into two perpendicular components—friction parallel to the road (this must supply the centripetal force), and the vertical normal force (which must equal the system’s weight). (a) Show that $\theta $ (as defined in the figure) is related to the speed $v$ and radius of curvature $r$ of the turn in the same way as for an ideally banked roadway—that is, $\theta ={\text{tan}}^{\text{\u20131}}{v}^{2}/\mathrm{rg}$ (b) Calculate $\theta $ for a 12.0 m/s turn of radius 30.0 m (as in a race). Exercise 7A large centrifuge, like the one shown in Figure 10(a), is used to expose aspiring astronauts to accelerations similar to those experienced in rocket launches and atmospheric reentries. (a) At what angular velocity is the centripetal acceleration $\mathrm{10}\phantom{\rule{0.25em}{0ex}}g$ if the rider is 15.0 m from the center of rotation? (b) The rider’s cage hangs on a pivot at the end of the arm, allowing it to swing outward during rotation as shown in Figure 10(b). At what angle $\theta $ below the horizontal will the cage hang when the centripetal acceleration is $\mathrm{10}\phantom{\rule{0.25em}{0ex}}g$ ? (Hint: The arm supplies centripetal force and supports the weight of the cage. Draw a free body diagram of the forces to see what the angle $\theta $ should be.) Show/Hide Solution Solutiona) 2.56 rad/s b) $\mathrm{5.71\xba}$ Exercise 8Integrated Concepts If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve (a real problem on icy mountain roads). (a) Calculate the ideal speed to take a 100 m radius curve banked at 15.0º. (b) What is the minimum coefficient of friction needed for a frightened driver to take the same curve at 20.0 km/h? Show/Hide Solution Solutiona) 16.2 m/s b) 0.234 Exercise 9Modern roller coasters have vertical loops like the one shown in Figure 11. The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top will be greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats. What is the speed of the roller coaster at the top of the loop if the radius of curvature there is 15.0 m and the downward acceleration of the car is 1.50 g? Exercise 10Unreasonable Results (a) Calculate the minimum coefficient of friction needed for a car to negotiate an unbanked 50.0 m radius curve at 30.0 m/s. (b) What is unreasonable about the result? (c) Which premises are unreasonable or inconsistent? Show/Hide Solution Solutiona) 1.84 b) A coefficient of friction this much greater than 1 is unreasonable . c) The assumed speed is too great for the tight curve.
