Calculate coeffective of friction on a car tire.Calculate appropriate rate and angle of a auto on a rotate.

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Any force or combicountry of pressures can reason a centripetal or radial acceleration. Just a couple of examples are the anxiety in the rope on a tether round, the force of Earth’s gravity on the Moon, friction in between roller skates and also a rink floor, a banked roadway’s pressure on a vehicle, and forces on the tube of a spinning centrifuge.

Any net force causing unidevelop circular movement is referred to as a centripetal force. The direction of a centripetal force is toward the facility of curvature, the same as the direction of centripetal acceleration. According to Newton’s second legislation of motion, net force is mass times acceleration: net F = ma. For unidevelop circular movement, the acceleration is the centripetal acceleration—ac. Thus, the magnitude of centripetal force Fc is Fc = mac.

By utilizing the expressions for centripetal acceleration ac from a_c=fracv^2r;a_c=romega^2\, we get 2 expressions for the centripetal pressure Fc in terms of mass, velocity, angular velocity, and radius of curvature: extF_c=mfracv^2r; extF_c=mromega^2\.

You might usage whichever before expression for centripetal pressure is more convenient. Centripetal force Fc is always perpendicular to the path and pointing to the facility of curvature, bereason ac is perpendicular to the velocity and pointing to the center of curvature.

Keep in mind that if you deal with the first expression for r, you get displaystyler=fracmv^2 extF_c\.

This suggests that for a provided mass and also velocity, a big centripetal pressure causes a little radius of curvature—that is, a tight curve.

Figure 1. The frictional force supplies the centripetal pressure and also is numerically equal to it. Centripetal force is perpendicular to velocity and also causes uniform circular movement. The larger the Fc, the smaller the radius of curvature r and also the sharper the curve. The second curve has actually the exact same v, however a larger Fc produces a smaller sized r′.

Example 1. What Coreliable of Friction Do Car Tires Need on a Flat Curve?

Calculate the centripetal pressure exerted on a 900 kg car that negotiates a 500 m radius curve at 25.0 m/s.Assuming an unbanked curve, uncover the minimum static coefficient of friction, in between the tires and the road, static friction being the factor that keeps the vehicle from slipping (view Figure 2).Strategy and Solution for Part 1

We know that extF_c=fracmv^2r\. Thus,

displaystyle extF_c=fracmv^2r=fracleft(900 ext kg ight)left(25.0 ext m/s ight)^2left(500 ext m ight)=1125 ext N\.

Strategy for Part 2

Figure 2 mirrors the pressures acting on the vehicle on an unbanked (level ground) curve. Friction is to the left, maintaining the vehicle from slipping, and bereason it is the just horizontal force acting on the auto, the friction is the centripetal pressure in this situation. We recognize that the maximum static friction (at which the tires roll however do not slip) is μsN, where μs is the static coeffective of friction and also N is the normal pressure. The normal force equates to the car’s weight on level ground, so that N=mg. Hence the centripetal force in this instance is

Fc = fμsN = μsmg.

Now we have a partnership in between centripetal pressure and the coeffective of friction. Using the initially expression for Fc from the equation

egincases extF_c=mfracv^2r\ extF_c=mromega^2endcases, ext mfracv^2r=mu_smg\

We fix this for μs, noting that mass cancels, and obtain


Systems for Part 2

Substituting the knowns,

displaystylemu_s=fracleft(25.0 ext m/s ight)^2left(500 ext m ight)left(9.80 ext m/s^2 ight)=0.13\.

(Due to the fact that coefficients of friction are approximate, the answer is given to only 2 digits.)


We can also deal with Part 1 utilizing the initially expression inegincases extF_c=mfracv^2r\ extF_c=mromega^2endcases\ bereason m,v, and also r are offered. The coeffective of friction discovered in Part 2 is much smaller than is typically discovered between tires and roadways. The car will certainly still negotiate the curve if the coefficient is better than 0.13, bereason static friction is a responsive pressure, being able to assume a value less than yet no more than μsN. A greater coreliable would certainly additionally permit the automobile to negotiate the curve at a greater rate, however if the coreliable of friction is much less, the safe speed would certainly be much less than 25 m/s. Keep in mind that mass cancels, implying that in this example, it does not matter how greatly loaded the auto is to negotiate the revolve. Mass cancels because friction is assumed proportional to the normal force, which consequently is proportional to mass. If the surface of the road were banked, the normal pressure would be much less as will certainly be debated below.

Figure 2. This auto on level ground is relocating away and also turning to the left. The centripetal pressure causing the automobile to rotate in a circular path is as a result of friction between the tires and the road. A minimum coreliable of friction is essential, or the automobile will relocate in a larger-radius curve and also leave the roadmeans.

Let us currently take into consideration banked curves, wright here the slope of the road helps you negotiate the curve. See Figure 3. The higher the angle θ, the quicker you can take the curve. Race tracks for bikes and also cars, for example, frequently have steeply banked curves. In an “ideally banked curve,” the angle θ is such that you have the right to negotiate the curve at a specific speed without the help of friction between the tires and also the road. We will certainly derive an expression for θ for an ideally banked curve and also take into consideration an instance regarded it.

For best banking, the net exterior 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 pressure and the weight of the car, respectively. In instances in which forces are not parallel, it is the majority of convenient to take into consideration components alengthy perpendicular axes—in this situation, the vertical and horizontal directions.

Figure 3 shows a complimentary body diagram for a automobile on a frictionmuch less banked curve. If the angle θ is ideal for the rate and radius, then the net external pressure will equal the necessary centripetal force. The only 2 exterior forces acting on the vehicle are its weight w and also the normal pressure of the road N. (A frictionless surface have the right to only exert a pressure perpendicular to the surface—that is, a normal pressure.) These two pressures should include to give a net outside force that is horizontal towards the center of curvature and also has magnitude mv2/r. Since this is the important pressure and also it is horizontal, we usage a coordinate system through vertical and also horizontal axes. Only the normal pressure has actually a horizontal component, and also so this should equal the centripetal force—that is,

Nsin heta=fracmv^2r\.

Since the vehicle does not leave the surconfront of the road, the net vertical pressure have to be zero, interpretation that the vertical components of the 2 exterior forces should be equal in magnitude and oppowebsite in direction. From the number, we see that the vertical component of the normal pressure is N cos θ, and the just various other vertical pressure is the car’s weight. These have to be equal in magnitude; for this reason, N cos θ = mg.

Now we can integrate the last two equations to get rid of N and also get an expression for θ, as wanted. Solving the second equation for N=fracmgcos heta\ , and substituting this right into the initially yields

displaystyleeginarray\mgfracsin hetacos heta=fracmv^2r\mg anleft( heta ight)=fracmv^2r\ an heta=fracv^2rgendarray\

Taking the inverse tangent gives

heta= an^-1left(fracv^2rg ight)\ (ideally banked curve, no friction).

This expression can be understood by considering just how θ relies on v and r. A big θ will be obtained for a huge v and a small r. That is, roads must be steeply banked for high speeds and sharp curves. Friction helps, because it permits you to take the curve at better or lower rate than if the curve is frictionmuch less. Keep in mind that θ does not depend on the mass of the vehicle.

Figure 3. The auto on this banked curve is moving amethod and also turning to the left.

Example 2. What Is the Ideal Speed to Take a Steeply Banked Tight Curve?

Curves on some test tracks and race, such as the Daytona Internationwide Speedmeans in Florida, are extremely steeply banked. This banking, through the assist of tire friction and exceptionally secure car configurations, permits the curves to be taken at very high rate. To show, calculate the rate at which a 100 m radius curve banked at 65.0° have to be pushed if the road is frictionmuch less.


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


Starting with

an heta=fracv^2rg\, we get v = (rg tan θ)1/2.

Noting that tan 65.0º = 2.14, we obtain

eginarray\v=left^1/2\ ext =45.8endarray\


This is simply around 165 km/h, continual with a really steeply banked and also fairly sharp curve. Tire friction allows a auto to take the curve at significantly higher speeds.

Calculations equivalent to those in the preceding examples deserve to be perdeveloped for a hold of amazing instances in which centripetal pressure is involved—a number of these are presented in this chapter’s Problems and also Exercises.

Take-Home Experiment

Ask a frifinish or family member to swing a golf club or a tennis racquet. Take proper measurements to estimate the centripetal acceleration of the end of the club or racquet. You may choose to perform this in sluggish activity.

PhET Explorations: Gravity and also Orbits

Move the sun, earth, moon and area station to view exactly how it affects their gravitational pressures and also orbital routes. Visualize the sizes and also distances between various heavenly bodies, and revolve off gravity to check out what would certainly occur without it!


Click the picture to downpack the simulation. Run using Java.

Section Summary

Centripetal force Fc is any kind of force leading to unidevelop circular motion. It is a “center-seeking” force that always points toward the facility of rotation. It is perpendicular to direct velocity v and has actually magnitude Fc = mac, which deserve to additionally be expressed as

egincases extF_c=mfracv^2r\ extor\ extF_c=mromega^2endcases\

Conceptual Questions

If you wish to alleviate the stress and anxiety (which is regarded centripetal force) on high-speed tires, would you use large- or small-diameter tires? Exordinary.Define centripetal pressure. Can any kind of kind of pressure (for example, anxiety, gravitational force, friction, and so on) be a centripetal force? Can any type of combicountry of pressures be a centripetal force?If centripetal force is directed toward the center, why execute you feel that you are ‘thrown’ away from the facility as a car goes approximately a curve? Exsimple.Race car vehicle drivers on a regular basis cut corners as presented in Figure 7. Exordinary exactly how this enables the curve to be taken at the greatest rate.

Figure 7. Two routes roughly a race track curve are presented. Race automobile motorists will certainly take the inside path (dubbed cutting the corner) whenever before possible bereason it permits them to take the curve at the greatest speed.

A number of amusement parks have actually rides that make vertical loops favor the one displayed in Figure 8. For safety and security, the cars are attached to the rails in such a means that they cannot autumn off. If the auto goes over the optimal at just the ideal rate, gravity alone will supply the centripetal force. What various other pressure acts and what is its direction if: (a) The automobile goes over the optimal at quicker than this speed? (b) The vehicle goes over the height at sreduced than this speed?

Figure 8. Amusement rides via a vertical loop are an example of a form of curved motion.

What is the direction of the pressure exerted by the car on the passenger as the auto goes over the top of the amusement ride pictured in Figure 8 under the adhering to circumstances: (a) The automobile goes over the height at such a speed that the gravitational force is the only pressure acting? (b) The auto goes over the height quicker than this speed? (c) The car goes over the optimal sreduced than this speed?As a skater forms a circle, what pressure is responsible for making her turn? Use a complimentary body diagram in your answer.Suppose a boy is riding on a merry-go-round at a distance around halfway between its center and edge. She has actually a lunch box resting on wax paper, so that tbelow is extremely little bit friction between it and the merry-go-round. Which path presented in Figure 9 will the lunch box take as soon as she lets go? The lunch box leaves a trail in the dust on the merry-go-round. Is that trail straight, curved to the left, or curved to the right? Explain your answer.

Figure 9. A kid riding on a merry-go-round releases her lunch box at point P. This is a view from over the clockwise rotation. Assuming it slides via negligible friction, will certainly it follow path A, B, or C, as regarded from Earth’s frame of reference? What will certainly be the shape of the path it leaves in the dust on the merry-go-round?

Do you feel yourself thrvery own to either side as soon as 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?Suppose a mass is relocating in a circular route on a frictionless table as displayed in figure. In the Earth’s framework of reference, tright here is no centrifugal force pulling the mass amethod from the centre of rotation, yet there is a very real force stretching the string attaching the mass to the nail. Using ideas regarded centripetal pressure and also Newton’s third legislation, describe what force stretches the string, identifying its physical origin.

Figure 10. A mass attached to a nail on a frictionmuch less table moves in a circular route. The pressure extending the string is actual and not fictional. What is the physical origin of the pressure on the string?

Problems & Exercises

(a) A 22.0 kg son is riding a playground merry-go-round that is rotating at 40.0 rev/min. What centripetal force need to she exert to stay on if she is 1.25 m from its center? (b) What centripetal pressure does she must remain on an amusement park merry-go-round that rotates at 3.00 rev/min if she is 8.00 m from its center? (c) Compare each pressure via her weight.Calculate the centripetal pressure on the end of a 100 m (radius) wind wind turbine blade that is rotating at 0.5 rev/s. Assume the mass is 4 kg.What is the right banking angle for a gentle rotate of 1.20 km radius on a highway via a 105 km/h speed limit (around 65 mi/h), assuming everyone travels at the limit?What is the right speed to take a 100 m radius curve banked at a 20.0° angle?(a) What is the radius of a bobsled rotate banked at 75.0° and also taken at 30.0 m/s, assuming it is ideally banked? (b) Calculate the centripetal acceleration. (c) Does this acceleration seem big to you?Part of riding a bicycle entails leaning at the correct angle when making a revolve, as checked out in Figure 4. To be stable, the pressure exerted by the ground have to be on a line going with the facility of gravity. The pressure on the bicycle wheel have the right to be resolved into two perpendicular components—friction parallel to the road (this must supply the centripetal force), and the vertical normal force (which have to equal the system’s weight). (a) Sjust how that θ (as characterized in the figure) is related to the speed v and radius of curvature r of the revolve in the very same means as for an ideally banked roadway—that is,  heta= an^-1fracv^2rg\; (b) Calculate θ for a 12.0 m/s turn of radius 30.0 m (as in a race).

Figure 6. 4. A bicyclist negotiating a rotate on level ground must lean at the correct angle—the ability to do this becomes instinctive. The pressure of the ground on the wheel needs to be on a line via the facility of gravity. The net external pressure on the mechanism is the centripetal pressure. The vertical component of the force on the wheel cancels the weight of the mechanism while its horizontal component should supply the centripetal pressure. This process produces a connection among the angle θ, the rate v, and the radius of curvature r of the rotate comparable to that for the right banking of roadmethods.

A large centrifuge, like the one shown in Figure 5a, is offered to reveal aspiring astronauts to accelerations comparable to those skilled in rocket launches and atmospheric reentries. (a) At what angular velocity is the centripetal acceleration 10 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, permitting it to swing outward throughout rotation as displayed in Figure 5b. At what angle θ listed below the horizontal will the cage hang as soon as the centripetal acceleration is 10 g? (Hint: The arm provides centripetal force and also supports the weight of the cage. Draw a cost-free body diagram of the forces to view what the angle θ must be.)

Figure 5. (a) NASA centrifuge offered to subject trainees to accelerations equivalent to those knowledgeable in rocket launches and also reentries. (credit: NASA) (b) Rider in cage mirroring exactly how the cage pivots external during rotation. This enables the total pressure exerted on the rider by the cage to be along its axis at all times.

Integrated Concepts. If a automobile takes a banked curve at much less than the right rate, friction is needed to store it from sliding toward the inside of the curve (a genuine trouble on icy hill roads). (a) Calculate the best rate to take a 100 m radius curve banked at 15.0º. (b) What is the minimum coreliable of friction required for a frightened driver to take the same curve at 20.0 km/h?Modern roller coasters have actually vertical loops favor the one presented in Figure 6. The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the peak will be higher than the acceleration as a result of gravity, keeping the passengers pressed firmly right into their seats. What is the speed of the roller coaster at the optimal of the loop if the radius of curvature there is 15.0 m and also the downward acceleration of the vehicle is 1.50 g?

Figure 6. Teardrop-shaped loops are provided in the latest roller coasters so that the radius of curvature slowly decreases to a minimum at the optimal. This implies that the centripetal acceleration builds from zero to a maximum at the optimal and gradually decreases aobtain. A circular loop would cause a jolting change in acceleration at entry, a disadvantage discovered lengthy earlier in railroad curve style. With a tiny radius of curvature at the optimal, the centripetal acceleration can more conveniently be kept better than g so that the passengers perform not shed call through their seats nor do they need seat belts to save them in area.

Unreasonable Results. (a) Calculate the minimum coeffective of friction needed for a automobile 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?


centripetal force: any type of net pressure bring about uniform circular motion

right banking: the sloping of a curve in a road, where the angle of the slope allows the vehicle to negotiate the curve at a specific rate without the aid of friction between the tires and the road; the net outside force on the vehicle equates to the horizontal centripetal force in the absence of friction

best speed: the maximum safe rate at which a car can revolve on a curve without the aid of friction between the tire and the road

ideal angle: the angle at which a vehicle have the right to turn safely on a steep curve, which is in propercentage to the best speed

banked curve: the curve in a road that is sloping in a manner that helps a automobile negotiate the curve

Selected Solutions to Problems & Exercises

1. (a) 483 N; (b) 17.4 N; (c) 2.24 times her weight, 0.0807 times her weight

3. 4.14º

5. (a) 24.6 m; (b) 36.6 m/s2; (c) ac = 3.73 g.This does not seem also big, yet it is clear that bobsledders feel most force on them going via sharply banked turns.

See more: Is The Process Of Receiving , Attending To, And Assigning Meaning To Verbal And Visual Stimuli.

7. (a) 2.56 rad/s; (b) 5.71º

8. (a) 16.2 m/s; (b) 0.234

10. (a) 1.84; (b) A coreliable of friction this a lot higher than 1 is unreasonable; (c) The assumed rate is too great for the tight curve.