Calculate coefficient that friction ~ above a auto tire.Calculate appropriate speed and angle of a vehicle on a turn.

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Any pressure or combination of forces can reason a centripetal or radial acceleration. Simply a few examples space the stress and anxiety in the rope ~ above a tether ball, the force of Earth’s gravity on the Moon, friction in between roller skates and a rink floor, a banked roadway’s pressure on a car, and forces on the tube of a spinning centrifuge.

Any net pressure causing uniform circular activity is dubbed a centripetal force. The direction of a centripetal force is toward the center of curvature, the very same as the direction the centripetal acceleration. Follow to Newton’s second law the motion, net force is mass time acceleration: network F = ma. Because that uniform one motion, the acceleration is the centripetal acceleration—ac. Thus, the magnitude of centripetal pressure Fc is Fc = mac.

By using the expressions for centripetal acceleration ac native a_c=\\fracv^2r;a_c=r\\omega^2\\\\, we gain two expressions because that the centripetal pressure Fc in regards to mass, velocity, angular velocity, and also radius of curvature: \\textF_c=m\\fracv^2r;\\textF_c=mr\\omega^2\\\\.

You might use whichever expression because that centripetal pressure is more convenient. Centripetal pressure Fc is constantly perpendicular to the path and pointing to the center of curvature, since ac is perpendicular come the velocity and pointing come the center of curvature.

Note that if you settle the first expression because that r, you gain \\displaystyler=\\fracmv^2\\textF_c\\\\.

This suggests that for a given mass and also velocity, a huge centripetal force reasons a small radius that curvature—that is, a tight curve.


Figure 1. The frictional pressure supplies the centripetal force and is numerically same to it. Centripetal force is perpendicular come velocity and causes uniform circular motion. The bigger the Fc, the smaller the radius the curvature r and also the trickster the curve. The 2nd curve has the very same v, but a larger Fc to produce a smaller r′.


Example 1. What Coefficient that Friction Do automobile Tires need on a flat Curve?

Calculate the centripetal pressure exerted on a 900 kg vehicle that negotiates a 500 m radius curve in ~ 25.0 m/s.Assuming an unbanked curve, find the minimum static coefficient that friction, in between the tires and the road, revolution friction being the reason that keeps the auto from slipping (see figure 2).Strategy and also Solution for component 1

We recognize that \\textF_c=\\fracmv^2r\\\\. Thus,

\\displaystyle\\textF_c=\\fracmv^2r=\\frac\\left(900\\text kg\\right)\\left(25.0\\text m/s\\right)^2\\left(500\\text m\\right)=1125\\text N\\\\.

Strategy for part 2

Figure 2 shows the pressures acting ~ above the automobile on one unbanked (level ground) curve. Friction is to the left, maintaining the car from slipping, and because the is the only horizontal pressure acting on the car, the friction is the centripetal force in this case. We know that the maximum static friction (at i m sorry the tires roll but do no slip) is μsN, wherein μs is the revolution coefficient the friction and N is the regular force. The normal force equates to the car’s weight on level ground, so that N=mg. Hence the centripetal force in this case is

Fc = fμsN = μsmg.

Now we have actually a relationship between centripetal force and also the coefficient that friction. Making use of the an initial expression because that Fc from the equation

\\begincases\\textF_c=m\\fracv^2r\\\\\\textF_c=mr\\omega^2\\endcases,\\text m\\fracv^2r=\\mu_smg\\\\

We fix this because that μs, noting the mass cancels, and also obtain

\\displaystyle\\mu_s=\\fracv^2rg\\\\.

Solution for part 2

Substituting the knowns,

\\displaystyle\\mu_s=\\frac\\left(25.0\\text m/s\\right)^2\\left(500\\text m\\right)\\left(9.80\\text m/s^2\\right)=0.13\\\\.

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

Discussion

We could also solve component 1 making use of the an initial expression in\\begincases\\textF_c=m\\fracv^2r\\\\\\textF_c=mr\\omega^2\\endcases\\\\ because m,v, and also r space given. The coefficient the friction uncovered in component 2 is much smaller than is frequently found in between tires and also roads. The vehicle will still negotiate the curve if the coefficient is better than 0.13, because static friction is a responsive force, gift able to i think a value much less than however no an ext than μsN. A greater coefficient would certainly also permit the auto to negotiate the curve in ~ a higher speed, but if the coefficient the friction is less, the safe speed would be much less than 25 m/s. Keep in mind that mass cancels, implying the in this example, the does not issue how heavily loaded the vehicle is to negotiate the turn. Fixed cancels since friction is presume proportional to the regular force, which in turn is proportional to mass. If the surface ar of the roadway were banked, the normal pressure would be much less as will be discussed below.


Figure 2. This car on level ground is moving away and turning to the left. The centripetal force causing the auto to revolve in a circular course is due to friction in between the tires and the road. A minimum coefficient of friction is needed, or the automobile will move in a larger-radius curve and leave the roadway.


Let us now consider banked curves, whereby the slope of the roadway helps you negotiate the curve. See figure 3. The better the edge θ, the much faster you deserve to take the curve. Gyeongju tracks for bikes as well as cars, because that example, often have steeply banked curves. In an “ideally banked curve,” the edge θ is such the you have the right to negotiate the curve at a specific speed there is no the aid of friction in between the tires and the road. We will derive one expression because that θ for an ideally banked curve and consider an example related come it.

For best banking, the net exterior force amounts to the horizontal centripetal force in the absence of friction. The contents of the normal pressure N in the horizontal and also vertical directions must equal the centripetal force and the load of the car, respectively. In instances in which forces are no parallel, that is most convenient to take into consideration components follow me perpendicular axes—in this case, the vertical and also horizontal directions.

Figure 3 shows a free body diagram because that a car on a frictionless banked curve. If the edge θ is best for the speed and also radius, then the net outside force will equal the necessary centripetal force. The only two exterior forces acting on the car are its weight w and also the normal pressure of the road N. (A frictionless surface deserve to only exert a force perpendicular come the surface—that is, a typical force.) these two pressures must add to provide a net outside force the is horizontal toward the center of curvature and has magnitude mv2/r. Because this is the crucial force and it is horizontal, we usage a coordinate device with vertical and also horizontal axes. Just the normal force has a horizontal component, and so this need to equal the centripetal force—that is,

N\\sin\\theta=\\fracmv^2r\\\\.

Because the automobile does no leave the surface of the road, the net vertical pressure must it is in zero, an interpretation that the vertical contents of the two exterior forces need to be equal in magnitude and also opposite in direction. From the figure, we check out that the vertical component that the normal force is N cos θ, and also the only other vertical pressure is the car’s weight. These must be equal in magnitude; thus, N cos θ = mg.

Now we can integrate the last two equations to get rid of N and get an expression for θ, as desired. Addressing the second equation because that N=\\fracmg\\cos\\theta\\\\ , and also substituting this right into the an initial yields

\\displaystyle\\beginarray\\\\mg\\frac\\sin\\theta\\cos\\theta=\\fracmv^2r\\\\mg\\tan\\left(\\theta\\right)=\\fracmv^2r\\\\\\tan\\theta=\\fracv^2rg\\endarray\\\\

Taking the station tangent gives

\\theta=\\tan^-1\\left(\\fracv^2rg\\right)\\\\ (ideally banked curve, no friction).

This expression can be understood by considering just how θ relies on v and r. A large θ will certainly be obtained for a large v and a small r. The is, roads should be steeply banked for high speeds and also sharp curves. Friction helps, since it allows you to take it the curve at better or lower speed than if the curve is frictionless. Keep in mind that θ does not count on the mass of the vehicle.


Figure 3. The automobile on this banked curve is relocating away and transforming to the left.


Example 2. What Is the ideal Speed to take it a Steeply Banked chop Curve?

Curves on part test tracks and also race smashville247.net, such as the Daytona worldwide Speedway in Florida, are very steeply banked. This banking, v the help of tire friction and really stable car configurations, permits the curve to it is in taken at very high speed. Come illustrate, calculation the speed at which a 100 m radius curve banked in ~ 65.0° must be propelled if the road is frictionless.

Strategy

We very first note the all terms in the expression because that the right angle that a banked curve other than for speed are known; thus, we require only rearrange that so the speed appears on the left-hand side and then substitute recognized quantities.

Solution

Starting with

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

Noting the tan 65.0º = 2.14, we obtain

\\beginarray\\\\v=\\left<\\left(100\\text m\\right)\\left(9.80\\text m/s^2\\right)\\left(2.14\\right)\\right>^1/2\\\\\\text =45.8\\endarray\\\\

Discussion

This is just around 165 km/h, regular with a an extremely steeply banked and also rather sharp curve. Tire friction enables a automobile to take it the curve at significantly greater speeds.

Calculations comparable to those in the preceding examples can it is in performed for a host of interesting cases in i beg your pardon centripetal pressure is involved—a number of these room presented in this chapter’s Problems and Exercises.


Take-Home Experiment

Ask a girlfriend or relative to waver a golf society or a tennis racquet. Take suitable measurements to calculation the centripetal acceleration the the end of the club or racquet. Friend may select to carry out this in slow motion.


PhET Explorations: Gravity and Orbits

Move the sun, earth, moon and room station come see just how it affects their gravitational forces and orbital paths. Visualize the sizes and also distances between different heavenly bodies, and turn off heaviness to see what would occur without it!


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Section Summary

Centripetal force Fc is any type of force bring about uniform circular motion. It is a “center-seeking” force that always points towards the facility of rotation. It is perpendicular to direct velocity v and has magnitude Fc = mac, which can also be express as

\\begincases\\textF_c=m\\fracv^2r\\\\\\textor\\\\\\textF_c=mr\\omega^2\\endcases\\\\


Conceptual Questions

If you great to minimize the stress (which is regarded centripetal force) ~ above high-speed tires, would you use large- or small-diameter tires? Explain.Define centripetal force. Can any type of force (for example, tension, gravitational force, friction, and also so on) it is in a centripetal force? have the right to any mix of pressures be a centripetal force?If centripetal force is directed toward the center, why do you feel the you room ‘thrown’ far from the center as a automobile goes roughly a curve? Explain.Race automobile drivers routinely cut corners as shown in number 7. Describe how this allows the curve to be taken in ~ the greatest speed.

Figure 7. 2 paths about a gyeongju track curve space shown. Race auto drivers will take the inside route (called cutting the corner) whenever possible because it allows them to take it the curve in ~ the highest speed.


A variety of amusement parks have actually rides that make vertical loops like the one displayed in figure 8. For safety, the cars space attached come the rails in together a means that lock cannot autumn off. If the car goes end the top at just the best speed, gravity alone will supply the centripetal force. What other pressure acts and also what is its direction if: (a) The auto goes over the height at faster than this speed? (b) The auto goes over the top at slower than this speed?

Figure 8. Amusement rides through a vertical loop are an instance of a type of curved motion.


What is the direction of the force exerted by the vehicle on the passenger together the auto goes over the optimal of the amusement journey pictured in figure 8 under the complying with circumstances: (a) The car goes over the optimal at such a rate that the gravitational pressure is the only pressure acting? (b) The vehicle goes end the top quicker than this speed? (c) The vehicle goes over the peak slower 보다 this speed?As a skater develops a circle, what pressure is responsible because that making her turn? use a complimentary body chart in your answer.Suppose a kid is riding ~ above a merry-go-round in ~ a distance about halfway between its center and edge. She has a lunch box relaxing on wax paper, so that there is very small friction in between it and also the merry-go-round. I beg your pardon path displayed in number 9 will the lunch box take when she lets go? The lunch box leaves a trace in the dust ~ above the merry-go-round. Is that trail straight, bent to the left, or bent to the right? define your answer.

Figure 9. A kid riding on a merry-go-round releases she lunch box at allude P. This is a watch from over the clockwise rotation. Assuming that slides through negligible friction, will certainly it follow course A, B, or C, as regarded from Earth’s structure of reference? What will be the shape of the route it leaves in the dust ~ above the merry-go-round?


Do you feel yourself thrown to one of two people side as soon as you negotiate a curve the is ideally banked for your car’s speed? What is the direction of the pressure exerted on friend by the auto seat?Suppose a massive is moving in a circular path on a frictionless table as presented in figure. In the Earth’s structure of reference, there is no centrifugal pressure pulling the mass far from the center of rotation, yet there is a very real force stretching the string it is registered the mass come the nail. Using principles related to centripetal force and also Newton’s 3rd law, define what pressure stretches the string, identifying its physical origin.

Figure 10. A massive attached come a pond on a frictionless table move in a circular path. The force stretching the cable is real and not fictional. What is the physical beginning of the pressure on the string?


Problems & Exercises

(a) A 22.0 kg son is speak a playground merry-go-round the is rotating in ~ 40.0 rev/min. What centripetal force must she exert to continue to be on if she is 1.25 m from its center? (b) What centripetal force does she have to stay on an amusement park merry-go-round the rotates in ~ 3.00 rev/min if she is 8.00 m from its center? (c) to compare each force with her weight.Calculate the centripetal pressure on the end of a 100 m (radius) wind turbine blade the is rotating in ~ 0.5 rev/s. Assume the mass is 4 kg.What is the best banking angle for a gentle turn of 1.20 kilometres radius ~ above a highway v a 105 km/h speed limit (about 65 mi/h), assuming anyone travels in ~ the limit?What is the appropriate speed to take a 100 m radius curve banked in ~ a 20.0° angle?(a) What is the radius of a bobsled revolve banked in ~ 75.0° and taken at 30.0 m/s, assuming the is ideally banked? (b) calculation the centripetal acceleration. (c) does this acceleration seem huge to you?Part of riding a bicycle requires leaning at the correct angle once making a turn, as watched in figure 4. To it is in stable, the pressure exerted by the ground should be top top a heat going v the facility of gravity. The pressure on the bicycle wheel deserve to be resolved into two perpendicular components—friction parallel come the road (this need to supply the centripetal force), and also the vertical normal pressure (which need to equal the system’s weight). (a) show that θ (as defined in the figure) is related to the speed v and radius the curvature r of the rotate in the same way as for an ideally banked roadway—that is, \\theta=\\tan^-1\\fracv^2rg\\\\; (b) Calculate θ because that a 12.0 m/s turn of radius 30.0 m (as in a race).

Figure 6. 4. A bicyclist negotiating a turn on level ground have to lean in ~ the exactly angle—the ability to carry out this i do not care instinctive. The force of the floor on the wheel requirements to it is in on a line v the facility of gravity. The net external force on the device is the centripetal force. The vertical component the the force on the wheel cancels the load of the system while that is horizontal component have to supply the centripetal force. This process produces a relationship amongst the edge θ, the rate v, and also the radius that curvature r the the turn similar to that for the appropriate banking of roadways.


A huge centrifuge, prefer the one presented in number 5a, is offered to reveal aspiring astronauts come accelerations similar to those competent in rocket launches and atmospheric reentries. (a) in ~ what angular velocity is the centripetal acceleration 10 g if the driver is 15.0 m from the center of rotation? (b) The rider’s cage hangs top top a pivot at the end of the arm, allowing it to swing outward throughout rotation as shown in number 5b. At what angle θ listed below the horizontal will certainly the cage hang once the centripetal acceleration is 10 g? (Hint: The arm provides centripetal force and also supports the weight of the cage. Draw a complimentary body chart of the pressures to see what the angle θ have to be.)

Figure 5. (a) NASA centrifuge used to topic trainees come accelerations comparable to those knowledgeable in rocket launches and also reentries. (credit: NASA) (b) driver in cage showing exactly how the cage pivots outward during rotation. This permits the full force exerted on the rider by the cage to be follow me its axis at every times.


Integrated Concepts. If a vehicle takes a banked curve at much less than the appropriate speed, friction is needed to keep it from sliding towards the inside of the curve (a real problem on icy mountain roads). (a) calculation the appropriate speed to take a 100 m radius curve banked at 15.0º. (b) What is the minimum coefficient the friction needed for a frightened driver to take the very same curve at 20.0 km/h?Modern roller coasters have actually vertical loops like the one shown in figure 6. The radius of curvature is smaller sized at the peak than top top the political parties so that the downward centripetal acceleration at the optimal will be greater than the acceleration as result of gravity, keeping the passengers pressed firmly into their seats. What is the speed of the roller coaster in ~ the optimal of the loop if the radius the curvature over there is 15.0 m and also the bottom acceleration that the car is 1.50 g?

Figure 6. Teardrop-shaped loops are provided in the latest roller coasters so the the radius of curvature progressively decreases come a minimum at the top. This means that the centripetal acceleration build from zero to a maximum in ~ the top and gradually reduce again. A one loop would cause a jolting readjust in acceleration in ~ entry, a disadvantage uncovered long back in railroad curve design. With a small radius that curvature in ~ the top, the centripetal acceleration can more easily it is in kept better than g so that the passengers carry out not lose contact with your seats nor carry out they need seat belts to keep them in place.


Unreasonable Results. (a) calculate the minimum coefficient that friction needed for a automobile to negotiate an unbanked 50.0 m radius curve at 30.0 m/s. (b) What is unreasonable around the result? (c) i beg your pardon premises space unreasonable or inconsistent?

Glossary

centripetal force: any net pressure causing uniform circular motion

ideal banking: the sloping that a curve in a road, where the edge of the slope enables the car to negotiate the curve at a details speed there is no the aid of friction between the tires and also the road; the net exterior force ~ above the vehicle equals the horizontal centripetal force in the lack of friction

ideal speed: the preferably safe speed at i m sorry a auto can revolve on a curve there is no the help of friction in between the tire and the road

ideal angle: the angle at i m sorry a automobile can revolve safely on a steep curve, i m sorry is in proportion to the appropriate speed

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


Selected services to problems & Exercises

1. (a) 483 N; (b) 17.4 N; (c) 2.24 times she 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 too large, but it is clear the bobsledders feeling a many of pressure on them going v sharply banked turns.

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7. (a) 2.56 rad/s; (b) 5.71º

8. (a) 16.2 m/s; (b) 0.234

10. (a) 1.84; (b) A coefficient that friction this much higher than 1 is unreasonable; (c) The assumed rate is too good for the chop curve.