without friction what net force is needed to maintain a 1000 kg car in uniform motion for 30 minutes

half dozen Applications of Newton's Laws

six.3 Centripetal Strength

Learning Objectives

Past the end of the department, you will be able to:

• Explain the equation for centripetal acceleration
• Apply Newton's second police force to develop the equation for centripetal forcefulness
• Use round motion concepts in solving problems involving Newton's laws of motion

In Motility in Two and Three Dimensions, nosotros examined the basic concepts of circular motion. An object undergoing circular motion, like one of the race cars shown at the beginning of this chapter, must exist accelerating because information technology is changing the direction of its velocity. We proved that this centrally directed acceleration, called centripetal dispatch , is given by the formula

where v is the velocity of the object, directed forth a tangent line to the curve at any instant. If nosotros know the angular velocity

, so we can utilize

Angular velocity gives the charge per unit at which the object is turning through the curve, in units of rad/s. This acceleration acts forth the radius of the curved path and is thus besides referred to equally a radial acceleration.

An dispatch must be produced by a strength. Any forcefulness or combination of forces tin can crusade a centripetal or radial acceleration. Merely a few examples are the tension in the rope on a tether ball, the forcefulness of World'due south gravity on the Moon, friction betwixt roller skates and a rink floor, a banked roadway'south strength on a automobile, and forces on the tube of a spinning centrifuge. Whatever net force causing uniform circular movement is called a centripetal force. The direction of a centripetal force is toward the center of curvature, the same equally the direction of centripetal dispatch. According to Newton's 2d law of movement, net force is mass times acceleration:

For uniform round motion, the acceleration is the centripetal acceleration: _.

Thus, the magnitude of centripetal forcefulness

is

By substituting the expressions for centripetal acceleration

nosotros go two expressions for the centripetal force

in terms of mass, velocity, angular velocity, and radius of curvature:

Yous may use whichever expression for centripetal force is more convenient. Centripetal force

is always perpendicular to the path and points to the centre of curvature, because

is perpendicular to the velocity and points to the center of curvature. Note that if y'all solve the get-go expression for r, you get

This implies that for a given mass and velocity, a large centripetal strength causes a small radius of curvature—that is, a tight curve, as in (Figure).

Effigy 6.20 The frictional forcefulness supplies the centripetal strength and is numerically equal to information technology. Centripetal forcefulness is perpendicular to velocity and causes uniform circular motion. The larger the

the smaller the radius of curvature r and the sharper the bend. The second curve has the same v, but a larger

produces a smaller r′.

Example

What Coefficient of Friction Exercise Cars Need on a Flat Curve?

(a) Summate the centripetal force exerted on a 900.0-kg car that negotiates a 500.0-m radius curve at 25.00 1000/south. (b) Assuming an unbanked curve, observe the minimum static coefficient of friction between the tires and the road, static friction existence the reason that keeps the machine from slipping ((Effigy)).

Figure six.21 This car on level basis is moving away and turning to the left. The centripetal strength causing the motorcar to turn in a circular path is due to friction between the tires and the route. A minimum coefficient of friction is needed, or the motorcar volition move in a larger-radius bend and leave the roadway.

Strategy

1. We know that

   

   Thus,

   

2. (Figure) shows the forces interim on the car on an unbanked (level ground) bend. Friction is to the left, keeping the automobile from slipping, and considering it is the only horizontal force acting on the machine, 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

   

   where

   

   is the static coefficient of friction and Due north is the normal strength. The normal force equals the car's weight on level ground, so

   

   Thus the centripetal force in this situation is

   

   Now we accept a relationship between centripetal forcefulness and the coefficient of friction. Using the equation

   

   we obtain

   

   Nosotros solve this for

   

   noting that mass cancels, and obtain

   

   Substituting the knowns,

   

   (Considering coefficients of friction are guess, the reply is given to only ii digits.)

Significance

The coefficient of friction found in (Figure)(b) is much smaller than is typically found between tires and roads. The car withal negotiates the bend if the coefficient is greater than 0.xiii, because static friction is a responsive strength, able to presume a value less than but no more than

A higher coefficient would also allow the car to negotiate the bend at a higher speed, but if the coefficient of friction is less, the safe speed would be less than 25 thousand/southward. Annotation that mass cancels, implying that, in this example, it does not thing how heavily loaded the car is to negotiate the turn. Mass cancels because friction is assumed proportional to the normal force, which in plough is proportional to mass. If the surface of the road were banked, the normal force would be less, as discussed next.

Check Your Understanding

A car moving at 96.8 km/h travels around a round curve of radius 182.9 thou on a flat country road. What must be the minimum coefficient of static friction to go along the motorcar from slipping?

[reveal-reply q="694795″]Prove Solution[/reveal-answer]
[hidden-reply a="694795″]0.40[/hidden-answer]

Banked Curves

Let usa at present consider banked curves, where the slope of the road helps you negotiate the curve ((Effigy)). The greater the angle

, the faster yous can take the curve. Race tracks for bikes as well as cars, for example, often take steeply banked curves. In an "ideally banked curve," the angle

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

for an ideally banked curve and consider an example related to it.

Figure half dozen.22 The motorcar on this banked curve is moving abroad and turning to the left.

For ideal banking, the cyberspace external force equals the horizontal centripetal strength in the absenteeism of friction. The components of the normal force N in the horizontal and vertical directions must equal the centripetal strength and the weight of the car, respectively. In cases in which forces are not parallel, it is about user-friendly to consider components along perpendicular axes—in this case, the vertical and horizontal directions.

(Figure) shows a gratuitous-body diagram for a machine on a frictionless banked bend. If the angle

is ideal for the speed and radius, then the net external force equals the necessary centripetal force. The just two external forces acting on the car are its weight

and the normal forcefulness of the route

(A frictionless surface can merely exert a force perpendicular to the surface—that is, a normal force.) These two forces must add together to give a internet external forcefulness that is horizontal toward the center of curvature and has magnitude

Because this is the crucial strength 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,

Considering the car does not leave the surface of the road, the net vertical force must be zero, pregnant that the vertical components of the two external forces must be equal in magnitude and opposite in direction. From (Effigy), we run into that the vertical component of the normal force is

and the only other vertical strength is the car'south weight. These must be equal in magnitude; thus,

At present we can combine these ii equations to eliminate N and get an expression for

, as desired. Solving the second equation for

and substituting this into the commencement yields

Taking the inverse tangent gives

This expression can be understood by considering how

depends on five and r. A large

is obtained for a large five and a pocket-size r. That is, roads must exist steeply banked for high speeds and sharp curves. Friction helps, considering information technology allows you to take the bend at greater or lower speed than if the bend were frictionless. Note that

does not depend on the mass of the vehicle.

Instance

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

Curves on some test tracks and race courses, such as Daytona International Speedway in Florida, are very steeply banked. This banking, with the assist of tire friction and very stable automobile configurations, allows the curves to be taken at very high speed. To illustrate, summate the speed at which a 100.0-m radius bend banked at

should be driven if the road were frictionless.

Strategy

We first note that all terms in the expression for the ideal bending of a banked curve except for speed are known; thus, we need merely rearrange information technology so that speed appears on the left-paw side then substitute known quantities.

Solution

Starting with

nosotros get

Noting that

we obtain

Significance

This is just about 165 km/h, consequent with a very steeply banked and rather abrupt bend. Tire friction enables a vehicle to take the bend at significantly college speeds.

Airplanes besides make turns by banking. The lift forcefulness, due to the forcefulness of the air on the wing, acts at right angles to the wing. When the aeroplane banks, the pilot is obtaining greater lift than necessary for level flight. The vertical component of lift balances the plane's weight, and the horizontal component accelerates the aeroplane. The cyberbanking bending shown in (Figure) is given by

. We analyze the forces in the same way we care for the example of the motorcar rounding a banked curve.

Figure half-dozen.23 In a banked turn, the horizontal component of lift is unbalanced and accelerates the plane. The normal component of lift balances the plane's weight. The cyberbanking angle is given by

. Compare the vector diagram with that shown in (Effigy).

Join the ladybug in an exploration of rotational motion. Rotate the merry-go-circular to change its angle or choose a constant angular velocity or angular acceleration. Explore how round motility relates to the problems's xy-position, velocity, and acceleration using vectors or graphs.

A circular motility requires a strength, the so-called centripetal forcefulness, which is directed to the axis of rotation. This simplified model of a carousel demonstrates this force.

Inertial Forces and Noninertial (Accelerated) Frames: The Coriolis Force

What practice taking off in a jet aeroplane, turning a corner in a car, riding a merry-go-round, and the circular move of a tropical whirlwind take in common? Each exhibits inertial forces—forces that merely seem to arise from motion, because the observer's frame of reference is accelerating or rotating. When taking off in a jet, virtually people would agree it feels as if you are being pushed back into the seat as the plane accelerates downward the rails. All the same a physicist would say that you tend to remain stationary while the seat pushes forward on you. An even more common experience occurs when you brand a tight curve in your car—say, to the right ((Figure)). You experience as if you are thrown (that is, forced) toward the left relative to the car. Once again, a physicist would say that you lot are going in a straight line (think Newton's start police) but the machine moves to the correct, non that you are experiencing a force from the left.

Figure 6.24 (a) The car driver feels herself forced to the left relative to the car when she makes a right turn. This is an inertial force arising from the apply of the machine equally a frame of reference. (b) In Earth's frame of reference, the driver moves in a direct line, obeying Newton's outset constabulary, and the car moves to the right. In that location is no force to the left on the driver relative to Earth. Instead, at that place is a force to the right on the car to brand information technology turn.

We can reconcile these points of view by examining the frames of reference used. Let us concentrate on people in a car. Passengers instinctively utilize the machine as a frame of reference, whereas a physicist might employ Globe. The physicist might brand this choice because World is almost an inertial frame of reference, in which all forces have an identifiable concrete origin. In such a frame of reference, Newton's laws of move take the form given in Newton'south Laws of Motion. The machine is a noninertial frame of reference because information technology is accelerated to the side. The force to the left sensed past machine passengers is an inertial force having no physical origin (it is due purely to the inertia of the passenger, non to some physical cause such as tension, friction, or gravitation). The automobile, as well as the driver, is actually accelerating to the right. This inertial force is said to exist an inertial forcefulness because it does non have a physical origin, such as gravity.

A physicist will choose whatever reference frame is most convenient for the situation existence analyzed. There is no problem to a physicist in including inertial forces and Newton's second law, as usual, if that is more convenient, for example, on a merry-go-round or on a rotating planet. Noninertial (accelerated) frames of reference are used when it is useful to practice so. Different frames of reference must be considered in discussing the motion of an astronaut in a spacecraft traveling at speeds near the speed of light, as you will appreciate in the study of the special theory of relativity.

Let u.s.a. now have a mental ride on a merry-become-round—specifically, a rapidly rotating playground merry-go-round ((Effigy)). Yous take the merry-get-round to be your frame of reference because you rotate together. When rotating in that noninertial frame of reference, you feel an inertial strength that tends to throw you off; this is often referred to every bit a centrifugal force (non to exist confused with centripetal forcefulness). Centrifugal force is a ordinarily used term, only information technology does not really exist. You must hang on tightly to counteract your inertia (which people often refer to as centrifugal force). In Globe's frame of reference, at that place is no force trying to throw y'all off; nosotros emphasize that centrifugal forcefulness is a fiction. You must hang on to make yourself get in a circle because otherwise yous would get in a straight line, right off the merry-get-round, in keeping with Newton's first constabulary. Simply the strength yous exert acts toward the center of the circle.

Figure vi.25 (a) A rider on a merry-go-round feels as if he is existence thrown off. This inertial force is sometimes mistakenly called the centrifugal force in an effort to explain the rider's motion in the rotating frame of reference. (b) In an inertial frame of reference and according to Newton'south laws, information technology is his inertia that carries him off (the unshaded rider has

and heads in a directly line). A force,

, is needed to crusade a round path.

This inertial result, carrying you abroad from the heart of rotation if in that location is no centripetal force to cause circular motion, is put to proficient employ in centrifuges ((Figure)). A centrifuge spins a sample very rapidly, as mentioned earlier in this chapter. Viewed from the rotating frame of reference, the inertial forcefulness throws particles outward, hastening their sedimentation. The greater the angular velocity, the greater the centrifugal force. But what actually happens is that the inertia of the particles carries them along a line tangent to the circle while the test tube is forced in a round path past a centripetal force.

Figure 6.26 Centrifuges use inertia to perform their task. Particles in the fluid sediment settle out because their inertia carries them away from the center of rotation. The big angular velocity of the centrifuge quickens the sedimentation. Ultimately, the particles come into contact with the test tube walls, which then supply the centripetal strength needed to brand them move in a circumvolve of constant radius.

Let us at present consider what happens if something moves in a rotating frame of reference. For instance, what if you slide a ball directly away from the middle of the merry-go-circular, as shown in (Figure)? The ball follows a direct path relative to World (assuming negligible friction) and a path curved to the right on the merry-get-round's surface. A person standing next to the merry-go-round sees the ball moving straight and the merry-go-circular rotating underneath information technology. In the merry-get-circular'southward frame of reference, nosotros explain the apparent curve to the correct past using an inertial force, called the Coriolis force, which causes the ball to bend to the right. The Coriolis force can be used by anyone in that frame of reference to explain why objects follow curved paths and allows united states to apply Newton's laws in noninertial frames of reference.

Figure six.27 Looking downward on the counterclockwise rotation of a merry-go-circular, we see that a brawl slid straight toward the edge follows a path curved to the correct. The person slides the ball toward betoken B, starting at bespeak A. Both points rotate to the shaded positions (A' and B') shown in the time that the ball follows the curved path in the rotating frame and a straight path in Earth's frame.

Up until now, we have considered Earth to be an inertial frame of reference with little or no worry about furnishings due to its rotation. Nonetheless such effects practise be—in the rotation of atmospheric condition systems, for example. Most consequences of Earth's rotation tin can exist qualitatively understood past analogy with the merry-go-round. Viewed from in a higher place the North Pole, Globe rotates counterclockwise, as does the merry-go-round in (Figure). As on the merry-go-round, whatever motion in Earth'due south Northern Hemisphere experiences a Coriolis force to the right. Just the contrary occurs in the Southern Hemisphere; there, the strength is to the left. Considering Earth'south athwart velocity is small, the Coriolis force is unremarkably negligible, merely for large-scale motions, such as current of air patterns, it has substantial effects.

The Coriolis force causes hurricanes in the Northern Hemisphere to rotate in the counterclockwise management, whereas tropical cyclones in the Southern Hemisphere rotate in the clockwise direction. (The terms hurricane, typhoon, and tropical storm are regionally specific names for cyclones, which are storm systems characterized by low pressure centers, stiff winds, and heavy rains.) (Figure) helps show how these rotations take place. Air flows toward whatsoever region of low pressure, and tropical cyclones contain particularly low pressures. Thus winds flow toward the center of a tropical cyclone or a low-pressure weather system at the surface. In the Northern Hemisphere, these inward winds are deflected to the right, as shown in the figure, producing a counterclockwise apportionment at the surface for low-pressure zones of whatever type. Depression pressure at the surface is associated with rise air, which too produces cooling and cloud formation, making low-pressure patterns quite visible from infinite. Conversely, wind circulation effectually high-pressure zones is clockwise in the Southern Hemisphere simply is less visible because high pressure level is associated with sinking air, producing articulate skies.

Figure 6.28 (a) The counterclockwise rotation of this Northern Hemisphere hurricane is a major result of the Coriolis forcefulness. (b) Without the Coriolis force, air would flow direct into a low-pressure level zone, such equally that found in tropical cyclones. (c) The Coriolis force deflects the winds to the correct, producing a counterclockwise rotation. (d) Wind flowing away from a high-pressure level zone is likewise deflected to the right, producing a clockwise rotation. (e) The reverse direction of rotation is produced past the Coriolis force in the Southern Hemisphere, leading to tropical cyclones. (credit a and credit e: modifications of work by NASA)

The rotation of tropical cyclones and the path of a brawl on a merry-go-round tin just likewise be explained past inertia and the rotation of the system underneath. When noninertial frames are used, inertial forces, such as the Coriolis force, must be invented to explain the curved path. In that location is no identifiable concrete source for these inertial forces. In an inertial frame, inertia explains the path, and no force is found to exist without an identifiable source. Either view allows us to draw nature, but a view in an inertial frame is the simplest in the sense that all forces have origins and explanations.

Summary

• Centripetal force

  

  is a "center-seeking" force that e'er points toward the center of rotation. Information technology is perpendicular to linear velocity and has the magnitude

  

• Rotating and accelerated frames of reference are noninertial. Inertial forces, such as the Coriolis strength, are needed to explain move in such frames.

Conceptual Questions

If you wish to reduce the stress (which is related to centripetal strength) on high-speed tires, would you use large- or small-diameter tires? Explain.

Ascertain centripetal force. Can whatever type of strength (for example, tension, gravitational force, friction, and so on) be a centripetal force? Can any combination of forces be a centripetal forcefulness?

[reveal-respond q="fs-id1165039453744″]Show Solution[/reveal-answer]

[hidden-answer a="fs-id1165039453744″]

Centripetal force is defined every bit any net forcefulness causing uniform circular motion. The centripetal force is not a new kind of strength. The label "centripetal" refers to any force that keeps something turning in a circle. That forcefulness could be tension, gravity, friction, electrical attraction, the normal force, or any other force. Any combination of these could be the source of centripetal force, for case, the centripetal force at the top of the path of a tetherball swung through a vertical circumvolve is the result of both tension and gravity.

[/subconscious-answer]

If centripetal strength is directed toward the center, why practise you feel that you are 'thrown' away from the heart as a car goes effectually a curve? Explicate.

Race car drivers routinely cut corners, every bit shown beneath (Path 2). Explain how this allows the curve to exist taken at the greatest speed.

[reveal-reply q="329939″]Show Solution[/reveal-answer]
[hidden-answer a="329939″]The driver who cuts the corner (on Path two) has a more than gradual curve, with a larger radius. That ane will be the meliorate racing line. If the commuter goes as well fast around a corner using a racing line, he will still slide off the rail; the key is to stay at the maximum value of static friction. So, the driver wants maximum possible speed and maximum friction. Consider the equation for centripetal force:

where 5 is speed and r is the radius of curvature. And so by decreasing the curvature (1/r) of the path that the automobile takes, we reduce the amount of force the tires have to exert on the road, meaning we can now increase the speed, 5. Looking at this from the point of view of the driver on Path one, we can reason this mode: the sharper the turn, the smaller the turning circumvolve; the smaller the turning circle, the larger is the required centripetal force. If this centripetal strength is not exerted, the outcome is a sideslip.[/subconscious-answer]

Many amusement parks have rides that brand vertical loops like the i shown beneath. For safety, the cars are fastened to the rails in such a mode that they cannot fall off. If the car goes over the top at just the correct speed, gravity alone volition supply the centripetal force. What other strength acts and what is its direction if:

(a) The motorcar goes over the top at faster than this speed?

(b) The car goes over the peak at slower than this speed?

What causes water to be removed from dress in a spin-dryer?

[reveal-answer q="fs-id1165039477270″]Show Solution[/reveal-answer]

[hidden-answer a="fs-id1165039477270″]

The barrel of the dryer provides a centripetal force on the clothes (including the water aerosol) to continue them moving in a circular path. Every bit a water droplet comes to one of the holes in the butt, it will movement in a path tangent to the circle.

[/subconscious-respond]

As a skater forms a circle, what forcefulness is responsible for making his turn? Use a gratis-body diagram in your reply.

Suppose a child is riding on a merry-become-circular at a distance almost halfway between its heart and edge. She has a lunch box resting on wax paper, so that there is very little friction between information technology and the merry-go-round. Which path shown below volition the dejeuner box accept when she lets get? The tiffin box leaves a trail in the dust on the merry-get-round. Is that trail straight, curved to the left, or curved to the right? Explicate your answer.

[reveal-answer q="60053″]Show Solution[/reveal-answer]
[hidden-respond a="60053″]If in that location is no friction, so in that location is no centripetal force. This ways that the lunch box volition move along a path tangent to the circle, and thus follows path B. The grit trail volition be direct. This is a result of Newton's outset law of motion.[/hidden-answer]

Do you experience yourself thrown to either side when yous negotiate a curve that is ideally banked for your auto's speed? What is the direction of the force exerted on you by the motorcar seat?

Suppose a mass is moving in a circular path on a frictionless table as shown below. In Earth'southward frame of reference, there is no centrifugal forcefulness pulling the mass away from the center of rotation, yet at that place is a forcefulness stretching the string attaching the mass to the blast. Using concepts related to centripetal force and Newton'south third law, explicate what force stretches the string, identifying its physical origin.

[reveal-reply q="965193″]Show Solution[/reveal-answer]
[hidden-reply a="965193″]There must be a centripetal force to maintain the circular motion; this is provided past the nail at the center. Newton'southward third law explains the miracle. The action strength is the force of the string on the mass; the reaction force is the force of the mass on the string. This reaction force causes the string to stretch.[/subconscious-reply]

When a toilet is flushed or a sink is drained, the h2o (and other material) begins to rotate about the drain on the mode down. Assuming no initial rotation and a flow initially directly straight toward the bleed, explain what causes the rotation and which direction it has in the Northern Hemisphere. (Note that this is a small effect and in most toilets the rotation is caused past directional h2o jets.) Would the direction of rotation reverse if h2o were forced up the bleed?

A automobile rounds a curve and encounters a patch of ice with a very low coefficient of kinetic fiction. The motorcar slides off the road. Describe the path of the car as it leaves the route.

[reveal-answer q="fs-id1165039077662″]Evidence Solution[/reveal-respond]

[hidden-answer a="fs-id1165039077662″]

Since the radial friction with the tires supplies the centripetal force, and friction is nearly 0 when the car encounters the ice, the car will obey Newton's commencement constabulary and go off the road in a direct line path, tangent to the bend. A common misconception is that the car volition follow a curved path off the road.

[/hidden-answer]

In one amusement park ride, riders enter a big vertical barrel and stand up against the wall on its horizontal floor. The butt is spun upwards and the floor drops away. Riders feel equally if they are pinned to the wall past a force something like the gravitational force. This is an inertial forcefulness sensed and used by the riders to explain events in the rotating frame of reference of the barrel. Explain in an inertial frame of reference (Globe is nearly one) what pins the riders to the wall, and identify all forces acting on them.

Ii friends are having a conversation. Anna says a satellite in orbit is in free fall because the satellite keeps falling toward Earth. Tom says a satellite in orbit is non in free autumn because the acceleration due to gravity is not

. Who practise you agree with and why?

[reveal-reply q="fs-id1165039083736″]Show Solution[/reveal-answer]

[hidden-answer a="fs-id1165039083736″]

Anna is correct. The satellite is freely falling toward Earth due to gravity, even though gravity is weaker at the altitude of the satellite, and g is not

. Free fall does not depend on the value of g; that is, you could experience gratuitous fall on Mars if you lot jumped off Olympus Mons (the tallest volcano in the solar system).
[/hidden-answer]

A nonrotating frame of reference placed at the middle of the Sun is very most an inertial one. Why is it non exactly an inertial frame?

Problems

(a) A 22.0-kg child is riding a playground merry-get-round that is rotating at 40.0 rev/min. What centripetal force is exerted if he is 1.25 m from its heart? (b) What centripetal force is exerted if the merry-get-round rotates at 3.00 rev/min and he is 8.00 m from its center? (c) Compare each forcefulness with his weight.

[reveal-respond q="fs-id1165039026811″]Bear witness Solution[/reveal-answer]

[hidden-answer a="fs-id1165039026811″]

a. 483 Northward; b. 17.4 N; c. two.24, 0.0807

[/subconscious-reply]

Summate the centripetal strength on the end of a 100-grand (radius) air current turbine blade that is rotating at 0.five rev/s. Assume the mass is iv kg.

What is the ideal banking angle for a gentle turn of 1.twenty-km radius on a highway with a 105 km/h speed limit (about 65 mi/h), assuming everyone travels at the limit?

[reveal-respond q="fs-id1165039344901″]Evidence Solution[/reveal-reply]

[hidden-answer a="fs-id1165039344901″]

[/hidden-answer]

What is the platonic speed to take a 100.0-m-radius curve banked at a

angle?

(a) What is the radius of a bobsled turn banked at

and taken at 30.0 m/s, bold it is ideally banked? (b) Summate the centripetal acceleration. (c) Does this acceleration seem large to you?

[reveal-answer q="fs-id1165039104209″]Bear witness Solution[/reveal-answer]

[hidden-answer a="fs-id1165039104209″]

a. 24.six m; b.

c. 3.73 times grand
[/hidden-answer]

Function of riding a cycle involves leaning at the correct bending when making a turn, as seen below. To be stable, the force exerted by the ground must be on a line going through the center of gravity. The strength on the bicycle bicycle tin can exist resolved into two perpendicular components—friction parallel to the road (this must supply the centripetal strength) and the vertical normal force (which must equal the system's weight). (a) Show that

(as defined equally shown) is related to the speed five and radius of curvature r of the turn in the same fashion every bit for an ideally banked roadway—that is,

(b) Calculate

for a 12.0-chiliad/due south turn of radius 30.0 m (equally in a race).

If a car takes a banked curve at less than the ideal speed, friction is needed to continue it from sliding toward the inside of the curve (a problem on icy mountain roads). (a) Summate the ideal speed to accept a 100.0 chiliad radius curve banked at

. (b) What is the minimum coefficient of friction needed for a frightened driver to have the aforementioned curve at 20.0 km/h?

[reveal-reply q="fs-id1165038980331″]Show Solution[/reveal-reply]

[subconscious-answer a="fs-id1165038980331″]

a. xvi.2 1000/southward; b. 0.234

[/hidden-reply]

Modernistic roller coasters accept vertical loops like the one shown hither. The radius of curvature is smaller at the acme than on the sides then that the downwardly centripetal acceleration at the top will be greater than the dispatch due to gravity, keeping the passengers pressed firmly into their seats. (a) What is the speed of the roller coaster at the tiptop of the loop if the radius of curvature there is 15.0 grand and the downward acceleration of the machine is one.50 g? (b) How high above the top of the loop must the roller coaster start from residual, assuming negligible friction? (c) If it actually starts 5.00 k college than your reply to (b), how much energy did it lose to friction? Its mass is

.

A kid of mass forty.0 kg is in a roller coaster car that travels in a loop of radius 7.00 m. At point A the speed of the automobile is 10.0 thousand/due south, and at bespeak B, the speed is 10.5 m/s. Presume the child is not property on and does not wear a seat belt. (a) What is the force of the machine seat on the child at point A? (b) What is the forcefulness of the car seat on the child at point B? (c) What minimum speed is required to continue the child in his seat at indicate A?

[reveal-answer q="484990″]Prove Solution[/reveal-answer]
[hidden-reply a="484990″]a. 179 N; b. 290 N; c. 8.3 chiliad/s[/hidden-reply]

In the unproblematic Bohr model of the ground state of the hydrogen atom, the electron travels in a circular orbit around a stock-still proton. The radius of the orbit is

and the speed of the electron is

The mass of an electron is

. What is the forcefulness on the electron?

Railroad tracks follow a circular curve of radius 500.0 m and are banked at an angle of

. For trains of what speed are these tracks designed?

[reveal-reply q="fs-id1165039111532″]Show Solution[/reveal-answer]

[hidden-reply a="fs-id1165039111532″]

xx.7 grand/south

[/hidden-answer]

The CERN particle accelerator is circular with a circumference of 7.0 km. (a) What is the acceleration of the protons

that move around the accelerator at

of the speed of light? (The speed of light is

) (b) What is the forcefulness on the protons?

A machine rounds an unbanked curve of radius 65 m. If the coefficient of static friction between the road and motorcar is 0.lxx, what is the maximum speed at which the car traverse the bend without slipping?

[reveal-answer q="fs-id1165039269152″]Evidence Solution[/reveal-answer]

[hidden-answer a="fs-id1165039269152″]

21 m/s

[/hidden-answer]

A banked highway is designed for traffic moving at 90.0 km/h. The radius of the curve is 310 m. What is the angle of banking of the highway?

Glossary

banked curve

bend in a road that is sloping in a manner that helps a vehicle negotiate the curve

centripetal force

any net force causing uniform circular move

Coriolis strength

inertial force causing the apparent deflection of moving objects when viewed in a rotating frame of reference

ideal banking

sloping of a bend in a road, where the angle of the gradient allows the vehicle to negotiate the curve at a certain speed without the aid of friction between the tires and the road; the internet external forcefulness on the vehicle equals the horizontal centripetal force in the absence of friction

inertial forcefulness

force that has no physical origin

noninertial frame of reference

accelerated frame of reference

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Source: https://opentextbc.ca/universityphysicsv1openstax/chapter/6-3-centripetal-force/

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