Refer Again to the Situation in Question 21 What Is the Maximum Speed of the Block

6 Applications of Newton's Laws

6.3 Centripetal Force

Learning Objectives

Past the stop of the section, you will exist able to:

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

In Motion in 2 and Three Dimensions, we examined the basic concepts of circular motion. An object undergoing circular movement, like one of the race cars shown at the beginning of this chapter, must be accelerating because it is changing the direction of its velocity. We proved that this centrally directed acceleration, called centripetal acceleration , is given past the formula

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

, then we can use

Athwart velocity gives the rate at which the object is turning through the bend, in units of rad/s. This acceleration acts along the radius of the curved path and is thus also referred to as a radial dispatch.

An dispatch must be produced by a force. Any force or combination of forces tin cause a centripetal or radial dispatch. Just a few examples are the tension in the rope on a tether brawl, the force of Earth'southward gravity on the Moon, friction between roller skates and a rink flooring, 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 chosen a centripetal strength. The direction of a centripetal force is toward the center of curvature, the same every bit the management of centripetal acceleration. According to Newton'southward second police of motion, net strength is mass times acceleration:

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

Thus, the magnitude of centripetal strength

is

By substituting the expressions for centripetal acceleration

we get ii expressions for the centripetal force

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

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

is always perpendicular to the path and points to the center of curvature, considering

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

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

Figure half-dozen.20 The frictional force supplies the centripetal strength and is numerically equal to information technology. Centripetal forcefulness is perpendicular to velocity and causes compatible circular movement. The larger the

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

produces a smaller r′.

Example

What Coefficient of Friction Do Cars Demand on a Flat Bend?

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

Figure 6.21 This car on level ground is moving abroad and turning to the left. The centripetal force causing the car to turn in a circular path is due to friction between the tires and the road. A minimum coefficient of friction is needed, or the car will motion in a larger-radius curve and leave the roadway.

Strategy

1. Nosotros know that

   

   Thus,

   

2. (Figure) shows the forces acting on the machine on an unbanked (level ground) curve. Friction is to the left, keeping the automobile from slipping, and because it is the merely horizontal forcefulness acting on the car, the friction is the centripetal force in this instance. We know that the maximum static friction (at which the tires roll but practice non sideslip) is

   

   where

   

   is the static coefficient of friction and N is the normal force. The normal force equals the car's weight on level footing, so

   

   Thus the centripetal strength in this situation is

   

   Now we have a relationship between centripetal force and the coefficient of friction. Using the equation

   

   we obtain

   

   We solve this for

   

   noting that mass cancels, and obtain

   

   Substituting the knowns,

   

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

Significance

The coefficient of friction found in (Effigy)(b) is much smaller than is typically found betwixt tires and roads. The car still negotiates the bend if the coefficient is greater than 0.13, considering static friction is a responsive force, able to presume a value less than but no more

A higher coefficient would also let the motorcar to negotiate the curve at a college speed, but if the coefficient of friction is less, the safe speed would be less than 25 k/s. Note that mass cancels, implying that, in this example, information technology does non thing how heavily loaded the motorcar is to negotiate the turn. Mass cancels because friction is assumed proportional to the normal forcefulness, which in turn is proportional to mass. If the surface of the road were banked, the normal force would be less, as discussed adjacent.

Check Your Understanding

A car moving at 96.8 km/h travels around a circular curve of radius 182.9 m on a apartment land route. What must exist the minimum coefficient of static friction to go along the car from slipping?

[reveal-answer q="694795″]Show Solution[/reveal-answer]
[subconscious-answer a="694795″]0.40[/subconscious-answer]

Banked Curves

Let united states of america now consider banked curves, where the gradient of the road helps yous negotiate the bend ((Effigy)). The greater the bending

, the faster yous tin take the curve. Race tracks for bikes too every bit cars, for example, often have steeply banked curves. In an "ideally banked curve," the angle

is such that you can negotiate the curve at a sure speed without the aid of friction between the tires and the road. We will derive an expression for

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

Figure 6.22 The car on this banked bend is moving away and turning to the left.

For ideal banking, the net external forcefulness 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 strength and the weight of the car, respectively. In cases in which forces are not parallel, it is most user-friendly to consider components along perpendicular axes—in this example, the vertical and horizontal directions.

(Figure) shows a costless-body diagram for a automobile on a frictionless banked curve. If the bending

is ideal for the speed and radius, then the net external forcefulness equals the necessary centripetal strength. The only 2 external forces acting on the car are its weight

and the normal force of the road

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

Considering this is the crucial strength and information technology is horizontal, nosotros use a coordinate system with vertical and horizontal axes. Only the normal force has a horizontal component, so this must equal the centripetal strength, that is,

Because the machine does not leave the surface of the route, the net vertical forcefulness must exist nothing, meaning that the vertical components of the 2 external forces must be equal in magnitude and opposite in direction. From (Effigy), we see that the vertical component of the normal force is

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

At present nosotros can combine these two equations to eliminate North and get an expression for

, as desired. Solving the second equation for

and substituting this into the first yields

Taking the inverse tangent gives

This expression tin be understood by because how

depends on five and r. A large

is 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 bend at greater or lower speed than if the curve were frictionless. Note that

does non depend on the mass of the vehicle.

Case

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

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

should be driven if the route were frictionless.

Strategy

We first annotation that all terms in the expression for the ideal angle of a banked bend except for speed are known; thus, we demand just rearrange it then that speed appears on the left-hand side and then substitute known quantities.

Solution

Starting with

we get

Noting that

nosotros obtain

Significance

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

Airplanes likewise make turns by banking. The elevator strength, due to the force of the air on the wing, acts at right angles to the wing. When the airplane banks, the airplane pilot is obtaining greater lift than necessary for level flight. The vertical component of elevator balances the airplane's weight, and the horizontal component accelerates the airplane. The banking angle shown in (Figure) is given by

. Nosotros analyze the forces in the aforementioned way we care for the instance of the car rounding a banked bend.

Effigy 6.23 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 banking bending is given past

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

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

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

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

What do taking off in a jet airplane, turning a corner in a car, riding a merry-go-round, and the circular motion of a tropical cyclone have in common? Each exhibits inertial forces—forces that merely seem to arise from motion, considering the observer'south frame of reference is accelerating or rotating. When taking off in a jet, most people would concur it feels as if you are existence pushed back into the seat as the airplane accelerates down the track. Yet a physicist would say that you lot tend to remain stationary while the seat pushes forrard on y'all. An even more mutual feel occurs when you make a tight bend in your car—say, to the right ((Figure)). You lot feel as if you are thrown (that is, forced) toward the left relative to the automobile. Once more, a physicist would say that y'all are going in a straight line (recall Newton'southward first police) but the motorcar moves to the right, not that you are experiencing a strength from the left.

Figure half dozen.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 use of the car as a frame of reference. (b) In Earth's frame of reference, the driver moves in a straight line, obeying Newton's starting time law, and the motorcar moves to the right. There is no force to the left on the driver relative to Earth. Instead, there is a strength to the right on the car to get in turn.

Nosotros can reconcile these points of view by examining the frames of reference used. Let the states concentrate on people in a machine. Passengers instinctively utilize the car as a frame of reference, whereas a physicist might use Earth. The physicist might make this option because Globe is nearly an inertial frame of reference, in which all forces take an identifiable physical origin. In such a frame of reference, Newton'due south laws of move take the form given in Newton's Laws of Motion. The car is a noninertial frame of reference because it is accelerated to the side. The force to the left sensed past auto passengers is an inertial forcefulness 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 car, as well every bit the driver, is actually accelerating to the right. This inertial force is said to be an inertial force because it does not take a concrete origin, such as gravity.

A physicist volition choose whatever reference frame is nigh convenient for the situation beingness analyzed. In that location is no problem to a physicist in including inertial forces and Newton's 2nd police, 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 do so. Dissimilar frames of reference must be considered in discussing the motion of an astronaut in a spacecraft traveling at speeds nearly the speed of light, equally you will appreciate in the report of the special theory of relativity.

Let u.s.a. now accept a mental ride on a merry-get-circular—specifically, a speedily rotating playground merry-go-round ((Figure)). Y'all take the merry-go-round to exist your frame of reference because you rotate together. When rotating in that noninertial frame of reference, you feel an inertial force that tends to throw you off; this is frequently referred to equally a centrifugal force (non to be confused with centripetal force). Centrifugal forcefulness is a commonly used term, merely it does not actually be. You lot must hang on tightly to counteract your inertia (which people often refer to as centrifugal force). In Earth's frame of reference, in that location is no force trying to throw you off; we emphasize that centrifugal force is a fiction. You must hang on to make yourself go in a circle because otherwise you would go in a straight line, right off the merry-go-round, in keeping with Newton'southward get-go law. Simply the forcefulness you exert acts toward the center of the circumvolve.

Figure 6.25 (a) A passenger on a merry-go-round feels as if he is existence thrown off. This inertial force is sometimes mistakenly called the centrifugal strength in an try to explain the passenger's motility in the rotating frame of reference. (b) In an inertial frame of reference and according to Newton's laws, it is his inertia that carries him off (the unshaded rider has

and heads in a direct line). A force,

, is needed to crusade a circular path.

This inertial result, carrying you away from the center of rotation if there is no centripetal force to crusade circular motility, is put to expert use in centrifuges ((Figure)). A centrifuge spins a sample very quickly, every bit mentioned earlier in this affiliate. Viewed from the rotating frame of reference, the inertial force throws particles outward, hastening their sedimentation. The greater the angular velocity, the greater the centrifugal force. Just what really 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 circular path by a centripetal force.

Figure half-dozen.26 Centrifuges employ inertia to perform their task. Particles in the fluid sediment settle out considering their inertia carries them away from the center of rotation. The large angular velocity of the centrifuge quickens the sedimentation. Ultimately, the particles come into contact with the test tube walls, which then supply the centripetal force needed to brand them move in a circle of constant radius.

Permit u.s.a. now consider what happens if something moves in a rotating frame of reference. For example, what if you slide a ball straight abroad from the center of the merry-get-round, equally shown in (Effigy)? The ball follows a straight path relative to Earth (assuming negligible friction) and a path curved to the right on the merry-go-circular's surface. A person continuing next to the merry-go-round sees the ball moving straight and the merry-become-circular rotating underneath it. In the merry-go-circular'southward frame of reference, we explain the apparent curve to the right by using an inertial forcefulness, called the Coriolis force, which causes the brawl to bend to the correct. The Coriolis strength tin can be used by anyone in that frame of reference to explain why objects follow curved paths and allows united states to utilize Newton's laws in noninertial frames of reference.

Figure six.27 Looking down on the counterclockwise rotation of a merry-go-round, we run into that a ball slid direct toward the edge follows a path curved to the right. The person slides the ball toward point B, starting at betoken A. Both points rotate to the shaded positions (A' and B') shown in the fourth dimension that the ball follows the curved path in the rotating frame and a straight path in Earth'southward frame.

Up until now, we have considered Globe to be an inertial frame of reference with trivial or no worry about effects due to its rotation. Yet such furnishings do exist—in the rotation of weather systems, for case. About consequences of Earth's rotation tin be qualitatively understood by analogy with the merry-get-round. Viewed from above the Due north Pole, World rotates counterclockwise, equally does the merry-go-round in (Figure). Equally on the merry-go-circular, any move in Earth'due south Northern Hemisphere experiences a Coriolis force to the right. Just the opposite occurs in the Southern Hemisphere; there, the force is to the left. Because Earth'due south angular velocity is small, the Coriolis force is usually negligible, simply for big-calibration 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 management. (The terms hurricane, typhoon, and tropical storm are regionally specific names for cyclones, which are storm systems characterized by low pressure centers, strong winds, and heavy rains.) (Effigy) helps show how these rotations have place. Air flows toward any region of low pressure, and tropical cyclones incorporate particularly low pressures. Thus winds flow toward the center of a tropical cyclone or a depression-pressure weather arrangement at the surface. In the Northern Hemisphere, these inward winds are deflected to the right, equally shown in the figure, producing a counterclockwise apportionment at the surface for low-pressure zones of any type. Depression pressure at the surface is associated with rising air, which besides produces cooling and cloud formation, making depression-force per unit area patterns quite visible from space. Conversely, current of air circulation effectually loftier-pressure level zones is clockwise in the Southern Hemisphere just is less visible considering loftier force per unit area is associated with sinking air, producing articulate skies.

Figure 6.28 (a) The counterclockwise rotation of this Northern Hemisphere hurricane is a major consequence of the Coriolis force. (b) Without the Coriolis force, air would menstruum straight into a low-pressure zone, such equally that found in tropical cyclones. (c) The Coriolis force deflects the winds to the right, producing a counterclockwise rotation. (d) Wind flowing abroad from a high-pressure zone is also deflected to the right, producing a clockwise rotation. (due east) The reverse management of rotation is produced by 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 ball on a merry-go-round can but likewise be explained by inertia and the rotation of the system underneath. When noninertial frames are used, inertial forces, such equally the Coriolis force, must exist invented to explicate the curved path. There is no identifiable physical source for these inertial forces. In an inertial frame, inertia explains the path, and no force is found to be without an identifiable source. Either view allows usa to describe nature, only a view in an inertial frame is the simplest in the sense that all forces take origins and explanations.

Summary

• Centripetal force

  

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

  

• Rotating and accelerated frames of reference are noninertial. Inertial forces, such as the Coriolis forcefulness, are needed to explicate movement in such frames.

Conceptual Questions

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

Define centripetal strength. Tin can whatever type of forcefulness (for example, tension, gravitational force, friction, so on) be a centripetal strength? Tin any combination of forces exist a centripetal forcefulness?

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

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

Centripetal force is defined as any net forcefulness causing uniform circular motion. The centripetal force is not a new kind of forcefulness. The label "centripetal" refers to whatsoever forcefulness that keeps something turning in a circle. That force could be tension, gravity, friction, electrical allure, the normal forcefulness, or any other force. Whatsoever combination of these could exist the source of centripetal force, for instance, the centripetal strength at the acme of the path of a tetherball swung through a vertical circle is the result of both tension and gravity.

[/hidden-answer]

If centripetal strength is directed toward the center, why practise yous experience that you are 'thrown' away from the center every bit a machine goes around a curve? Explain.

Race machine drivers routinely cut corners, as shown beneath (Path two). Explain how this allows the curve to exist taken at the greatest speed.

[reveal-answer q="329939″]Show Solution[/reveal-reply]
[hidden-answer a="329939″]The driver who cuts the corner (on Path 2) has a more gradual curve, with a larger radius. That one will be the meliorate racing line. If the driver goes too fast effectually a corner using a racing line, he will still slide off the runway; the central 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 forcefulness:

where v is speed and r is the radius of curvature. So by decreasing the curvature (1/r) of the path that the auto takes, we reduce the amount of forcefulness the tires have to exert on the road, meaning nosotros can now increase the speed, v. Looking at this from the signal of view of the driver on Path i, we can reason this way: the sharper the plow, the smaller the turning circle; the smaller the turning circle, the larger is the required centripetal force. If this centripetal force is not exerted, the result is a sideslip.[/hidden-answer]

Many amusement parks accept rides that brand vertical loops like the one shown below. For prophylactic, the cars are attached to the rails in such a way that they cannot fall off. If the car goes over the superlative at just the right speed, gravity lonely will supply the centripetal strength. What other strength 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?

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

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

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

The barrel of the dryer provides a centripetal force on the dress (including the h2o droplets) to proceed them moving in a round path. As a water droplet comes to i of the holes in the butt, information technology volition movement in a path tangent to the circle.

[/hidden-answer]

Equally a skater forms a circle, what force is responsible for making his turn? Utilize a gratuitous-trunk diagram in your respond.

Suppose a child is riding on a merry-go-round at a altitude about halfway between its centre and edge. She has a lunch box resting on wax paper, so that at that place is very piddling friction between it and the merry-go-round. Which path shown below volition the tiffin box accept when 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? Explicate your reply.

[reveal-reply q="60053″]Prove Solution[/reveal-answer]
[hidden-answer a="60053″]If there is no friction, so there is no centripetal forcefulness. This means that the lunch box will move along a path tangent to the circle, and thus follows path B. The dust trail will be straight. This is a result of Newton's first law of motion.[/hidden-answer]

Exercise you feel yourself thrown to either side when you negotiate a curve that is ideally banked for your automobile's speed? What is the direction of the force exerted on you past the car seat?

Suppose a mass is moving in a circular path on a frictionless table as shown beneath. In World's frame of reference, at that place is no centrifugal force pulling the mass away from the middle of rotation, however there is a strength stretching the string attaching the mass to the smash. Using concepts related to centripetal force and Newton'south 3rd law, explain what strength stretches the string, identifying its physical origin.

[reveal-answer q="965193″]Show Solution[/reveal-answer]
[hidden-answer a="965193″]There must be a centripetal force to maintain the circular motility; this is provided by the blast at the center. Newton'south third law explains the phenomenon. The activity force is the forcefulness of the cord on the mass; the reaction force is the strength of the mass on the string. This reaction force causes the string to stretch.[/hidden-reply]

When a toilet is flushed or a sink is tuckered, the h2o (and other fabric) begins to rotate near the drain on the manner down. Assuming no initial rotation and a menses initially directly straight toward the drain, explain what causes the rotation and which direction it has in the Northern Hemisphere. (Annotation that this is a small-scale effect and in most toilets the rotation is acquired past directional h2o jets.) Would the direction of rotation reverse if water were forced up the bleed?

A car rounds a bend and encounters a patch of ice with a very low coefficient of kinetic fiction. The machine slides off the route. Depict the path of the car as it leaves the road.

[reveal-answer q="fs-id1165039077662″]Show 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 auto will obey Newton's showtime constabulary and go off the road in a straight line path, tangent to the bend. A mutual misconception is that the machine will follow a curved path off the route.

[/hidden-answer]

In one amusement park ride, riders enter a large vertical barrel and stand confronting the wall on its horizontal floor. The butt is spun upwardly and the flooring drops abroad. Riders feel as if they are pinned to the wall by a forcefulness something like the gravitational strength. This is an inertial strength sensed and used past the riders to explicate events in the rotating frame of reference of the barrel. Explain in an inertial frame of reference (Earth is nearly ane) what pins the riders to the wall, and identify all forces acting on them.

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

. Who do you agree with and why?

[reveal-answer 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, fifty-fifty though gravity is weaker at the distance of the satellite, and m is non

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

A nonrotating frame of reference placed at the heart of the Sunday is very nearly an inertial one. Why is it not exactly an inertial frame?

Issues

(a) A 22.0-kg child is riding a playground merry-go-round that is rotating at xl.0 rev/min. What centripetal forcefulness is exerted if he is ane.25 m from its center? (b) What centripetal force is exerted if the merry-go-round rotates at 3.00 rev/min and he is eight.00 one thousand from its center? (c) Compare each force with his weight.

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

[hidden-reply a="fs-id1165039026811″]

a. 483 Northward; b. 17.4 N; c. 2.24, 0.0807

[/hidden-answer]

Calculate the centripetal strength on the finish of a 100-thou (radius) wind turbine bract that is rotating at 0.5 rev/s. Assume the mass is four kg.

What is the ideal banking angle for a gentle plow of 1.20-km radius on a highway with a 105 km/h speed limit (nearly 65 mi/h), assuming everyone travels at the limit?

[reveal-reply q="fs-id1165039344901″]Prove Solution[/reveal-answer]

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

[/hidden-answer]

What is the ideal speed to take a 100.0-m-radius bend banked at a

angle?

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

and taken at thirty.0 m/s, assuming information technology is ideally banked? (b) Calculate the centripetal acceleration. (c) Does this acceleration seem large to you?

[reveal-answer q="fs-id1165039104209″]Show Solution[/reveal-respond]

[subconscious-respond a="fs-id1165039104209″]

a. 24.half dozen m; b.

c. 3.73 times g
[/hidden-respond]

Office of riding a bicycle involves leaning at the correct angle when making a turn, as seen below. To be stable, the force exerted by the footing must be on a line going through the center of gravity. The strength on the bicycle wheel tin exist 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) Prove that

(as divers as shown) is related to the speed v and radius of curvature r of the turn in the same way every bit for an ideally banked roadway—that is,

(b) Calculate

for a 12.0-m/southward turn of radius 30.0 m (as in a race).

If a automobile takes a banked bend at less than the ideal speed, friction is needed to go on information technology from sliding toward the inside of the curve (a trouble on icy mountain roads). (a) Calculate the ideal speed to take a 100.0 m radius curve banked at

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

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

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

a. 16.two g/s; b. 0.234

[/hidden-respond]

Modernistic roller coasters have vertical loops like the one shown here. The radius of curvature is smaller at the top than on the sides so that the downwards centripetal acceleration at the top will be greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats. (a) What is the speed of the roller coaster at the top of the loop if the radius of curvature in that location is fifteen.0 1000 and the downwards acceleration of the auto is 1.50 grand? (b) How high in a higher place the elevation of the loop must the roller coaster beginning from remainder, assuming negligible friction? (c) If it really starts 5.00 m college than your reply to (b), how much energy did it lose to friction? Its mass is

.

A child of mass xl.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 car is 10.0 chiliad/s, and at point B, the speed is 10.5 chiliad/due south. Assume the child is non belongings on and does non wearable a seat chugalug. (a) What is the forcefulness of the car seat on the kid at bespeak A? (b) What is the force of the motorcar seat on the child at point B? (c) What minimum speed is required to proceed the child in his seat at betoken A?

[reveal-reply q="484990″]Show Solution[/reveal-answer]
[hidden-reply a="484990″]a. 179 N; b. 290 N; c. 8.3 m/south[/subconscious-respond]

In the simple Bohr model of the ground state of the hydrogen atom, the electron travels in a circular orbit around a fixed proton. The radius of the orbit is

and the speed of the electron is

The mass of an electron is

. What is the force on the electron?

Railroad tracks follow a round curve of radius 500.0 chiliad and are banked at an angle of

. For trains of what speed are these tracks designed?

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

[hidden-answer a="fs-id1165039111532″]

20.7 k/s

[/subconscious-answer]

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

that motility effectually the accelerator at

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

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

A car rounds an unbanked bend of radius 65 yard. If the coefficient of static friction between the road and car is 0.70, what is the maximum speed at which the car traverse the curve without slipping?

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

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

21 yard/south

[/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

curve 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 forcefulness

inertial strength causing the credible deflection of moving objects when viewed in a rotating frame of reference

platonic cyberbanking

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

inertial force

forcefulness 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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