People are wild about entertainment parks. Each solar day, we flock by the millions to the nearest park, paying a sizable hunk of money to expect in long lines for a short threescore-second ride on our favorite roller coaster. The thought prompts one to consider what is it nigh a roller coaster ride that provides such widespread excitement among then many of us and such dreadful fear in the rest? Is our excitement nigh coasters due to their loftier speeds? Absolutely not! In fact, it would be foolish to spend so much time and money to ride a selection of roller coasters if it were for reasons of speed. It is more than likely that most of us sustain higher speeds on our ride along the interstate highway on the style to the entertainment park than we practise once we enter the park. The thrill of roller coasters is not due to their speed, but rather due to their accelerations and to the feelings of weightlessness and weightiness that they produce. Roller coasters thrill u.s.a. because of their power to accelerate usa downward one moment and upwards the next; leftwards one moment and rightwards the next. Roller coasters are near dispatch; that's what makes them thrilling. And in this part of Lesson 2, we will focus on the centripetal acceleration experienced by riders within the circular-shaped sections of a roller coaster rails. These sections include the clothoid loops (that we will approximate as a circle), the sharp 180-degree banked turns, and the small dips and hills found along otherwise straight sections of the track.
The Physics of Roller Coaster Loops
The almost obvious section on a roller coaster where centripetal acceleration occurs is within the then-called clothoid loops . Roller coaster loops assume a tear-dropped shape that is geometrically referred to every bit a clothoid. A clothoid is a section of a screw in which the radius is constantly changing. Unlike a round loop in which the radius is a constant value, the radius at the bottom of a clothoid loop is much larger than the radius at the pinnacle of the clothoid loop. A mere inspection of a clothoid reveals that the amount of curvature at the bottom of the loop is less than the corporeality of curv
ature at the top of the loop. To simplify our assay of the physics of clothoid loops, nosotros will approximate a clothoid loop every bit being a serial of overlapping or adjoining circular sections. The radius of these circular sections is decreasing equally one approaches the meridian of the loop. Furthermore, we volition limit our analysis to two points on the clothoid loop - the top of the loop and the bottom of the loop. For this reason, our analysis will focus on the two circles that can be matched to the curvature of these two sections of the clothoid. The diagram at the right shows a clothoid loop with two circles of unlike radius inscribed into the height and the bottom of the loop. Note that the radius at the bottom of the loop is significantly larger than the radius at the top of the loop.
As a roller coaster rider travels through a clothoid loop, she experiences an dispatch due to both a change in speed and a change in direction. A rightward moving passenger gradually becomes an upwards moving passenger, so a leftward moving rider, then a downward moving rider, before finally condign a rightward-moving rider once once again. There is a continuous change in the direction of the rider equally she moves through the clothoid loop. And as learned in Lesson one, a change in direction is one feature of an accelerating object. In addition to changing directions, the rider also changes speed. Every bit the rider begins to ascend (climb upward) the loop, she begins to slow down. As energy principles would suggest, an increase in superlative (and in turn an increase in potential energy) results in a decrease in kinetic free energy and speed. And conversely, a decrease in superlative (and in turn a decrease in potential energy) results in an increase in kinetic energy and speed. And then the passenger experiences the greatest speeds at the lesser of the loop - both upon entering and leaving the loop - and the everyman speeds at the elevation of the loop.
This change in speed every bit the rider moves through the loop is the second aspect of the dispatch that a rider experiences. For a passenger moving through a round loop with a constant speed, the acceleration can be described as existence centripetal or towards the center of the circle. In the case of a rider moving through a noncircular loop at non-constant speed, the acceleration of the rider has two components. There is a component that is directed towards the middle of the circle ( ac ) and attributes itself to the direction change; and there is a component that is directed tangent ( at ) to the rail (either in the opposite or in the aforementioned direction every bit the car's direction of motion) and attributes itself to the car'southward change in speed. This tangential component would be directed opposite the management of the car'due south motion as its speed decreases (on the ascent towards the top) and in the same direction every bit the car'southward move as its speed increases (on the descent from the elevation). At the very tiptop and the very lesser of the loop, the acceleration is primarily directed towards the center of the circle. At the acme, this would be in the downward direction and at the bottom of the loop it would be in the upward management.
Force Assay of a Coaster Loop
We learned in Lesson 1 that the inward acceleration of an object is caused by an inwards net strength. Circular motion (or merely motion along a curved path) requires an inwards component of internet force. If all the forces that deed upon the object were added together every bit vectors, then the cyberspace force would be directed inward. Neglecting friction and air resistance, a roller coaster car volition experience 2 forces: the force of gravity (Fgrav) and the normal strength (Fnorm). The normal forcefulness is directed in a direction perpendicular to 
the track and the gravitational force is e'er directed downwards. We will concern ourselves with the relative magnitude and direction of these ii forces for the pinnacle and the bottom of the loop. At the bottom of the loop, the track pushes upward upon the auto with a normal strength. However, at the peak of the loop the normal forcefulness is directed downwards; since the track (the supplier of the normal force) is above the car, information technology pushes downward upon the automobile. The complimentary-body diagrams for these two positions are shown in the diagrams at the right.
The magnitude of the force of gravity acting upon the rider (or car) can easily be establish using the equation Fgrav = m•thou where one thousand = acceleration of gravity (9.8 1000/s2). The magnitude of the normal force depends on two factors - the speed of the car, the radius of the loop and the mass of the passenger. As depicted in the gratuitous trunk diagram, the magnitude of Fnorm is always greater at the bottom of the loop than it is at the top. The normal strength must e'er be of the advisable size to combine with the Fgrav in such a way to produce the required in or centripetal internet force. At the bottom of the loop, the Fgrav points outwards abroad from the center of the loop. The normal force must exist sufficiently large to overcome this Fgrav and supply some excess strength to result in a net inward force. In a sense, Fgrav and Fnorm are in a tug-of-war; and Fnorm must win by an amount equal to the net forcefulness. At the meridian of the loop, both Fgrav and Fnorm are directed inwards. The Fgrav is constitute in the usual way (using the equation Fgrav = thousand•m). Once more than the Fnorm must provide sufficient force to produce the required inward or centripetal internet force.
Before in Lesson 2, the use of Newton's second law and gratis-trunk diagrams to solve circular motion diagrams was illustrated. It was emphasized at that fourth dimension that whatsoever given physical state of affairs could be analyzed in terms of the individual forces that are interim upon an object. These individual forces must add up as vectors to the net force. Furthermore, the net force must exist equal to the mass times the acceleration. The process of conducting a force analysis of a physical situation was first introduced in Unit 2 of The Physics Classroom. Now we will investigate the use of these central principles in the analysis of situations involving the move of objects in circles. We volition utilize the basic trouble-solving approach that was introduced earlier in Lesson 2. This approach can be summarized as follows.
Suggested Method of Solving Circular Motion Bug - From the verbal clarification of the concrete situation, construct a complimentary-body diagram. Represent each force past a vector pointer and label the forces according to blazon.
- Identify the given and the unknown information (express in terms of variables such as m= , a= , v= , etc.).
- If whatever of the private forces are directed at angles to the horizontal and vertical, then use vector principles to resolve such forces into horizontal and vertical components.
- Determine the magnitude of whatever known forces and label on the costless-body diagram.
(For example, if the mass is given, then the Fgrav tin can exist determined. And as some other example, if there is no vertical acceleration, so it is known that the vertical forces or force components residuum, allowing for the possible decision of 1 or more than of the individual forces in the vertical direction.) - Use circular motion equations to decide any unknown data.
(For example, if the speed and the radius are known, then the acceleration tin can be determined. And equally another case, if the period and radius are known, then the acceleration can be determined.) - Employ the remaining information to solve for the requested data.
- If the problem requests the value of an individual force, then use the kinematic information (R, T and 5) to determine the dispatch and the Fnet; then use the free-body diagram to solve for the individual force value.
- If the problem requests the value of the speed or radius, then use the values of the private forces to determine the net strength and dispatch; then use the acceleration to determine the value of the speed or radius.
Combine a strength analysis with the to a higher place method to solve the post-obit roller coaster problem.
Sample Roller Coaster Problem Anna Litical is riding on The Demon at Keen America. Anna experiences a downward acceleration of 15.half-dozen chiliad/south2 at the top of the loop and an upward acceleration of 26.iii grand/s2 at the bottom of the loop. Use Newton's 2nd law to make up one's mind the normal strength acting upon Anna's 864 kg roller coaster machine.
Steps 1 and 2 involve the construction of a free body diagram and the identification of known and unknown quantities. This is shown in beneath.
| | Given Info: m = 864 kg atop = 15.6 m/s2 , downwards bottom = 26.iii m/s2 , up Find: Fnorm at top and lesser |
Pace 3 of the suggested method would non apply to this problem since in that location are no forces directed "at angles" (that is, all the forces are either horizontally or vertically directed). Step 4 of the suggested method involves the determination of whatsoever known forces. In this case, the forcefulness of gravity tin can be determined from the equation Fgrav = m • g . Using a g value of ix.8 m/s2, the strength of gravity acting upon the 864-kg car is approximately 8467 N. Step v of the suggested method would be used if the dispatch were not given. In this instance, the acceleration is known. If the acceleration were not known, and then it would accept to exist calculated from speed and radius information.
Step 6 of the suggested method involves the determination of an individual strength - the normal force. This volition involve a two-step process: first the net strength (magnitude and direction) must be determined; and so the net strength must exist used with the free body diagram to determine the normal force. This ii-step procedure is shown below for the tiptop and the bottom of the loop.
| Bottom of Loop Fnet = g * a Fnet = (864 kg) * (26.iii grand/sii, up) Fnet = 22 723 Due north, upwards From FBD:  Fnorm must be greater than the Fgrav by 22723 N in order to supply a net upwards force of 22723 North. Thus, Fnorm = Fgrav + Fcyberspace Fnorm = 31190 N | Top of Loop Finternet = 1000 * a Fnet = (864 kg) * (15.6 one thousand/due southtwo, downward) Fnet = 13478 Due north, downward From FBD:  Fnorm and Fgrav together must combine together (i.e., add upward) to supply the required inwards cyberspace force of 13478 North. Thus, Fnorm = Fnet - Fgrav Fnorm = 5011 North |
Sensations of Weightlessness
Observe that the normal force is greater at the bottom of the loop than it is at the top of the loop. This becomes a reasonable fact when round motion principles are considered. At all points along the loop - which we will refer to equally round in shape - there must be some in component of cyberspace force. When at the elevation of the loop, the gravitational strength is directed inwards (downward) and so at that place is less of a need for a normal force in order to see the net centripetal forcefulness requirement. When at the lesser of the loop, the gravitational force is directed outwards (downward) and so now there is a need for a big upwards normal force in society to encounter the centripetal force requirement. This principle is oftentimes demonstrated in a physics class using a bucket of water tied to a string. The water is spun in a vertical circle. Equally the h2o traces out its round path, the tension in the cord is continuously irresolute. The tension force in this demonstration is analogous to the normal strength for a roller coaster rider. At the top of the vertical circle, the tension force is very small; and at the bottom of the vertical circumvolve, the tension force is very large. (You might attempt this action yourself exterior with a small plastic bucket half-filled with water. Requite extra circumspection to stay articulate of all people, windows, trees and overhead ability lines. Echo enough cycles to discover the noticeable departure in tension strength when the bucket is at the top and the bottom of the circle.)
If you accept ever been on a roller coaster ride and traveled through a loop, then yous have likely experienced this small normal force at the top of 
the loop and the large normal force at the bottom of the loop. The normal forcefulness provides a feel for a person'due south weight. (Every bit will be discussed later in Lesson four, we can never experience our weight; we tin simply feel other forces that act as a event of contact with other objects.) The more than you lot weigh, the more normal forcefulness that y'all will experience when at rest in your seat. But if you board a roller coaster ride and accelerate through circles (or clothoid loops), then you will experience a normal strength that is constantly changing and dissimilar from that which you are accepted to. This normal force provides a sensation or feeling of weightlessness or weightiness. When at the top of the loop, a rider will feel partially weightless if the normal forces become less than the person'south weight. And at the bottom of the loop, a rider volition experience very "weighty" due to the increased normal forces. It is of import to realize that the strength of gravity and the weight of your body are non changing. Only the magnitude of the supporting normal force is changing! (The miracle of weightlessness will exist discussed in much more detail later in Lesson 4.)
There is some interesting history (and physics) behind the gradual usage of clothoid loops in roller coaster rides. In the early days of roller coaster loops, circular loops were used. There were a multifariousness of problems, some of which resulted in fatalities, as the result of the employ of these circular loops. Coaster cars entering round loops at high speeds encountered excessive normal forces that were capable of causing whiplash and cleaved bones. Efforts to correct the problem by lowering entry speeds resulted in the inability of cars to make it around the entire loop without falling out of the loop when reaching the top. The decrease in speeds as the cars ascended the large circular loop resulted in coaster cars turning into projectile cars (a state of affairs known to be non good for business organization). The solution to the problem involved using low entry speeds and a loop with a sharper curvature at the top than at the bottom. Since clothoid loops take a continually changing radius, the radius is big at the bottom of the loop and shortened at the elevation of the loop. The effect is that coaster cars tin enter the loops at loftier speeds; however due to the large radius, the normal forces do non exceed 3.v Thou's. At the pinnacle of the loop, the radius is small thus allowing a lower speed car to still maintain contact with the runway and successfully brand it through the loop. The clothoid loop is a testimony to an engineer's application of the centripetal dispatch equation - a = v2/R. Now that's physics for amend living!
Physics of Coaster Dips and Hills
The above word and force analysis applies to the circular-like motility of a roller coaster car in a clothoid loop. The 2nd section along a roller coaster track where circular motility is experienced is forth the minor dips and hills. These sections of track are frequently plant almost the end of a roller coaster ride and involve a serial of minor hills followed by a abrupt drib. Riders often feel heavy as they arise the hill (forth regions A and Eastward in the diagram below). Then near the crest of the hill (regions B and F), their up motion makes them feel as though they will fly out of the car; often times, it is but the safety chugalug that prevents such a mishap. Every bit the motorcar begins to descend the precipitous drop, riders are momentarily in a land of gratuitous autumn (along regions C and G in the diagram beneath). And finally as they accomplish the bottom of the sharp dip (regions D and H), there is a large upwards force that slows their downwards move. The cycle is often repeated mercilessly, churning the riders' stomachs and mixing the afternoon's cotton wool candy into a slurry of ... . These small dips and hills combine the physics of round motion with the physics of projectiles in order to produce the ultimate thrill of dispatch - speedily changing magnitudes and directions of acceleration. The diagram below shows the various directions of accelerations that riders would experience forth these hills and dips.
Strength Analysis of Coaster Hills
At diverse locations along these hills and dips, riders are momentarily traveling along a circular shaped arc. The arc is part of a circumvolve - these circles have been inscribed on the to a higher place diagram in blue. In each of these regions there is an inward component of acceleration (as depicted past the black arrows). This inward acceleration demands that there besides exist a force directed towards the center of the circumvolve. In region A, the centripetal strength is supplied by the track pushing normal to the track surface. Along region B, the centripetal strength is supplied past the force of gravity and possibly fifty-fifty the safety machinery/bar. At especially loftier speeds, a safety bar must supply fifty-fifty extra downwardly force in order to pull the riders downward and supply the remaining centripetal force required for circular motility. In that location are also wheels on the car that are normally tucked nether the track and pulled downward by the rails. Along region D, the centripetal forcefulness is one time more supplied past the normal force of the track pushing upwards upon the car.
The magnitude of the normal forces along these diverse regions is dependent upon how sharply the track is curved along that region (the radius of the circumvolve) and the speed of the car. These two variables touch on the acceleration according to the equation
a = v2 / R and in plow affect the net force. As suggested past the equation, a large speed results in a large acceleration and thus increases the demand for a large net forcefulness. And a large radius (gradually curved) results in a pocket-sized acceleration and thus lessens the need for a big net force. The human relationship between speed, radius, acceleration, mass and cyberspace force can be used to make up one's mind the magnitude of the seat force (i.eastward., normal forcefulness) upon a roller coaster rider at diverse sections of the track. The sample trouble below illustrates these relationships. In the process of solving the problem, the aforementioned problem-solving strategy enumerated above will be utilized.
Sample Roller Coaster Trouble Anna Litical is riding on The American Eagle at Great America. Anna is moving at xviii.9 m/south over the top of a hill that has a radius of curvature of 12.7 m. Employ Newton's second law to determine the magnitude of the applied force of the rails pulling down upon Anna'southward 621 kg roller coaster auto.
Steps 1 and 2 involve the construction of a free body diagram and the identification of known and unknown quantities. This is shown in below.
 | Given Info: | k = 621 kg | v = 18.9 m/s | | R = 12.7 m | | Find: Fapp at top of hill |
Step 3 of the suggested method would non apply to this trouble since there are no forces directed "at angles" (that is, all the forces are either horizontally or vertically directed). Stride 4 of the suggested method involves the conclusion of whatsoever known forces. In this case, the force of gravity tin can exist determined from the equation Fgrav = m * chiliad . So the force of gravity interim upon the 621-kg car is approximately 6086 N. Pace 5 of the suggested method involves the adding of the acceleration from the given values of the speed and the radius. Using the equation given in Lesson 1, the acceleration can exist calculated as follows
a = vtwo / R a = (18.ix m/due south)ii / (12.7 m)
a = 28.1 chiliad/s 2
Step 6 of the suggested method involves the determination of an individual force - the applied force. This will involve a 2-step process: first the net force (magnitude and direction) must be determined; so the cyberspace force must be used with the free trunk diagram to make up one's mind the applied strength. This ii-step process is shown beneath.
| Finternet = chiliad • a Fcyberspace = (621 kg) • (28.ane m/s2, downward) Fnet = 17467 Due north, downwards As shown in FBD at right: Fapp = Finternet - Fgrav Fnorm = 11381 North |  Fapp and Fgrav must combine together (i.e., add together up) to supply the required downwards net forcefulness of 17467 N. |
This same method could be applied for whatever region of the track in which roller coaster riders momentarily experience circular motion.
We Would Like to Advise ...
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Check Your Understanding
one. Anna Litical is riding on The Shock Wave at Neat America. Anna experiences a downward acceleration of 12.five thousand/due south2 at the top of the loop and an upward acceleration of 24.0 yard/due southtwo at the bottom of the loop. Use Newton's second police to determine the normal force acting upon Anna's fifty-kg torso at the top and at the bottom of the loop.
2. Noah Formula is riding a roller coaster and encounters a loop. Noah is traveling 6 m/s at the acme of the loop and 18.0 m/s at the lesser of the loop. The top of the loop has a radius of curvature of iii.2 m and the bottom of the loop has a radius of curvature of 16.0 m. Use Newton'due south 2nd constabulary to determine the normal forcefulness acting upon Noah's 80-kg body at the elevation and at the bottom of the loop.
3. Noah Formula is riding an one-time-fashioned roller coaster. Noah encounters a small colina having a radius of curvature of 12.0 k. At the crest of the hill, Noah is lifted off his seat and held in the machine by the condom bar. If Noah is traveling with a speed of 14.0 k/south, then use Newton's 2d law to determine the strength applied by the rubber bar upon Noah's 80-kg trunk.
4. Anna Litical is riding a "woody" roller coaster. Anna encounters the bottom of a small dip having a radius of curvature of 15.0 k. At the bottom of this dip Anna is traveling with a speed of 16.0 m/s and experiencing a much larger than usual normal force. Utilize Newton'due south 2nd law to make up one's mind the normal force acting upon Anna's 50-kg body.
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