Thorpe Park Essays

Thorpe Park Essays Thorpe Park Essay Thorpe Park Essay My aim is to investigate the different ways in which physics are involved to create thrilling yet safe rides. Some examples of the physics used in a theme park, which I intend to explore in this piece of coursework, are the following:* The sensation of weightlessness* Large accelerations of the body* Large decelerations of the body* Changes in kinetic and potential energy* Newtons laws of motion* Power and work done* Lifting forces and stopping forcesI have chosen the ride called detonator. I must find out two pieces of information needed before I carry out my investigation and carry out my calculations. I will firstly need to estimate the mass of the carriage and the height (displacement) of the tower.Estimating the massThere are twelve seats, each holding a person who on average has a mass of 90kg. There is a large carriage with a mass of 1500kg although this is an estimate. The total mass equals 2580kg, this is an estimate this is not very accurate as not every rider is identical and has a mass of 90kg some may have a higher or less weight. This will affect my results slightly.Estimating the heightTo estimate the height of the detonator I will use similar triangles to accomplish this.Two triangles are similar if they have the same shape. They dont have to be the same size. For the smaller triangle, I will evaluate the ratio of the side with length h to the side with length l. This is the same as, with respect to the larger triangle, the ratio of the side with length H to the side with length L. That is h/l = H/LI used two rulers to record the measurements of l and h, the two rulers being 90 degrees parallel to one another. The value of L, length, was recorded by the distance from where I was to the point at where the ride is situated. I recorded the values for:* h = 29.1cm = 0.29meters,* l = 37.1cm = 0.37meters,* L = (97 paces at 0.39) + 6m = 43.83meters.I am now able to calculate the height of the detonator (H):h/l = H/L h/41.8 = 0.29/0.37 h= (43.83x 0.29) /0.37Height = 34.35 meters.When I calculated the height of the ride I did not consider where the carriage started, therefore the displacement is not from the top of the tower. My calculation is not very accurate. I will now subtract 3.27 meters for the length of the carriage. So my new height is 31.08 meters.Average VelocitiesOnce I have estimated the height of the tower I am now able to calculate the average velocities of various stages in the ride. Velocity is the speed at which the carriage changes position. A higher velocity means an object is travelling faster between two given positions. I will include the average velocity of the carriage on its journey up to the top of the ride, the average velocity of the carriage on its journey down the ride, the average velocity of the whole ride, the average velocity of the free fall distance part of the ride. I will also estimate the average velocity of the braking distance. I used a stopwatch and was able to time how long the carriage t ook to complete all five stages of the ride. With this value I was able to calculate the velocity with the simple formula s=d/t. My average velocities are unfair, as I have not taken into consideration air resistance or friction, which would slow the carriage down slightly. Although in most cases this is small enough to be neglected.I got the free fall distance by estimating the displacement of the ride from where the carriage started at the top of the tower until the while plates located on the side of the tower, which are used to slow the carriage down. This is the displacement that the carriage is travelling whilst in free fall. I think this is an inaccurate measurement as I used my observation. I latter went onto the Thorpe park website and found that the free fall distance is actually 29.5meters. I estimated 27 meters and was only 2.5 meters too short, I think that is only a small inaccuracy.I went on to the Internet and found the actual average free fall velocity of the ride t o be 15ms-1, with my value at 11.64ms-1. This is only a slight inaccuracy and is a fairly reasonable result. The inaccuracy may have been caused by the many possible human errors or slight inaccurate results collected by using the stopwatch. An example would be the delay or quickness of my reactions to stop the watch when the required destination is met.TimeI recorded the times with a standard digital stopwatch. I recorded three sets of results for the five times and worked out an average for a more accurate set of results. This was a reliable but inaccurate method as many human or technical errors where possible, causing an unfair set of results.Free fallGalileo first introduced the concept known as free fall. According to legend, Galileo dropped balls of different mass from the Leaning Tower of Pisa to help support his ideas. These classic experiments led to the finding that all objects free fall at the same rate, regardless of their mass.An object in the state of free fall is onl y influenced by the force of gravity. The object has a downward acceleration toward the centre of the earth (9.81ms-2), the source of gravity.I will use a constant acceleration formula to investigate whether the detonator has an acceleration of -9.81ms-2, as Galileo predicted. The displacement of the free fall ride is 27meters as I previously estimated (see page 1). The time of the free fall part of the ride is 2.32 seconds as I also previously recorded.A=?S=27metersT= 2.32 secondsU= 0ms-1V=S= ut + 1/2at2S = 1/2 at2a= 2s/t2a = 2x 272.322a= 10.03 ms-2This is near the value of acceleration due to gravity. I think this is a reasonable result due to the limitations. Firstly the displacement was just an estimate and may not be the exact value, this would affect the result, as it may be a greater or less value. The time is not the exact value of the free fall ride as I used my observation to stop the watch when I thought the carriage reached the white plates, when it started to decelerate , this may not be an accurate value. With such a small inaccuracy I think that I can safely draw the conclusion that the carriage free falls at the same acceleration as that of the gravitational pull of the earth, as predicted above. I have taken into consideration inaccuracies. Another ride that relies on free fall is called tidal wave. This is a roller coaster, which descends down a slope with a component of the gravitational pull of the earth. As further work I could work out the acceleration of this ride to investigate the component of gravity.Newton later took Galileos discoveries concerning mechanics and incorporated these principles into his laws of motion. Newtons first law applied very much to the ride which states; every body remains at rest or in uniform motion in a straight lie unless acted upon by a force. It is this first law of motion, which applies to the ride, firstly when the ride is at the top and is stationary. Whilst at the top there are two forces acting upon t he seat of the carriage. The forces act in such a way that they counteract to each other. As the rider sits in their seat at this moment, the seat pushes upward with a force equal in strength and opposite in direction to the force of gravity, which is causing the passengers weight. These two forces are said to balance each other, causing the rider to remain at rest. If the seat is suddenly pulled out from under the rider, then he experiences an unbalanced force. There is no longer an upward seat force to balance the downward pull of gravity, so he accelerates uniformly to the ground, until the carriage decelerates. It is at this point where a force is applied to slow the carriage down bringing it to a controlled stop.The detonator can give its passengers the sensation of free-fall. It consists of three distinct parts: ascending to the top, momentary suspension, and then the quick downward plunge.Ascending to the topIn the first part of the ride, force applied to the car lifts it to the top of the tower. The amount of force depends on the mass of the car and the passengers within the car. Motors create this upward force. There are built-in safety allowances for variables concerning the mass of the riders. F=ma, F=2580x-9.81, F= 25309.8N. This may not be a very accurate result as the mass of the carriage is just an estimate and not a true value. Whilst the carriage is gaining height it is also gaining gravitational energy. The carriage also contains little kinetic energy on its journey.Momentary suspensionWhilst the car is stationary at the top of the tower the gravitational potential energy is at its greatest. Gravitational potential energy is the energy possessed by the carriage because of its height above the ground. The amount of GPE depends upon its mass and its height. I have worked the GPE with the following formulae:Gain in gravitational potential energy = mass x gravity x height.2580(kg) x 9.81(ms-1) x 31.08(m)= 786628NThe value for the GPE may be inacc urate as I estimated the height and this might not be the exact value. Also the mass of the carriage it an estimate and is not identical for every ride, so it varies for every ride.With the GPE I am now able to calculate the minimum power used by the motor to lift the carriage. I will use the formula: power=Energy Transferred/Time.P=786628(N)/17.82(s)P = 44143 wattsThis value may be inaccurate as the GPE I calculated is not an accurate value, also the time may be inaccurate as I used a stopwatch and inaccuracy could have occurred. This is the minimum power used, as the motor is not 100% efficient due to friction, increase/decrease in mass due to passenger mass, air resistance, also loss of power due to heat e.t.c.Newtons third law of motion states that; To every action there is an equal and opposite reaction. A 6okg man would experience a 0N resultant force acting on his seat whilst at rest, and a 589N reaction force acting on his seat. F=ma, F=609.81=589N.Quick downward plungeFrom the top of the tower the car is swiftly accelerated towards the ground under the influence of the earths gravitational field only, as I previously calculated. It is at this point, which the car gains kinetic energy and looses potential energy. A unique magnetic braking system helps bring the ride to a controlled stop. The faster the carriage moves, the larger resistive forces become.F=maF=25809.81=25309.8NWe measure the energy an object has by the work it can do. The work is done when the carriage is moved by the force of gravity. I will now calculate the work done:work done = force x displacement in direction of forcewd=25310x27wd= 683370j this is not very accurate as I did not consider air resistance or friction. Also the force involves an estimate of the mass of the carriage, which is not identical in all the rides.Kinetic energyThe kinetic energy of the carriage is the energy possessed because of its motion. The kinetic energy is greatest when it is at a minimum height. I have w orked out the kinetic energy possessed by the carriage whilst in free-fall.Loss of Gravitational Potential Energy = Gain in kinetic EnergyMgh = 1/2mv2gh = 1/2 v29.81 x 31.08 = 1/2 v2304.8 = 1/2 v2609 = vv = 25ms-1The main limitation in this calculation is the fact that the height of the ride is an estimate, which may be inaccurate. Also I did not consider the air resistance or friction of the carriage and tower, which would slow the carriage down. In most cases this is small enough to be neglected.I will now use the constant acceleration formulas to calculate the final velocity just after the free-fall part of the ride, just before the deceleration.A= -9.8ms-2S= 27metersT= 2.32sU= 0ms-1V= ?I will now calculate the rate of deceleration as the carriage comes to a stop. I will use the constant acceleration formulas. Displacement is 4.08 meters as I estimated (page 1), the time for the carriage to stop during deceleration is 2.32 seconds as I also recorded; the initial velocity is 23.3m s-1 as I calculated above.A=?S= 4.08mT= 1.7sU= 23.3ms-1V= 0V2 = u2 + 2asv2_u 2 = a2sa = 02 23.3224.08a=66.5ms-2=6.8gsI looked up onto the Internet and found out that the carriage is changing its speed at a rate of 5gs. My calculation was 6.8gs, (6.8 5 =) I was just 1.8gs out. There are many factors, which could have caused this inaccuracy. Firstly the displacement of the deceleration is an estimate, I measured the distance that the carriage travels whilst decelerating the same way I measured the height, by using similar triangles (see page 1). This was an inaccurate measurement as I latter found out on the Internet the displacement to be 5.0 meters, I was 0.92 meters out. This affected my calculation significantly. Secondly the time was such a small measurement that it was really hard for me to get the exact time whilst the carriage is decelerating as I used a manually operated digital stop watch and my reactions were not fast enough. Also I did not take into account air resistanc e or friction of the carriage, which would slow the carriage down. This is usually a small amount as the mass of the carriage is high but would affect my result slightly.WeightlessnessThe detonator produces a sensation of weightlessness. These sensations result when riders no longer feel an external force acting upon their bodies. At the top of the tower of a free-fall ride, a 60kg rider would feel 589N of force from the seat pushing as an external force upon their body. The rider feels their normal weight. Yet as, she falls from the tower, the seat has fallen from under her. She no longer feels the external force of the seat and subsequently has a brief sensation of weightlessness, in this ride 2.32 seconds of weightlessness. The rider has not lost any weight, but she feels as though she has because of the absence of the seat force.Discussion of future developmentsIt is hard to find any future developments, which would improve the ride in any way because top physics have been used to design this ride. Maybe you could increase the duration of the free fall part of the ride, making a longer sensation of weightlessness. Also increasing the height of the ride would increase the gravitational potential energy and as the carriage descends it will loose more kinetic energy therefore going faster. This too would increase the sensation if weightlessness.ImprovementsI think that this investigation was overall very successful. I estimated the height of the detonator to be 31.08meters high. I later researched on the Internet and found the actual height to be 30.48. My estimate was only 0.6 meters out and was a very pleasing result. To improve this investigation I would spend a bit more time in the theme park to gather and check through the calculations and to make sure they are accurate.