Design a Rollercoaster

Transcription

1 Design a Rollercoaster This activity has focussed on understanding circular motion, applying these principles to the design of a simple rollercoaster. I hope you have enjoyed this activity. Here is my own rollercoaster design, complete with the metholodology for calculating the energy, speed, velocity and acceleration at points around the track. Throughout I will use free-body diagrams to analyse the forces acting on the carriage. Newton s Second Law of motion applied to circular motion gives the following equations relating speed v, period T, radius r and acceleration a. and v = 2 π r T, (1) r. (2) We also know that the mass of the carriage is 800 kg, and that acceleration due to gravity is 9.8ms 2. Figure 1: The initial ascent and descent of my rollercoaster design. F g denotes the direction of the gravitational force, and F n denotes the direction of the normal force, at three points along the track. 1

2 The rollercoaster has to start with an uphill climb, and I have set the height of my hill to 30m. The carriage is stationary at the top of this hill, so the initial energy of the carriage is coming only from the gravitational potential energy gained during the hill climb. In fact, the frictionless track ensures that energy is conserved throughout the rollercoaster ride. I will call a change in gravitational potential energy E gpe. So, E gpe = mgh = = J. (3) The carriage is stationary, so the speed and acceleration are zero. Additionally, the only force acting on the carriage is gravity, which gives a component F g = mg downwards. The normal force F n opposes the gravitational force, and in this case the forces must balance, F g = F n = mg = = 7840 N. (4) My rollercoaster will now descend, on a slope of angle 40 degrees, down the hill (height = 30m). I ll examine the energy and forces half way down the slope, so after the rollercoaster has dropped 15m. The change in gravitational potential energy from the top of the slope is, E gpe = mgh = = J. (5) This gravitational potential energy is converted to kinetic energy, so I can calculate the speed of the carriage, E kinetic = 1 2 mv2 = J, (6) giving v = 17.1ms 1. The gravitational force F grav is still acting downwards, but now the normal force F n is acting perpendicular to the track. Therefore, to determine the net force and acceleration, I can resolve forces along the slope, and perpendicular to it. The net force F net in the direction perpendicular to the slope is the combination of F n and a component of the gravitational force, F net ma perp = F n mg cos θ, (7) where θ is the angle of incline and a perp is the acceleration in the direction perpendicular to the slope, which is zero. Therefore, F n = mg cos θ = cos (40 ) = N. (8) In the direction parallel to the slope, the only contribution is from the gravitational force. This time, the acceleration a par is non-zero, as the carriage is accelerating down the slope. Therefore, and F net ma par = mg sin θ = sin (40 ) = N, (9) a par = g sin θ = 9.8 sin (40 ) = 6.3 m s 2. (10) I can perform exactly the same calculations to determine the energy, forces, speed and acceleration at the base of the slope. By now the carriage has descended 30m, and the conservation of energy means that all the gravitational potential energy has been converted to kinetic energy, as the carriage is now at the same height as the starting point. So, the change in gravitational potential energy is simply that described by Equation 3 and the kinetic energy is, E kinetic = 1 2 mv2 = J, (11) and v = 24.2 m s 1. The forces and acceleration of the carriage have not changed from the previous position, as the carriage is at the bottom of the slope (still at angle 40 ), and its weight and normal force remain the same. The free body diagram is exactly equivalent to the previous position. Now I quite fancy including a loop-the-loop, and as explained in Lecture 2b of the suggested resource, this is known as a clothoid loop, and can be approximated by considering 3 circles. I have set my clothoid loop as shown in Figures 2 and 3. 2

3 Figure 2: The loop-the-loop section of my rollercoaster. As with the clothoid loop example, this is built from circles. The two largest circles (dashed lines) have radius 15m, and I have designed the loop so that the radius of the smaller circle (dot-dashed line) is 7.5m, and it sits between the centres of the large circles. I have designed the loop so that it is built from sections of three circles, as described in the example of clothoid loops. I have set the 2 large circles to have a radius of 15m, and the smaller circle to have a radius of 7.5m. I have also set the smaller circle to pass through the centres of the large circles (the reason will become clear in the next section). I will analyse the forces and energy at the start, top and end of the loop. At the first position, there has been no change in gravitational potential energy, and the frictionless track means that we can assume the carriage is still travelling at v = 24.2 m s 1 (Equation 11). In the first position shown in Figure 3, the carriage is moving along the circumference of one of the largest circles. The gravitational force acts downwards, and the normal force acts perpendicular to the track. The circular motion equations allow the calculation of the centripetal acceleration, r = = g (!) (12) This is at the limit of 4g stipulated by the activity, so the loop could not be any smaller. Considering the forces acting on the carriage, F net ma = F n mg, (13) giving F net = = 31360N. Rearranging allows the calculation of F n, F n = F net + mg = = N. (14) I can now consider the position at the top of the loop. The carriage has now gained gravitational potential energy, at the loss of kinetic energy. Here is where the loop design is important, as I need to calculate the vertical height gained at the top of the loop. Due to the way I have constructed the loop, this is simply m = 22.5m. Therefore, the conservation of energy stipulates, E kinetic + E gpe = 1 2 mv2 + mgh J. (15) The gravitational potential energy is simply = J, giving kinetic energy E kinetic = = 58800J. Therefore, the speed of the carriage at the top of the 3

4 Figure 3: The loop-the-loop track is now shown in red, and the forces acting on the carriage at three points along the loop are shown. loop is 12.1ms 1, and this can be used to determine the centripetal acceleration, where the radius of the circle in question is now 7.5m, Considering the net centripetal force, we have r = = 19.6 m s 2 2 g. (16) F net = F n + F g mv2 r = Again mg = 7840N, giving F n = = 7840N. = N. (17) Once the carriage descends to the bottom of the loop, the gravitational potential energy is converted back to kinetic energy, and Equation 11 again applies to give v = 24.2ms 1. The free body diagram is exactly the same as the position going into the loop, meaning that the centripetal acceleration is again equal to 39.0ms 2. To return to the beginning of the track, I will include a banked turn, a straight section and a final banked turn. Both banked turns will have a radius of 25m (see Figure 4). Evaluating the conservation of energy for the banked curve, there is no change in gravitational potential energy, and so we still have v = 24.2ms 1. Evaluating the forces on the banked curve, there is a net centripetal force acting horizontally into the centre of the circle, and we can resolve the forces in the horizontal and vertical directions. The gravitational force does not contribute to the horizontal force, but there is a component from the normal force. Horizontally, F net = ma horizontal mv2 r I can first calculate the centripetal acceleration, = F n sin θ (18) and then the net force, a horizontal = v2 r = = 23.5 m s g, (19) F net = ma horizontal = = N, (20) 4

5 and finally the normal force by rearranging Equation 18, F n = F net sin θ = = N. (21) sin (20 ) Figure 4: A free-body diagram of the banked curve section of track. The radius of the circle is 25m, and the track is banked at an angle of 20. The conservation of energy and lack of friction mean that the carriage will travel along the straight track at a constant velocity of 24.2ms 1, before undergoing the same forces as just described for the final banked curve. Then it is a question of applying the brakes to stop the rollercoaster back at its starting point. Your rollercoaster will probably be more complicated than this, but this gives you an indication of how to make the calculations for a range of track sections. Going Further The situation becomes much more complicated when friction is involved. Whereas before we could rely on the conservation of energy to provide the kinetic energy required to get around the track, this is now affected by the inclusion of the frictional force. Including the friction force adds an extra force component in a direction opposite to the direction of motion and affects the calculated accelerations. However, the presence of friction means that a component of kinetic energy is lost as heat through the friction of the wheels. It is therefore no longer straightforward to determine the carriage speed. We can expect that the inclusion of the frictional force will necessitate modifications to the choice of track sections to ensure that the carriage is always at a sufficient speed to get around the track. You can re-analyse your free-body diagrams to see how your choice of track sections is affected when friction is incorporated. You may like to estimate the amount of energy lost through friction the work done due to friction is the integral of the frictional force between two points. In the simplest case (motion in a straight line), the work done is equal to F d where F is the applied force and d is the distance travelled. If the force is varying, it is necessary to integrate the force between 2 positions. For more information on work and force follow this link. 5