triggernel Posted November 22, 2009 Share Posted November 22, 2009 Could someone please help me figure out how TTD's launch, fits the Law of Conservation of Momentum? Here is a description of the law: http://www.stmary.ws/highschool/physics/home/notes/dynamics/momentum/LAWCONSERVATIONMOMENTUM.htm Thanks, a ton!!! Link to comment Share on other sites More sharing options...
Kennyweird Posted November 22, 2009 Share Posted November 22, 2009 What about the launch, in particular? Link to comment Share on other sites More sharing options...
triggernel Posted November 22, 2009 Author Share Posted November 22, 2009 For a project, I need an example of how any part of the ride relates to the law, and I fugured the launch would be the best place to start. Link to comment Share on other sites More sharing options...
Bolliger&Mabillard Posted November 22, 2009 Share Posted November 22, 2009 It's been a while since I actually had to apply priniples of physics to something... Right before the train is launched, this is a display of stored energy. It's energy that will be used for something. Once the train launches, this is a display of kinetic energy. The kinetic energy in the catch car is transferred to the train and the kinetic energy is what lifts the train up the spike. As the kinetic energy is used, potential energy is the next principle used. As the train makes it's descent, Gravity pulls the train down causing it to accelerate. This principle converts the potential energy back into kinetic energy, and will remain in this state of energy until acted upon from some outside force (In the case of TTD, the magnetic field from the brake fins) Link to comment Share on other sites More sharing options...
chemical_echo Posted November 22, 2009 Share Posted November 22, 2009 Did your teacher ask you to use TTD as an example, or is that your choice? The website you gave is focusing on momentum before and after collisions; something you don't really want to think about on TTD. You can always do it relating to Fabio and the Bird on Apollo's Chariot.... m1 = mass of Fabio v1 = velocity of Fabio right before impact m2 = mass of bird v2 = velocity of bird right before impact v1' = velocity of Fabio right after impact solve for v2' - the velocity of the bird right after impact. Link to comment Share on other sites More sharing options...
ParkTrips Posted November 22, 2009 Share Posted November 22, 2009 It really doesn't conserve momentum because Cedar Point does not operate in a vacuum (or closed system, as your book words it): there is friction between the wheels and track, air restisanc eetc Link to comment Share on other sites More sharing options...
Crazy_Behemoth_Lady_Jess Posted November 22, 2009 Share Posted November 22, 2009 I'm not sure if this also relates: "An object in motion has the tendency to stay in motion in the same speed and direction and objects at rest have a tendency to stay at rest unless acted upon by an unbalanced force." That is why when the train hits the brakes, you kind of get thrown forward abit. And to relate this to the conservation of momentum, the brakes kind of "take up" the kinetic energy from the train to bring it to a halt. The way you get thrown forward is because you still have that kinetic energy from going down the drop. I think it's like decceleration or something. Speaking of TTD and physics, I've been wondering about this for a while. How come on rides like TTD or KK with vertical drops, you are able to stay completely upright while going down the drop? While Dive Machines on the other hand, you tend to "fold over" meaning your head wants to touch your knees while being held on the brake before the drop facing downwards. I'm not really a math/science person, as you can tell from my avatar. Link to comment Share on other sites More sharing options...
Bolliger&Mabillard Posted November 22, 2009 Share Posted November 22, 2009 How come on rides like TTD or KK with vertical drops, you are able to stay completely upright while going down the drop? While Dive Machines on the other hand, you tend to "fold over" meaning your head wants to touch your knees while being held on the brake before the drop facing downwards. Rocket coasters carry over momentum from the kinetic energy from the launch. Before it's descent, it's still rolling forward, so as the train continues along the arch of the top hat, the forward momentum keeps you back. Being held in a stationary position however (in your example of a Dive machine) you come to a stop, which brings the object [the train] to rest. Since the object is in a position to release kinetic energy, that is what you feel, the potential to drop from a resting position. Link to comment Share on other sites More sharing options...
Kennyweird Posted November 22, 2009 Share Posted November 22, 2009 Here's a good example: The total momentum generated at launch is equal to the momentum of the train plus the momentum of the launch system (catch-car, cable, sheave wheels, capstan, etc.) Expressed as a formula: Mtotal = Mtrain + Mlaunchsystem You can then go on to say that once the train and catch-car separate, the launch system's momentum is dissipated through the use of induction brakes similar to the ones used to stop the train at the end of the ride. Other than that, I really can't think of any other TTD examples that relate directly to momentum. The examples other people have given relate more to energy, which is a slightly different concept. Link to comment Share on other sites More sharing options...
coasterfreak101 Posted November 22, 2009 Share Posted November 22, 2009 Did your teacher ask you to use TTD as an example, or is that your choice?The website you gave is focusing on momentum before and after collisions; something you don't really want to think about on TTD. You can always do it relating to Fabio and the Bird on Apollo's Chariot.... m1 = mass of Fabio v1 = velocity of Fabio right before impact m2 = mass of bird v2 = velocity of bird right before impact v1' = velocity of Fabio right after impact solve for v2' - the velocity of the bird right after impact. I vote for this one. (m1)(v1)+(m2)(v2)=(m1+m2)v'. As far as TTD goes, could you set 1/2mv^2 (kinetic energy) equal to mgh (gravitational potential energy) and show that the two are transferred to one another as the train climbs and falls? i.e. kinetic is maxed out after the launch, then zero at the very apex of the top hat, and then all that potential energy turns back into potential energy at the bottom of the drop. Link to comment Share on other sites More sharing options...
Kennyweird Posted November 23, 2009 Share Posted November 23, 2009 Did your teacher ask you to use TTD as an example, or is that your choice?The website you gave is focusing on momentum before and after collisions; something you don't really want to think about on TTD. You can always do it relating to Fabio and the Bird on Apollo's Chariot.... m1 = mass of Fabio v1 = velocity of Fabio right before impact m2 = mass of bird v2 = velocity of bird right before impact v1' = velocity of Fabio right after impact solve for v2' - the velocity of the bird right after impact. I vote for this one. (m1)(v1)+(m2)(v2)=(m1+m2)v'. As far as TTD goes, could you set 1/2mv^2 (kinetic energy) equal to mgh (gravitational potential energy) and show that the two are transferred to one another as the train climbs and falls? i.e. kinetic is maxed out after the launch, then zero at the very apex of the top hat, and then all that potential energy turns back into potential energy at the bottom of the drop. I think he wants an example of how TTD relates to the conservation of momentum, not energy. Besides, your description is incorrect: the train is still moving as it crests the hill, therefore the KE at the apex cannot be zero. Link to comment Share on other sites More sharing options...
the ghost Posted November 23, 2009 Share Posted November 23, 2009 God damn! You guys are all mathematically brilliant. I am in like an algebra 1 class, and I'm a senior. Oh well, Im also in college level english. Link to comment Share on other sites More sharing options...
Crazy_Behemoth_Lady_Jess Posted November 23, 2009 Share Posted November 23, 2009 God damn! You guys are all mathematically brilliant. I am in like an algebra 1 class, and I'm a senior. Oh well, Im also in college level english. Not me, like I said look at my avatar. I just have an occational interest in science/physics. Especially when it's related to something I like. Theme parks! Link to comment Share on other sites More sharing options...
Blythy Posted November 23, 2009 Share Posted November 23, 2009 conservation of momentum is related to a closed system. I.e. there's no energy that travels into or out of the group of objects under consideration. In TTD case, the system (in this case, the train) has a net transfer of energy into the system (i.e. the cable transfers energy to the train). The momentum of the train increases, as well as the momentum of the launch system. As there is a gain in both systems momentum, there is a net overall gain, and therefore no conservation. If you really want to apply conservation of momentum to a rollercoaster, consider a stalled train being stuck by a moving train. Then if you really want to impress, talk about the implications of this. (i.e. high accelerations, high jerk (how acceleration changes over time), bad for the train, bad for the people (will probably kill several people if it's above 30/40 mph) etc) Link to comment Share on other sites More sharing options...
Coasterlvr_nc Posted November 23, 2009 Share Posted November 23, 2009 ^ like mentioned above, TTD is mostly a system of Potential and kinetic energy and the conservation of energy. Not so much the conservation of momentum. Link to comment Share on other sites More sharing options...
hyyyper Posted November 23, 2009 Share Posted November 23, 2009 Like said before, conservation of momentum applies on collisions, not coasters (if you're lucky) Link to comment Share on other sites More sharing options...
MonkeyPants Posted November 23, 2009 Share Posted November 23, 2009 Agreed... I just took a quick glance, but I don't think TTD, or basically any coaster, is going to illustrate the principle on the webpage you linked to. That equation is relevant to collisions, which hopefully you don't have on coasters, at least not in normal operation. Now bumper cars would illustrate it quite nicely! Link to comment Share on other sites More sharing options...
lond Posted November 24, 2009 Share Posted November 24, 2009 Here are some measurements from TTD: Linear G: Positive / Negative G: Lateral G: // Marcus Link to comment Share on other sites More sharing options...
coasterfreak101 Posted November 24, 2009 Share Posted November 24, 2009 Did your teacher ask you to use TTD as an example, or is that your choice?The website you gave is focusing on momentum before and after collisions; something you don't really want to think about on TTD. You can always do it relating to Fabio and the Bird on Apollo's Chariot.... m1 = mass of Fabio v1 = velocity of Fabio right before impact m2 = mass of bird v2 = velocity of bird right before impact v1' = velocity of Fabio right after impact solve for v2' - the velocity of the bird right after impact. I vote for this one. (m1)(v1)+(m2)(v2)=(m1+m2)v'. As far as TTD goes, could you set 1/2mv^2 (kinetic energy) equal to mgh (gravitational potential energy) and show that the two are transferred to one another as the train climbs and falls? i.e. kinetic is maxed out after the launch, then zero at the very apex of the top hat, and then all that potential energy turns back into potential energy at the bottom of the drop. I think he wants an example of how TTD relates to the conservation of momentum, not energy. Besides, your description is incorrect: the train is still moving as it crests the hill, therefore the KE at the apex cannot be zero. Sorry, I meant that in regards to the vertical component. At the very top, the train has zero velocity in the y-direction because of the change in direction. Regardless, the law of conservation of momentum applies a bit more to collisions and such. You could maybe try the recent Blackpool Roller Coaster collision? All you need is the mass of a train and the average speed of the ride, and you could show that in the elastic collision that happened all of the momentum was conserved? Or you could go morbid and figure out how much the train on B:TR was slowed down when it hit that kid that jumped the fence. Either one works! Link to comment Share on other sites More sharing options...
triggernel Posted November 24, 2009 Author Share Posted November 24, 2009 Thansk guys. I turned in my project and just decided to go with Energy instead, thanks for the help. Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now