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A frictionless rollercoaster

  1. Apr 21, 2015 #1
    If a car falling from H rides downhill to h and then uphill back to H on a frictionless road and the hills all reach the same altitude H , in theory it can cover any distance, without an engine, right? Of course there shouldn't be sharp curves, something like this http://www.coasterimage.com/wp-content/uploads/2012/03/millenniumforce02wide.jpg [Broken], right? is there an equation for the best profile of the curve?

    Now, if we suppose drag and friction, I suppose that the less the difference H-h, and the less speed, the better, is my intuition right?

    I am sure drag increases with speed, what about friction?

    Thanks for your attention
     
    Last edited by a moderator: May 7, 2017
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  3. Apr 21, 2015 #2

    CWatters

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    Only if there are no losses (no drag, no friction etc) and that's impossible in practice.

    What do you mean by "best"? If there are no losses then the design is already optimum.

    Air drag is proportional to velocity squared so yes you want to keep the velocity as low as possible. You should also keep the velocity constant.

    Some other types of friction forces are either constant or proportional to velocity.
     
  4. Apr 21, 2015 #3
    Not completely impossible if it is on a mag.lev track. I was referring to an ideal situation: no friction no drag.

    Coming down to earth: not all curves are equal, if the hill it's too steep and the change of direction makes an acute " V "angle more momentum and energy are lost., right? Therefore there must be an optimal profile of the road that loses less momentum.

    Is friction (here static friction, right?) the same at any speed?

    If the car goes from A to B on a 1) straight road (ideal conditions) we need to accelerate it to v, if 2) the road is like a V, even if its length is greater, we need no fuel at all., right?

    Now , on a real road, in case 1) we need some more fuel now and then to compensate friction and drag, in case 2) we have more drag at the bottom , as v is greater, but yet, the fuel we need to get to B will be less than in case 1) right? Can we conclude that it is always better to take the downhill-uphill road and switch on the engine only after the bottom?

    What other factors can help that choice? surely the more massive is the car, the more the drag can be overcome, right?
     
  5. Apr 21, 2015 #4
    in theory it can cover any unlimitted distance if there is no friction, the equations you want are simply PE(U)=mgh, KE=(1/2)mv^2. At the top of the cost PE is max and therefore KE is minimum, at the bottom of the coster PE is minimum and KE is max. Maybe this can help?>>https://phet.colorado.edu/en/simulation/energy-skate-park
     
  6. Apr 21, 2015 #5

    jbriggs444

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    You have to tell us what makes one curve better than the next before we can help you figure out which curve is best.

    Do you, for instance, want to minimize travel time for a fixed horizontal distance and a fixed drop? Or do you want to maximize distance travelled for a fixed drop?
     
    Last edited by a moderator: May 7, 2017
  7. Apr 21, 2015 #6

    jbriggs444

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    If one assumes some air resistance (proportional to the square of velocity) and some ordinary friction (constant force) then although you want to minimize velocity, thus keeping air resistance low, you also want to minimize distance travelled, thus keeping the energy losses (total drag integrated over distance) low.

    For a non-zero starting velocity, this is not an easy optimization problem.
    For a zero starting velocity the solution seems trivial.
     
  8. Apr 21, 2015 #7
    One curve is better when it takes the care to the same height. I already noted that if the slope is too steep and the bottom there is an acute V angle lots of KE are lost.

    We wan to find the best conditions that allow in real life a car travel the max distance with the min petrol. The initial velocity is zero, what is not trivial is to calculate which curve and which difference in altitude give the highest mileage.
     
  9. Apr 21, 2015 #8

    CWatters

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    No there isn't. You said you are referring to a situation with no drag, no friction etc. That also applies to sharp twists and turns, steep drops, sharp bends etc etc. The mass of the coaster is constant so if the coaster were to "loose momentum" it's velocity and hence it's energy would also reduce. Where would the energy go if everything is lossless (no drag, no friction)? Just going around a curve won't loose you energy.

    On an ideal road (no drag, no friction) neither case 1 or 2 needs fuel. In case 1) the energy required to accelerate the car to v can be recovered using regenerative braking. Just as in case 2 the potential energy used to accelerate the car is recovered and converted back to potential energy.

    It's not trivial to work out which is best. Perhaps see this vid for the low drag case... Note they both get to the same height at the end...



    but the more mass you have to raise back up on route 2.
     
  10. Apr 21, 2015 #9

    CWatters

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    See also...
     
  11. Apr 21, 2015 #10

    A.T.

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    Maximal amount of vomit?
     
  12. Apr 21, 2015 #11

    A.T.

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    It still could go into dissipative deformation. The be lossless it needs to be perfecly rigid.
     
  13. Apr 21, 2015 #12
    If I remember my first year applied maths lectures from 1974 correctly (not guaranteed), the shape of the curve that gives you the minimum roll time, regardless of starting velocity, is the same type of shape that you get when you hang a rope from one point to the other. And I seem to remember being taught that the mathematical description of this shape is not simple, i.e.it is not a circular arc, parabola, hyperbola, ellipse, cosine, logarithmic, exponential, ...
     
  14. Apr 22, 2015 #13
    Yes, the catenary! Intuitivelly that does seem like the optimum profile for a car rolling from one hill to another.

    Geometrical properties
    Over any horizontal interval, the ratio of the area under the catenary to its length equals a, independent of the interval selected. The catenary is the only plane curve other than a horizontal line with this property. Also, the geometric centroid of the area under a stretch of catenary is the midpoint of the perpendicular segment connecting the centroid of the curve itself and the x-axis.

    I’m sure someone here can work out the math to prove this!:wink:
     
  15. Apr 22, 2015 #14

    CWatters

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    Well yes ok.

    Perhaps I'm misunderstanding the OP? He seemed to be suggesting that the sudden turns/drops/bends somehow cause a loss of energy even on an ideal track.
     
  16. Apr 22, 2015 #15

    jbriggs444

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    Sharp bends (i.e. infinite curvature) are problematic on an ideal track. If you want to conserve both energy and momentum, the object will bounce off the track. If you model the object as always moving parallel with the track, then energy is lost.

    The OP seems to understand this but has interpreted it to mean that this is a gradual phenomenon -- the tighter the curve, the more energy is lost. That is an error. For an ideal track there is no loss at all, no matter how high the [finite] curvature. Only for infinite curvature is there energy loss.
     
  17. Apr 24, 2015 #16
    But in a real (railway for example) track the radius of curvature is relevant isn't it? and also for a car on a road, right?

    Can someon tell me how to calculate the total air drag of a body with a constant acceleration, is it a difficult integral?

    suppose a rollercoaster is falling pulled by g (=10), starts falling at v = 10 m/s and drops 6m in height and reaches 21 m/s. Air resistance grows by the square of the speed , therefore the coefficient is 100 at the beginning and 441 at the end, if we consider an average speed of 15,5^2 = 241 are we far from the exact result. If the grade is relevant consider any value you like or just 5%. Does the result change with the slope?

    Thanks, you have been very helpful
     
  18. Apr 24, 2015 #17

    jbriggs444

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    It could be relevant in a number of ways. But what I can think of would involve engineering considerations rather than pure physics.

    The radius of curvature should not be any tighter than the radius of the wheels. Otherwise, the path traced out by the axles will have a sharp corner. Clearly, the tighter the radius of curvature, the higher the peak radial force/acceleration. At some point that will exceed the elastic limit of your materials and this will lead to energy loss. But the elasic limit is not basic physics. That's engineering and materials science. As radial acceleration and radial force increases, one might expect rolling resistance to increase proportionately. If so, the tightness of the ratio of angular change to momentum loss is fixed regardless of corner radius.

    The radius of curvature on the convex portion of the track should not be so tight that the car leaves the tracks, of course.

    Air resistance is generally approximated as a function of velocity, not acceleration. In the presence of air resistance, acceleration will generally not be constant. It is usually not an integral at all. It becomes a differential equation that may or may not be possible to solve analytically.

    Solving the differential equation numerically can be as easy as figuring out the drag at the beginning, determining position and velocity 0.1 seconds later and repeating until reaching the bottom of the stretch of track.
     
  19. Apr 24, 2015 #18
    Yes , it grows wih the square of velocity. Now, suppose a car (A = 2m^2) is travelling at 20 m/s the drag is proportional to 400, if it gets constant acceleration suppose that after 10 seconds it reaches 30 m/s. That means that a = 1m/s^2 and that it has travelled roughly 240 m with growing resistance, from 400 to 900. Right?

    Can someone tell me how to find the exact value of the energy lost? Neglecting air density (rho) , the formula considers a coefficient (1/6) *A(2m^2)*d(240m)* )*v^2(25?). If we consider the average speed 25m/s are we near the true value?
    How do I find the exact value of v? integral or differential equation?

    Thanks
     
  20. Apr 24, 2015 #19

    jbriggs444

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    I make it 250 meters (average speed of 25 meters per second over 10 seconds).

    The obvious approach would be to integrate power over time. Power is given by drag force times current velocity. So you are integrating kv3 with respect to time.

    By assumption, acceleration is 1 meter per second and is independent of air resistance. That's a questionable assumption. But it may be approximately correct. It certainly simplifies the problem greatly. If you start from rest at t=0 then v(t) = t and you are simply integrating kt3dt.
     
    Last edited: Apr 24, 2015
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