Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Sliding down a hill with no friction, non-constant incline

  1. May 17, 2009 #1
    How do i solve this? heres how far i've gotten

    suppose i have x[q], y[q]=S, the surface of a hill

    dy/dx=(dy/dt)/(dx/dt)

    atan(dy/dx)= the angle of incline at a location on the hill

    -GM*cos(pi/2-atan(dy/dx))=the tangeantal component of acceleration
    im ignoring the perpendicular component because theres no friction here.
    -GM*sin(atan(dy/dx)=d2(x^2+y^2)^.5/dt2

    i guess i could apply conservation of energy and get

    E=.5(dy/dt)^2+.5(dx/dt)^2+gm*INT(sin(atan(dy/dx)))d(x^2+y^2)^.5
    but dy/dx isnt neccessarily a function of x^2+y^2, so im kind of out of luck
    and even then i dont have a second conserved quantity(such as angular momentum like in the central force problem), which would be useful.
    i resolved the components again and got this system

    -gm*cos(atan(dy/dx))*sin(atan(dy/dx))=d2x/dt2
    -gm*(sin(atan(dy/dx))^2=d2y/dt2



    So, how do i solve this problem where the slope of the hill is not a constant?
    I want to find how fast an object will be, the time taken, etc, for an object to go from one position to another on the hill. how do i do this?
     
  2. jcsd
  3. May 20, 2009 #2
    I think the equations you obtained from acceleration component and the conservation of energy are equivalent. Probably you need a relation for the constraint. Is the equation for the surface S given? Or is there any dimension given.
     
  4. May 21, 2009 #3
    Hello okkvlt
    The jump from the q parameter to velocity quotient I do not understand. Does the q definion also allow bumps and more tantelizing loops? Is gravity in the y direction?
    greetings Janm
     
  5. May 22, 2009 #4
    oops, that was i typo, i meant (dy/dq)/(dx/dq)

    gravity is in the negative y direction.

    the resolved acceleration vector seems to tell me im right, because d2y/dt2 is always negative, and d2x/dt2 is positive only if the incline is negative and negative only if the incline is positive.

    im pretty sure it should work for any "roller coaster" type situation.

    how do i prove that the starting position doesnt matter if S is a cycloid? (tautochrome problem) i cant figure that out.
     
  6. May 22, 2009 #5
    Hello okkvlt
    Definition is almost complete. Is there a starting velocity at y=y_0?
    Thank you for understanding that a looping for a hill is rather difficult. I got enthousiastic from the q coordinates. Indeed with a free perameter for the curve even a loop is possible. Rather make that out of wood or metal.

    x(q),y(q) let us say q is the natural parameter for the curve in for instance number of meters: y(q) going down, x(q) going up...
    A bump is y(q) going up while x(q) remains going up!
    A looping is y(q) going up and x(q) changing direction going down, or I would rather say going back.

    Nice problem okkvlt.
    greetings Janm
     
  7. May 22, 2009 #6
    Here is a beautiful problem of a frictionless object sliding down an incline that Newton solved about 1696. Consider a frictionless object sliding from a point x=0, y=0 along a variable incline to a lower point x=2, y= -0.5 (dimensions in meters). What is the shape of the incline that has the shortest transit time? (This is not the tautochrone problem.)
     
  8. May 23, 2009 #7
    Hello Bob S
    An example not defining the shape of the "hill" but asking what would the shape need to be to accomplish fastest descent. Of coarse I am asking myself what is the difference with the tautochrone problem of which I am familiar. My first way to solve problems is intuitionistic and I can explain that as when the problem is solved it is just one of the things (like in mathematics) fixed never to be changed...
    What I also mean is that the way in finding solutions is also interesting. My intuition tells me that frictionless gives easier solutions but in fact is harder to solve. In differential calculus is spoken of umbilical points. On a perfect ball-surface all points are umbilical, while on an ellipsoid points can be identified as being of different sort by their specific Eulerian curvatures... Another example is the ring of 's Gravesande. A globe fits in a ring when it is cold, yet after it is heaten (and so expanded) it doesn't surpass the ring. When this experiment is practiced the amanuensis has the problem that it is often the other way around. The ball is rusty (or dirty in some other way) and does not fit the ring when it is cold. After heating (cleaning!) the ring does fit...

    Ok my intuistic thought about the Bob S example: Friction usualy is F=fN, f a specific value depending on the materials of 1 the gliding or rolling object and 2 the "hill"; N is the normal force. Ok frictionless means that f=0 but minimising the total effects of N could still be the way to solve the equation. If the hill would be a (singing) saw then the normal force would deform the track!
    Start with much vertical to get velocity end with much horizontal because there the velocity is the highest and bend the velocity by 90 degrees with a radius of curvature larger than that of the object rolling down the hill.

    Do you see Bob that in the intuitive proces problems come, which problably need to be defined before solving? Frictionless or not do we have a rolling stone or a sled coming down the hill?
    Greetings Janm
     
  9. May 24, 2009 #8
    Hello Bob S
    Did some calculation on this item. s=v_o*t+g*t^2/2=1/2, v_0=0 and so t=1/sqrt(g). V_e=gt=sqrt(g). That is for falling down a straight line. Some ingenius Fresnel bumper in the right angle corner bumps the object to the last 2 meter. That takes s=v_e*t +a*t^2/2=2, a=0 and so t=2/sqrt(g), that makes the total time 3/sqrt(g)=3/pi=21/22 secs.
    I am interested if this is more or less then the straight slope between the two points. Length of the straight slope = sqrt(4+1/4)=2*sqrt(1+1/16)=2*(1+1/32)=33/16. The cosine of the slope:cs = (1/2)/(33/16)=16*1/(2*33)=8/33. s=v_0*t+g*cs*t^2/2=33/16, v_0=0 and cs=8/33; so t^2=2*(33/16)/(8/33)=2*33^2/(16*8)=33^2/8^2, so t=33/8 sec>4 seconds.
    The Fresnel pooltable solution is faster, but I don't know if there is a brachistochrone solution which could be faster then the 21/22 secs!
    greetings Janm
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Sliding down a hill with no friction, non-constant incline
Loading...