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Velocity of a rollercoaster at the bottom of a curve

  1. Sep 4, 2012 #1
    How would you go about calculating the velocity of a rollercoaster once it reaches the bottom, specifically, something like this:

    http://www.joyrides.com/sfmm/photos/superman1.jpg [Broken]

    It's not hard to calculate the velocity it accumulates during the vertical part but how do you deal with the curved part?
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 4, 2012 #2


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    Use conservation of energy. There is some friction, but it tends to be pretty minor in that type of coasters, so if you assume all potential energy went into kinetic energy, you'll get a very good estimate.
  4. Sep 6, 2012 #3
    Is it possible to do it the hard way (without a computer) by taking the slope at each point (assuming the function of the curved part is known) and calculating the acceleration at each point, multiplying it by an infinitesimal change in time and summing it all up?
  5. Sep 6, 2012 #4
    Given a curve and a force, it is always possible to write down the differential equations of motion along that curve, and if that curve is nice enough, solve them analytically or numerically otherwise. But certain things, like the kinetic energy if the force is potential, could be obtained without doing any of that.
  6. Sep 6, 2012 #5
    ^ Yeah, I guess I'm more interested in how you'd use calculus/DEs to do a problem like this now. So, suppose we didn't know about energy, how would you set up a differential equation for this problem?

    I did one example where I used a part of the curve f(x)=1/x on the positive side but I only found the acceleration at the point x=0.5 using the derivative and some trigonometry.
  7. Sep 7, 2012 #6
    Suppose the curve is given implicitly by two equations: [tex]
    f(x, y, z) = 0
    \\ f(x, y, z) = 0
    [/tex] Then the equations of motion are [tex]
    m\ddot{x} = F_x + \lambda f_x + \mu g_x
    \\m\ddot{y} = F_y + \lambda f_y + \mu g_y
    \\m\ddot{z} = F_z + \lambda f_z + \mu g_z
    [/tex] Where [itex] F_x, \ F_y, \ F_z [/itex] are projections of the external force onto the coordinate axes, and [itex]f_x, \ f_y, \ f_z, \ g_x, \ g_y, \ g_z[/itex] are partial derivatives of [itex]f, \ g[/itex], and [itex] \lambda, \ \mu[/itex] are Lagrange multipliers (which can be used to determine the reaction force). Note that with the equations of the curve you have five equations, and you have five unknowns. This is, however, usually too cumbersome. What is done instead, the natural coordinate frame of the curve are used. At any point of the curve, there is a tangent vector, a normal vector, and a binormal vector. Together they are always mutually perpendicular. Because the curve is given, they can easily be computed at any point on the curve. So the motion is treated in this coordinate frame.[tex]
    m\frac {d^2s} {dt^2} = F_{\tau}
    \\ \frac {mv^2} {\rho} = F_n + N_n
    \\ 0 = F_b + N_b
    [/tex] where [itex] F_{\tau}, \ F_n, \ F_b[/itex] are projections of the external force onto the tangent, normal, and binormal vectors; [itex] N_n, \ N_b[/itex] are projections of the reaction force; [itex]s[/itex] is the distance along the curve; [itex]\rho[/itex] is the radius of curvature. In principle, you need only the first equation to integrate the system and obtain the law of motion. The other two are only required if you need to know the reaction force.
  8. Sep 12, 2012 #7
    Thanks voko...is that vector calculus?
  9. Sep 12, 2012 #8
    Vectors are used during the derivation of those equations. But vector calculus usually denotes a somewhat different subject.
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