Motion Along a Curve, Part 1

Earlier this week, I saw an intriguing problem suggested, one involving motion along a curve. The idea is this: in elementary mechanics, we learn about how an object will move when placed on an inclined surface. If we restrict motion to two dimensions, this surface can be represented by a line. With gravity as the propelling force, can we find a general approach to deriving its motion along an arbitrary curve? In other words, given a function f(x), can we find functions describing the position (and therefore velocity and acceleration) over time?

Here are my initial thoughts. One, I am sure that this type of problem has been analyzed before. That doesn’t matter; I am going to try figuring it out myself. Two, I think we can use conservation of energy to determine the object’s velocity at any point. However, that will depend on knowing its height, and I don’t know how this approach would handle the object flying up off a short hill, or dropping away from a cliff. So three, I am going to restrict the object’s motion to the curve — consider it a track, rather than a 1D surface. That means that four, I can use the derivative to find the direction, but that means the object can only move forward. Five, I have to ignore friction. My conservation of energy approach will require that all energy be potential or kinetic; I won’t know hot to deal with losses to friction. Six, I am using constant downward gravitation. Seven, I am starting with initial position x=0 and initial velocity of 0. I don’t believe this method requires it, but I am already concerned about the complexity of the math. And seven, I am planning to obtain a velocity equation and integrate to find position, but I don’t know if the equation will have an analytic solution (or if I will know how to integrate it).

These are some significant limitations, of course. If this approach works, I suspect that it could be expanded to deal with several of these. For instance, we could keep in terms for initial position and velocity. Gravity or whichever force could be represented by a vector field rather than a constant force; this would make the potential energy term much more complex. I also wonder if I could incorporate friction by adding in an extra term, but I think this would turn the equation into a more complex differential equation.

Let’s start with a simple case that we can solve using conventional means, so we’ll have an answer to check later. I can also use this to check my conservation of energy approach. Let’s have the object rolling down an incline of 30° (π/6), starting at a height of 1. The ball will roll down to the right (I find it more intuitive to imagine a rolling ball rather than a frictionless object sliding, especially if the surface is curved). The equation for this surface will be

$\displaystyle y(x) = -\frac{x\sqrt{3}}{3}+1 \, \bullet$

where the slope is -tan(π/6).

You can see that it would form a right triangle with height 1, length √3, and hypotenuse 2. The slope is therefore 1/√3, or √3/3. Perhaps it would have been better to use an incline of 60° (π/3) so that the slope would be -√3, but I like this one better. Let’s try the conservation of energy approach to see what the velocity would be when the ball reaches the bottom (that is, y = 0). The total energy E is the sum of the potential energy U and the kinetic energy K. This should remain constant, so

$\displaystyle E_i = E_f$

$\displaystyle U_i + K_i = U_f + K_f$

We know that potential energy is given by

$\displaystyle U = mgh$

assuming constant gravitation g (the acceleration due to gravity), with mass m and height h; and kinetic energy is given by

$\displaystyle K = \frac{1}{2}mv^2$

where v is the magnitude of the velocity vector v. I confess I am not very proficient with vectors, but clearly the ball is moving in two dimensions and we will need both components in the x and y directions. I am going to try to not be sloppy but to carefully think about what we mean by v or v every time I write it.

We assume that the initial velocity, and therefore kinetic energy, is zero.

$\displaystyle U_i + K_i = U_f + K_f$

$\displaystyle mgh_i + \frac{1}{2}m\cdot0^2 = mgh_f + \frac{1}{2}mv_f^2$

The height is y, and we can divide both sides by the nonzero mass m:

$\displaystyle gy_i = gy_f + \frac{1}{2}v_f^2$

Let’s solve for v:

$\displaystyle 2g(y_i-y_f)=v_f^2$

$\displaystyle v_f = \sqrt{2g(y_i-y_f)} \blacktriangleleft$

This should be the general case, so if this checks out, we can use this equation to develop the method. As a check, if the units of g are m/s² and y is in m, the units will be m/s, appropriate for velocity. Also note that this is simply the magnitude of the velocity vector (the speed); the actual vector could point in any direction if all we know is conservation of energy. That’s why I earlier constrained the ball to move along the curve — that will give us our direction. Plugging in our initial and final values of y, we get

$\displaystyle v_f = \sqrt{2g(1-0)} = \sqrt{2g}$

Next, to verify this solution, let’s solve using the traditional free-body approach. Gravity applies a force downwards of W=mg. This can be split into two components, one perpendicular to the surface mg cos(θ), and one along the surface mg sin(θ). The perpendicular force will be balanced out by the equal and opposite normal force, leaving a force of mg sin(θ). This is simple motion in one dimension, and let me introduce a variable s to represent its position along the surface (so I don’t generate confusion with our x’s and y’s). Think of it as placing a tape measure from the bottom of the incline up to the starting point. We know that for a given constant acceleration a,

$\displaystyle v = \int \! a \, dt = at + v_0$

$\displaystyle s = \int \! v \, dt = \int \! (at+v_0) \, dt = \frac{1}{2}at^2 + v_0 t + s_0$

By Newton’s second law, F = ma, so a = g sin(θ) and will be negative, since it’s pointing towards decreasing s (since this is motion in one dimension, we can use the sign to expression direction rather than need vectors). Initial velocity is zero, and initial position is 2 (the distance along the incline, if the bottom is at s=0).

$\displaystyle 0 = -\frac{1}{2}g \sin(\frac{\pi}{6}) \cdot t^2 + 0t + 2$

$\displaystyle \frac{1}{2} g \left( \frac{1}{2}\right)t^2 = 2$

$\displaystyle \frac{gt^2}{4} = 2$

$\displaystyle t^2 = \frac{8}{g}$

$\displaystyle t = \pm \sqrt{\frac{8}{g}}$

I can discard the negative solution, since I’m interested in the behavior starting from time = 0, not how it might have been launched before to come to a temporary stop at time = 0. Let’s plug this time to get our velocity from v = at (since the initial velocity is zero):

$\displaystyle v = at = -g\left(\frac{1}{2}\right)\sqrt{\frac{8}{g}} = - \sqrt{\frac{g^2}{4}}\sqrt{\frac{8}{g}} = -\sqrt{2g}$

This velocity is negative since it points downwards along the s direction, along the incline. Its magnitude will be the absolute value which is what we obtained earlier using the conservation of energy method. Armed with this success, and with some test values, we’ll be ready to work on the actual problem in the next post.

Edit: I added a graph.

4 thoughts on “Motion Along a Curve, Part 1”

Tacitus Umbra says:

31 October 2015 at 09:23 GMT-0600

I love this. Mechanics is my absolute favourite part of maths to work on and it’s really cool to do some exploratory mechanics as well. I didn’t have time upon first reading to follow through the working exactly and fully understand it but I look forward to doing this later. Also looking forward to seeing the next steps!
Darmok says:

31 October 2015 at 17:18 GMT-0600

Thanks! I love playing around with the equations and seeing what emerges from the math. Let me know if something doesn’t make sense or if you see any errors. I tend to basically just walk through my thought process, so it’s not always the most elegant path.
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4 thoughts on “Motion Along a Curve, Part 1”