Month: November 2015

Motion Along a Curve, Part 3

See Part 1 and Part 2, where I discuss trying to find equations of motion for a ball rolling along a track defined by an arbitrary function y(x). So far, I’ve worked out my general approach, and tested it on the very simple case of an inclined line. I do want to tackle some complex functions, but first I want to summarize the method, and to incorporate some changes I learned as I tried my first solution.

I had initially stated I wanted to start with x(0) = 0, to keep it simple. But I didn’t end up needing this restriction. My equation for velocity used the initial yi, which we got from plugging in the initial value of x. Also, we used the x(0) = 0 to find the constant after integrating. But these happened later in the process.

I also had started with an initial velocity of zero. That did make a difference. But looking back on part 1, I don’t think it would complicate the equation too much, and if it’s zero it will be an extra term to just drop out. Let’s go back to the conservation of energy equation, and keep vi in this time.

\displaystyle U_i + K_i = U + K

\displaystyle mgy_i + \frac{1}{2}m{v_i}^2 = mgy + \frac{1}{2}mv^2

\displaystyle gy_i - gy = \frac{1}{2}(v^2 - {v_i}^2)

\displaystyle 2g(y_i - y) = v^2 - {v_i}^2

\displaystyle 2g(y_i - y) + {v_i}^2 = v^2

\displaystyle v = \sqrt{2g(y_i-y)+{v_i}^2} \blacktriangleleft

I’m leaving the ± in this time. Strictly speaking, I’m not treating this as the magnitude of the vector, since magnitudes must be positive. Rather, I want to consider a velocity vector that can point ether forwards or backwards along the direction of the curve. I’m going to allow motion in both directions, not just forward.

Recall the graph showing components of the velocity vector:

Graph of arbitrary curve showing tangent line with right-triangle components
Graph generated in Python/Matplotlib.

Now I want to find the x-component. As I discovered last time, I don’t need to bother with the y-component — once I find an equation for x(t), I can use that directly to obtain y(t). The x-component will be

\displaystyle v_x = v \frac{1}{\sqrt{(y')^2+1}}

\displaystyle v_x = \pm \sqrt{\frac{2g(y_i - y) + {v_i}^2}{(y')^2+1}}

\displaystyle \frac{dx}{dt} = \pm \sqrt{\frac{2g(y_i - y) + {v_i}^2}{(y')^2+1}} \blacktriangleleft

Since I will have y and y′ in terms of x, I would need to rearrange to solve the differential equation. Let’s see how far I can take the general case:

\displaystyle \pm \sqrt{\frac{(y')^2+1}{2g(y_i-y)+{v_i}^2}} \, dx = dt

\displaystyle \pm \int \sqrt{\frac{(y')^2+1}{2g(y_i-y)+{v_i}^2}} \, dx = \int dt

\displaystyle \pm \int \sqrt{\frac{(y')^2+1}{2g(y_i-y)+{v_i}^2}} \, dx = t \blacktriangleleft

where I did not include a constant of integration on the right side, since it can be absorbed into the constant that the left integral will produce.

So, the general approach should be as follows: Given our equation y(x), find y′(x). Plug in those expressions, plug in the initial velocity, and plug in the initial height y[x(0)]. Integrate, and solve for x in terms of t to get x(t), then plug that into y(x) to get y(t). I’ll test if this approach can actually work in the next post.

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