Motion Along a Curve, Part 2

I would like to find equations for the position of a small ball rolling along a curve given by an equation, rather than a flat incline. In Part 1, I introduced the problem and set up a test case. Now let’s try to actually make some headway.

Last time, we showed that

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

where v is the magnitude of the velocity vector v, g is the acceleration due to gravity, and y is the height. I note that if y>yi, then the quantity inside the radical is negative and v will not be real — which makes sense, since the ball should never be able to roll higher than its starting position. (Recall that I am starting with an initial velocity of zero). Since we’re assuming that the ball rolls along the function, its direction will be given by the derivative of y with respect to x. I do see a potential problem with notation arising. The value of y depends on x (by definition, since that’s how I’m defining the curve). But x and y both depend on t if we’re thinking about the motion of the ball. I want to define our variables carefully. We should think of x as a function x(t), and y as a function of x, so y(x) = y[x(t)]. I am going to reserve the prime notation y′ specifically to refer to the derivative with respect to x.

\displaystyle y' = \frac{dy}{dx}

We know that

\displaystyle \mathbf{v} = \langle r, \angle\theta \rangle

The magnitude will be v. If we take the tangent line to the curve and sketch a triangle, we can see that for a change in x of 1, y changes by y′, so tan θ = y′/1. I illustrated this below with an arbitrary curve.

Graph of a function and its derivative, showing the x- and y- components

\displaystyle \mathbf{v} = \langle v, \angle \arctan y' \rangle

Let’s now find the component vectors vx and vy, where of course vx = r cos θ and vy = r sin θ.

\displaystyle \mathbf{v} = \langle v \cos \arctan y', v \sin \arctan y' \rangle

There’s actually no need for these arctangents. If I go back to the earlier sketched triangle, I can see that the hypotenuse is \sqrt{(y')^2 +1} and so I can read the cosine and sine directly off that triangle (1 over the hypotenuse and y’ over the hypotenuse, respectively):

\displaystyle \mathbf{v} = \left\langle v \frac{1}{\sqrt{(y')^2 + 1}}, v \frac{y'}{\sqrt{(y')^2 +1}} \right\rangle

\displaystyle \mathbf{v} = \frac{v}{\sqrt{(y')^2+1}}\langle 1, y' \rangle

Let’s substitute in our expression for the magnitude, v:

\displaystyle \mathbf{v} = \sqrt{\frac{2g(y_i-y)}{(y')^2+1}} \langle 1, y' \rangle \blacktriangleleft

I believe that this should be the general solution, given the constraints (ball must roll along the curve, uniform gravity, no friction, starting x and v of 0). Let’s pause to make some observations. The expression in the numerator is the magnitude of the velocity which we found in the last post. The potential/kinetic energy relationship is obvious, with the product of g and the change in height from potential energy, and the times two and the square root being inverses from kinetic energy. The <1, y′> ensures the proper direction, and the denominator adjusts for the length of that direction vector (it basically makes it a unit vector).

This equation is so far pretty straightforward. However, in order to find equations for x and y in terms of time t, we’d have to plug in the expression for y and y′, then integrate the velocity. Let’s try this with what should be a very simple example, the incline (line) from the prior entry:

Graph of y=-x sqrt(3)/3 + 1

\displaystyle y = -\frac{x\sqrt{3}}{3} + 1

\displaystyle y' = -\frac{\sqrt{3}}{3}

\displaystyle y_i = y(0) = 1

Plugging these into our equation for v gives:

\displaystyle \mathbf{v} = \sqrt{\frac{2g(y_i-y)}{(y')^2+1}} \langle 1, y' \rangle

\displaystyle \mathbf{v} = \sqrt{\frac{2g[1-(-\frac{x\sqrt{3}}{3}+1)]}{(-\frac{\sqrt{3}}{3})^2+1}} \left\langle 1, -\frac{\sqrt{3}}{3} \right\rangle

I am starting to regret setting my initial angle to π/6. There was no benefit to choosing a “nice” value for the angle; it would have been better to choose a slope with an integer value. However, it’s too late to go back and change it now. Let me see if I can simplify this.

\displaystyle \mathbf{v} = \sqrt{\frac{2g\frac{x\sqrt{3}}{3}}{\frac{1}{3}+1}} \left\langle 1, -\frac{\sqrt{3}}{3} \right\rangle

\displaystyle \mathbf{v} = \sqrt{\frac{2gx\sqrt{3}}{\frac{4}{3}\cdot 3}} \left\langle 1, -\frac{\sqrt{3}}{3} \right\rangle

\displaystyle \mathbf{v} = \sqrt{\frac{gx\sqrt{3}}{2}} \left\langle 1, -\frac{\sqrt{3}}{3} \right\rangle

Let’s split this into its component vectors, and switch to magnitudes:

\displaystyle \mathbf{v} = \mathbf{v_x} + \mathbf{v_y}

\displaystyle \mathbf{v} = v_x \boldsymbol{\hat{\i}} + v_y \boldsymbol{\hat{\j}}

\displaystyle v_x = \sqrt{\frac{gx\sqrt{3}}{2}}

and

\displaystyle v_y = -\sqrt{\frac{gx\sqrt{3}}{2}} \cdot\frac{\sqrt{3}}{3} = -\sqrt{\frac{gx\sqrt{3}}{6}}

Let’s now try integrating the first one:

\displaystyle \frac{dx}{dt} = \sqrt{\frac{gx\sqrt{3}}{2}}

\displaystyle \frac{1}{\sqrt{x}} \, dx = \sqrt{\frac{g\sqrt{3}}{2}} \, dt

\displaystyle \int x^{-\frac{1}{2}} dx = \int \sqrt{\frac{g\sqrt{3}}{2}} \, dt

\displaystyle \frac{x^{\frac{1}{2}}}{\frac{1}{2}} + C_1 = t \sqrt{\frac{g\sqrt{3}}{2}} + C_2

\displaystyle 2\sqrt{x} = t \sqrt{\frac{g\sqrt{3}}{2}} + C_3

\displaystyle \sqrt{x} = t \sqrt{\frac{g\sqrt{3}}{8}} + C_3

This is looking promising. Before I square both sides, let’s deal with the constant. Since we’re starting at a position x=0 at t=0,

\displaystyle 0 = 0 + C_3

so clearly the constant is zero. (This won’t be so simple for the y-component.)

\displaystyle \sqrt{x} = t \sqrt{\frac{g\sqrt{3}}{8}}

I will square both sides, realizing that I might be introducing extraneous solutions. In fact, t cannot be negative, since the square root of x is nonnegative.

\displaystyle x(t) = \frac{g\sqrt{3}}{8} t^2, \, t\geq 0 \blacktriangleleft

Before tackling the y-direction, let’s test this answer. Let’s plug in our value from the last post for t, for the time when it reaches the x-axis.

\displaystyle x\left(\sqrt{\frac{8}{g}}\right) = \frac{g\sqrt{3}}{8} \left(\sqrt{\frac{8}{g}}\right)^2 = \frac{g\sqrt{3}}{8} \cdot \frac{8}{g} = \sqrt{3}

This is indeed the value of x at the base of the triangle (recall height 1, width √3, hypotenuse 2 for a 30-60-90 triangle)! Now let’s find y. Let’s start with our equation for the y-component of the velocity:

\displaystyle v_y = -\sqrt{\frac{gx\sqrt{3}}{6}}

This one is going to be a little more complicated.

\displaystyle \frac{dy}{dt} = -\sqrt{\frac{gx\sqrt{3}}{6}}

I already see a complication. My equation has x, not y. We of course have our starting equation for y in terms of x, and it would be easy to solve that for x and substitute it in. However, if we want to try this on other functions, especially those that aren’t injective (that is, where multiple values of x might result in the same value of y — in a parabola, or sine curve for instance), then solving will be difficult. I think it would be better to try to use the chain rule.

\displaystyle \frac{dy}{dx} \frac{dx}{dt} = -\sqrt{\frac{gx\sqrt{3}}{6}}

We already found y′ above:

\displaystyle -\frac{\sqrt{3}}{3} \frac{dx}{dt} = -\sqrt{\frac{gx\sqrt{3}}{6}}

I just realized something. We already have \frac{dx}{dt} . It was the vx we started with to find x(t) above!

Wait. That expression is in terms of x, and we have no y’s left. Substituting that in will just result in an identity.

\displaystyle -\frac{\sqrt{3}}{3} \sqrt{\frac{gx\sqrt{3}}{2}} = -\sqrt{\frac{gx\sqrt{3}}{6}}

\displaystyle -\sqrt{\frac{gx\sqrt{3}}{6}} = -\sqrt{\frac{gx\sqrt{3}}{6}}

This is bad news, because I really don’t think that solving the starter equation for y will be a practical solution for most functions. But if we don’t have a y or a dy, then we’re not going to get an equation for y.

Scratch that. There’s no need to do any of this! We just derived an equation x(t). We already know y(x). So we can find y[x(t)]!

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

\displaystyle x(t) = \frac{g\sqrt{3}}{8}t^2

So, plugging in the equation for x in,

\displaystyle y(t) = -\frac{\sqrt{3}}{3} \left(\frac{g\sqrt{3}}{8}\right) t^2 + 1

\displaystyle y(t) = -\frac{g}{8}t^2 + 1, \, t\geq 0 \blacktriangleleft

This looks right. It’s of a similar form to our equation for x(t), but with a factor of -1 instead √3 (which follows from our 30-60-90 triangle, and y is decreasing. And it has a +1 to raise the initial height. Just to be sure, let’s check it with our test value for t:

\displaystyle y\left(\sqrt{\frac{8}{g}}\right) = -\frac{g}{8}\left(\sqrt{\frac{8}{g}}\right)^2 + 1 = -1 + 1 = 0

And that’s correct, because we specifically picked the test case for when the ball reaches the x-intercept. So our final solution is

\displaystyle x(t) = \frac{g\sqrt{3}}{8} t^2, \,t\geq0 \blacktriangleleft

\displaystyle y(t) = -\frac{g}{8}t^2 + 1, \, t\geq 0 \blacktriangleleft

The method works! In future explorations, it would be interesting to try this for a more complicated starter equation.

The restriction t≥0 is interesting. If it weren’t there, the equations for the motion of the ball would still be valid, since it would represent the ball being launched backwards up the slope, coming to rest for an instant at (0,1) at t=0, then rolling back down the slope. I’m not surprised that it is there, though, because my method explicitly assumed that the ball’s velocity vector points in the direction <1, y′>, which points in the forward x direction. Let’s trace back the restriction on t. It comes from where it times a constant equals √x, which must be positive. Why did we have √x, as opposed to ±√x? That came from the equation for the magnitude of the velocity v. That in turn came from the fact that K = ½mv2. Should we have kept the negative solution? No, because v represents the magnitude, and the magnitude is nonnegative. Still, a negative v would correspond to a vector pointing in the opposite direction, when put into our component equation. I believe that those solutions should be allowed. I could justify it by saying that v was not the magnitude, but rather ±magnitude, to allow the vector to point in either direction. Now that I think about it, this will occur for every problem we attempt with this method. Since the laws of physics and therefore the equations of motion should be symmetric over time, this means that if the ball is starting at rest at time = 0, we could “run the film in reverse” to see what path it took to get to the starting point. This is an interesting development, and I wonder if expanding the method in this manner would allow the ball to roll backwards at other times, too — say if it were to roll down a valley, then up the other side. If we allow these “negative magnitudes,” might we be able to get an equation which expresses the motion as it rolls back down into the valley and back up the original side?

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