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).

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

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.

Simple Harmonic Motion

I came across an interesting approach to a problem involving a spring the other day, and it made me wonder if I could derive the equations for simple harmonic motion. The basic principle of a spring is Hooke’s law, that the spring exerts a restoring force proportional to its displacement from equilibrium position: F = –kx, where k is the spring constant representing the stiffness of the spring. I wondered if I could derive the formulas for its motion, and where the sinusoidal components would appear from, since clearly there are no trigonometric functions in this basic equation. To begin with, from Newton’s second law, F = ma:

\displaystyle ma = -kx

\displaystyle a = -\frac{kx}{m}

We know that acceleration is the derivative of velocity and velocity is the derivative of position:

\displaystyle a = \frac{dv}{dt}


\displaystyle v = \frac{dx}{dt} .

We can rewrite these as a dt = dv and v dt = dx, and integrate. However, our acceleration equation has a in terms of position x, not time t, so I would prefer to integrate with respect to x (as at this point I don’t know how x depends on t — that’s what we’re trying to derive!).

\displaystyle a \, dt = dv

\displaystyle a \left( \frac{dx}{v} \right) = dv

\displaystyle a \, dx = v \, dv

This equation is well-known and could have been used as a starting point, but it’s nice to derive it from the most basic principles. We can now substitute our expression in for a and integrate both sides.

\displaystyle -\frac{kx}{m} \, dx = v \, dv

\displaystyle \int \! \left( -\frac{kx}{m} \right) \, dx = \int \! v \, dv

\displaystyle -\frac{k}{m} \int \! x \, dx = \int \! v \, dv

\displaystyle -\frac{k}{m} \left( \frac{x^2}{2} + C_1 \right) = \frac{v^2}{2} + C_2

Let’s solve this for v:

\displaystyle \frac{v^2}{2} + C_2 = -\frac{k}{m} \left( \frac{x^2}{2} + C_1 \right)

\displaystyle \frac{v^2}{2} + C_2 = -\frac{kx^2}{2m} + C_3

\displaystyle \frac{v^2}{2} = -\frac{kx^2}{2m} + C_4

\displaystyle v^2 = -\frac{kx^2}{m} + C_5

\displaystyle v = \pm \sqrt { -\frac{kx^2}{m} + C_5 }

There’s still no sign of sine appearing anywhere (I know; I couldn’t resist). I’m also a bit uncomfortable with the plus/minus sign, but velocity can be both positive and negative, so I don’t think that discarding the negative solution is justified. I want to integrate again to remove the velocity, but we face a similar problem as before: we have velocity in terms of position x, not time t. We can rearrange the differential again:

\displaystyle v = \frac{dx}{dt}

to get v on the dx side:

\displaystyle dt = \frac{dx}{v} .

Let’s substitute in our expression for v, and integrate:

\displaystyle dt = \frac{dx}{\pm \sqrt { -\frac{kx^2}{m} + C_5 } }

\displaystyle \int dt = \int \! \frac{dx}{\pm \sqrt { -\frac{kx^2}{m} + C_5 } }

This is not pretty, and I’m not sure how to integrate the right side. I believe I can pull out the plus/minus as a constant.

\displaystyle t + C_6 = \pm \int \! \frac{dx}{\sqrt { -\frac{k}{m} x^2 + C_5 } }

I had to consult a table of integrals for assistance with this integration. Of course, it can be solved with trignometric substitution! All of a sudden, we have a hint of where a sine function will appear. We need to get the radical in the form \sqrt{a^2 - b^2 x^2 } :

\displaystyle \sqrt{C_5 - \frac{k}{m} x^2 }

\displaystyle a = \sqrt{C_5} = C_7

\displaystyle b= \sqrt{\frac{k}{m}}

Now, we set

\displaystyle x = \frac{a}{b} \sin \theta :

\displaystyle x = C_7\sqrt{\frac{m}{k}}\sin \theta

Let’s now take our original integral back, and start substituting.

\displaystyle t + C_6 = \pm \int \! \frac{dx}{\sqrt { C_5 -\frac{k}{m} x^2 } }

\displaystyle t + C_6 = \pm \int \! \frac{dx}{\sqrt { {C_7}^2 -\frac{k}{m} x^2 } }

\displaystyle t + C_6 = \pm \int \! \frac{dx}{\sqrt { {C_7}^2 -\frac{k}{m} \left( C_7\sqrt{\frac{m}{k}}\sin \theta \right) ^2 } }

This expression is complicated, but I can already see how it will simplify out.

\displaystyle t + C_5 = \pm \int \! \frac{dx}{\sqrt{ {C_7}^2 - \frac{k}{m} \cdot {C_7}^2 \cdot \frac{m}{k} \cdot \sin^2 \theta} }

\displaystyle t + C_5 = \pm \int \! \frac{dx}{\sqrt{ {C_7}^2 - {C_7}^2 \sin^2 \theta } }

\displaystyle t + C_5 = \pm \int \! \frac{dx}{\sqrt{{C_7}^2 (1 - \sin^2 \theta)} }

\displaystyle t + C_5 = \pm \frac{1}{C_7} \int \! \frac{dx}{\sqrt{1 - \sin^2 \theta} }

Let’s convert the dx to a dθ:

\displaystyle x = C_7\sqrt{\frac{m}{k}}\sin \theta

\displaystyle \frac{dx}{d\theta} = C_7\sqrt{\frac{m}{k}}\cos \theta

\displaystyle dx = C_7\sqrt{\frac{m}{k}}\cos \theta \, d\theta

Substituting this back in yields:

\displaystyle t + C_5 = \pm \frac{1}{C_7} \int \! \frac{C_7\sqrt{\frac{m}{k}}\cos \theta \, d\theta }{\sqrt{1 - \sin^2 \theta} }

\displaystyle t + C_5 = \pm \frac{C_7}{C_7}\sqrt{\frac{m}{k}}\int \! \frac {\cos \theta}{\sqrt{1 - \sin^2 \theta}} d\theta

\displaystyle t + C_5 = \pm \sqrt{\frac{m}{k}}\int \! \frac{\cos \theta}{\sqrt{\cos^2 \theta}} d\theta

\displaystyle t + C_5 = \pm \sqrt{\frac{m}{k}}\int \! \frac{\cos \theta}{\cos \theta} d\theta

\displaystyle t + C_5 = \pm \sqrt{\frac{m}{k}}\int d\theta

\displaystyle t + C_5 = \pm \theta \sqrt{\frac{m}{k}} + C_8

\displaystyle t + C_9 = \pm \theta \sqrt{\frac{m}{k}}

Remembering our equation for x in terms of θ:

\displaystyle C_7\sqrt{\frac{m}{k}}\sin \theta = x

\displaystyle \sqrt{\frac{m}{k}}\sin \theta = C_{10}x

\displaystyle \sin \theta = C_{10}x\sqrt{\frac{k}{m}}

\displaystyle \theta = \arcsin \left(C_{10}x\sqrt{\frac{k}{m}} \right)

Finally, we can substitute this back in for θ and solve for x:

\displaystyle t + C_9 = \pm \left[ \arcsin \left(C_{10}x\sqrt{\frac{k}{m}} \right) \right] \sqrt{\frac{m}{k}}

\displaystyle \pm t\sqrt{\frac{k}{m}} + C_{11} = \arcsin \left(C_{10}x\sqrt{\frac{k}{m}} \right)

The plus-minus is driving me crazy, but I can’t justify dropping it — even though I want to.

\displaystyle \sin \left( \pm t\sqrt{\frac{k}{m}}+ C_{11} \right) = C_{10}x\sqrt{\frac{k}{m}}

Actually, I think I can get rid of it now! Since sin(-x) = -sin(x) = sin(x+π), I can incorporate the +π into the constant!

\displaystyle C_{10}x\sqrt{\frac{k}{m}} = \sin \left( t\sqrt{\frac{k}{m}} + C_{12} \right)

\displaystyle x = C_{13} \sin \left( t\sqrt{\frac{k}{m}} + C_{12} \right)

Let’s rewrite this equation using the conventional symbols for the constants.

\displaystyle x = A \sin \left( t\sqrt{\frac{k}{m}} - \varphi \right)

There it is. The basic sine wave equation is readily apparent, with the constants representing the amplitude and the phase shift, and the radical representing the angular frequency. The fourth possible constant (a “+ D”) is absent, and I wondered if I had accidentally dropped a constant somewhere — since that entire family of curves should be solutions. That is, x should be able to oscillate around 1, -5, whatever. But the reason for its absence is our starting point of Hooke’s law, F=-kx, which assumes that the equilibrium point is x=0. We would have had to start with F=-k(xxeq) to have had that last constant.

It’s pretty neat to see the equation emerge from just four basic equations: Hooke’s law F = –kx, Newton’s second law F=ma, and the definitions of velocity and acceleration v=x′ and a=v′!

The Nobel Prize in Physics 2007 Goes to Albert Fert and Peter Grünberg for the Discovery of Giant Magnetoresistance

Winning the Nobel Prize is one of the highest honors one can achieve. Winners bring their institutions and their countries prestige. I’d like to highlight this year’s prizewinners.

The Nobel Prize in Physics this year was awarded to French scientist Albert Fert and German scientist Peter Grünberg. They were recognized for their independent discovery of giant magnetoresistance. The concept’s a bit esoteric, but the Nobel Prize site,, has some nice introductory material. In fact, it’s really put together well and you are advised to browse through it for more information about any aspect of the Nobel Prizes.

I especially like their “speed read” summaries. The Physics entry is quite easy to understand and begins as follows:

The Giant within Small Devices

Lying at the heart of the computer which you are using to read this article is a memory retrieval system based on the discoveries for which the 2007 Nobel Prize in Physics was awarded to Albert Fert and Peter Grünberg. They discovered, quite independently, a new way of using magnetism to control the flow of electrical current through sandwiches of metals built at the nanotechnology scale.


And if you have time, you should definitely read a nice 7-page PDF explaining the concept for the layperson, using illustrations and easy-to-understand concepts. I won’t bother going into detail here since the site does such a nice job. There’s no excuse not to know the basics of this discovery!

You can also see videos of the announcement, or read the press release.