Parabolic Ramp

Consider a point mass m moving along a smooth parabolic ramp. Let the shape of the ramp be described by the equation

y(x)=\frac{1}{2} a x^2

where a is a positive constant. We will make the following assumptions; the mass experiences no frictional force through contact with the ramp; the mass is point-like, and therefore has a negligible moment of inertia; the mass never loses contact with the ramp; the ramp itself cannot move.

Our goal is to find the equation of motion describing the behaviour of the mass, first using Newtonian mechanics, and then using Lagrangian mechanics.

Newtonian Method

We will use the diagram below for reference:

Parabola Diagram

The angle \theta is defined to be that made between the x axis and the direction of the reaction force R. The angle \varphi is that subtended by the tangent to the parabola with the x axis, as pictured.

First we will write down the equations given by Newton’s second law:

m\ddot{x} =R \cos\theta

m\ddot{y}=R \sin\theta-mg

For sake of ease, we will choose to work with the x coordinate first. Consider the first of the two equations. We want to generate a differential equation containing only x and its time derivatives. This means finding expressions for both R and \theta in terms of x. We will eliminate R first, using the second equation:

m\ddot{y}=R \sin\theta-mg

\displaystyle R=\frac{m(\ddot{y}+g)}{\sin\theta}

Obeying the constraint that the mass cannot leave the parabola, \ddot{y} can be expressed as a function of x, making use of the product and chain rules:

y=\frac{1}{2} a x^2

\dot{y}=ax\dot{x}

\ddot{y}=a\dot{x}^2+ax\ddot{x}

\displaystyle R=\frac{m(a\dot{x}^2+ax\ddot{x}+g)}{\sin\theta}

On substituting, we get

\displaystyle m\ddot{x}=\frac{m(a\dot{x}^2+ax\ddot{x}+g)}{\sin\theta}\cos\theta

\displaystyle \ddot{x}=\frac{a\dot{x}^2+ax\ddot{x}+g}{\tan\theta}

Now we make use of the diagram to find an expression for \theta. We can see that

\theta=\varphi+\frac{\pi}{2}

Here we can use the following relation:

\tan(\varphi+\frac{\pi}{2})=-\cot\varphi

\ddot{x}=-(a\dot{x}^2+ax\ddot{x}+g)\tan\varphi

The angle \varphi is easy to relate to the gradient of the parabola’s slope:

\tan\varphi=\frac{dy}{dx}

\tan\varphi=\frac{d}{dx}(\frac{1}{2} a x^2)

\tan\varphi=ax

\ddot{x}=-(a\dot{x}^2+ax\ddot{x}+g)ax

Rearranging yields

(1+a^2 x^2)\ddot{x}+(a^2 \dot{x}^2+ag)x=0

Lagrangian Method

We will now use the Euler-Lagrange equation to deduce the same equation, choosing x as our generalised coordinate. The Lagrangian is given by the expression

L=T-U

where T is the system’s kinetic energy and U its potential energy. No diagram is needed to calculate these, only the constraint stated above. We will quickly run through the calculations:

T=\frac{1}{2} m \dot{x}^2 + \frac{1}{2} m \dot{y}^2

T=\frac{1}{2} m(\dot{x}^2+\dot{y}^2)

T=\frac{1}{2} m(\dot{x}^2+a^2 x^2 \dot{x}^2)

U=mgy

U=\frac{1}{2}mag x^2

L=\frac{1}{2}m(1+a^2 x^2)\dot{x}^2-\frac{1}{2}mag x^2

Now using the Euler-Lagrange equation:

\displaystyle \frac{d}{dt}\bigg(\frac{\partial L}{\partial \dot{x}}\bigg)=\frac{\partial L}{\partial x}

\frac{d}{dt}(m(1+a^2 x^2)\dot{x})=m a^2 x \dot{x}^2-magx

\frac{d}{dt}((1+a^2 x^2)\dot{x})=a^2 x \dot{x}^2-agx

2a^2 x \dot{x}^2+(1+a^2 x^2)\ddot{x}=a^2 x \dot{x}^2-agx

a^2 x \dot{x}^2+(1+a^2 x^2)\ddot{x}=-agx

(1+a^2 x^2)\ddot{x}+(a^2 \dot{x}^2+ag)x=0

We’ve recovered the same differential equation.

Solution

The equation above is classified as a second order non-linear ordinary homogeneous differential equationThe property about which we should be worried is its non-linearity; as such, it is inherently difficult to solve. We will therefore make the assumption that the mass’ range of motion is small. By this, we mean that its amplitude is small (meaning x is generally small) and by extension, it never moves particularly quickly (\frac{dx}{dt} is small). Let us assume it is sufficiently small that terms of order 2 in x and \frac{dx}{dt} can be neglected. In this case, the equation reduces to

\ddot{x}+agx=0

This is the familiar differential equation whose solutions are of the form

x(t)=A\cos\omega t+B\sin\omega t

where, in this case, \omega^2=ag. If we suppose that the mass is released from rest from a height h, the constants A and B take the values

\displaystyle A=\sqrt{\frac{2h}{a}}

B=0

Hence the motion of the mass is approximately simple harmonic, described by

x(t)=\sqrt{\frac{2h}{a}}\cos(t\sqrt{ag})

y(t)=h\,\text{cos}^2(t\sqrt{ag})

Comparison

I’ve used Wolfram Mathematica to provide a numerical solution for x(t). Here is an animation comparing the simple harmonic approximation (gold) with the true motion (brown) over roughly one period for h=\frac{1}{2} a. I have chosen the constants to have the value a=\frac{1}{2} and g=10.

Parabola Diagram

We see that our approximation moves a little faster than is realistic.

There are a few comments to be made about this exercise. First, it highlights the advantage of Lagrangian mechanics over Newtonian mechanics for a system such as this. For all intents and purposes, the reaction force R is a redundancy – this is the case in many systems. So too, in some sense, is the angle \theta. When formulating the Lagrangian, we only needed to consider a single equation.

It also demonstrates the importance of simple harmonic motion. Most any periodic motion, considered in the limit in which its amplitude is small, can be approximated as being simple harmonic.

A final point to consider: how would the mass behave if a were negative? The equation of motion is unchanged, but its solution must be subtly amended.

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