Homework 1 Problem 1

Problem

Use Newtonian mechanics to derive the equation of motion

(1)
\begin{align} \ddot x \left(1 + \left(f'(x)\right)^2\right) + \dot x^2 f'(x) f''(x) + g f'(x) = 0 \end{align}

for the “snowboarder” described in class.

Solution

To derive the equation using Newtonian mechanics, we first draw a free body diagram of the snowboarder of mass $m$ riding along the curve $f(x)$ under the influence of gravity $g$:

Homework1prob1

As seen in the slick Microsoft Paint free body diagram above, the gravitational force $mg$ can be split into components normal and tangent to the curve $f(x)$ at any given point. The normal component of gravity will be exactly canceled by the normal force $N$ exerted by the curve (hill) on the mass (snowboarder). That leaves the tangential force (call it $F_t$) to cause the mass to move along the curve according to Newton's second law,

(2)
\begin{align} F_t = \frac{d}{dt} (mv) \end{align}

Before we go any further, note that the angle $\theta$ can be derived as follows:

(3)
\begin{align} \tan \theta = \frac{\Delta y}{\Delta x} = f'(x) \end{align}
(4)
\begin{align} &\to \theta = \tan^{-1} (f'(x)) \end{align}

Also, note that the velocity $v$ of the mass can be calculated as follows:

(5)
\begin{align} v = \sqrt{\dot x^2 + \dot y^2} = \sqrt{\dot x^2 + \dot x^2 \left(f'(x)\right)^2} = \dot x \sqrt{1+\left(f'(x)\right)^2} \end{align}

Returning to the free body diagram, we can come up with an expression for the tangential force to use on the left side of Newton's second law.

(6)
\begin{align} F_t = - mg \sin \theta = - mg \sin(\tan^{-1}(f'(x))) = - mg \frac{f'(x)}{\sqrt{1+\left(f'(x)\right)^2}} \end{align}

The sign of the tangential force is negative because it acts “against” the slope of the hill.

We now plug everything in to Newton's second law and simplify:

(7)
\begin{align} F_t = \frac{d}{dt} (mv) \end{align}
(8)
\begin{align} - mg \frac{f'(x)}{\sqrt{1+\left(f'(x)\right)^2}} = \frac{d}{dt} (m \dot x \sqrt{1+\left(f'(x)\right)^2}) \end{align}
(9)
\begin{align} - mg \frac{f'(x)}{\sqrt{1+\left(f'(x)\right)^2}} = m \ddot x \sqrt{1+\left(f'(x)\right)^2} + m \dot x^2 \frac{f'(x) f''(x)}{\sqrt{1+\left(f'(x)\right)^2}} \end{align}
(10)
\begin{align} - g f'(x) = \ddot x \left( 1+\left(f'(x)\right)^2 \right) + \dot x^2 f'(x) f''(x) \end{align}

And finally:

(11)
\begin{align} \ddot x \left(1 + \left(f'(x)\right)^2\right) + \dot x^2 f'(x) f''(x) + g f'(x) = 0 \end{align}

Scott agrees!

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