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Problem 2.16 from Morin's Classical Mechanics: Given a semi-infinite stick (one that goes off to infinity in one direction), determine how its density should depend on position so that it has the following property: If the stick is cut at an arbitrary location, the remaining semi-infinite piece will balance on a support that is located a distance $l$ from the end.

I am having a hard time understanding part of the solution to the above worked problem (2.16) in Morin’s Classical Mechanics text. He arrives at the following expression for total torque applied to the balanced rod which makes sense to me:

$\tau = \int_{x_0}^\infty \rho(x)(x-(x_0+l))gdx=0$

He then argues that because $\tau = 0$ regardless of where the rod is cut (ie for all $x_0$), that $\tau’=0$, which I also understand.

In order to find a differential equation to solve for $\rho(x)$, he then replaces $x_0$ with $x_0 + dx_0$ to differentiate $\tau$ with respect to $x_0$ to arrive at:

$0 = \frac{d\tau}{dx_0} = gl\rho(x_0) - g\int_{x_0}^{\infty}\rho(x)dx$

What I don’t understand is

  1. how he can just replace $x_0$ with $x_0 + dx_0$,

  2. what he is expanding as a first order approximation, and

  3. why he has to do that in the first place.

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    $\begingroup$ Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead. $\endgroup$ Jun 26 at 16:55
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    $\begingroup$ In addition, please change the title to something more descriptive of the core of the question at hand. $\endgroup$ Jun 26 at 16:56
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    $\begingroup$ got it - changed $\endgroup$ Jun 26 at 19:52

2 Answers 2

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When differentiating an integral with respect to limits, according to the Wikipedia page on the Leibniz integral rule

\begin{equation} \frac{\text{d}}{\text{d}x}\left( \int_{a(x)}^{b(x)}f(x,t)\text{d}t \right ) = f(x,b(x))b'(x)-f(x,a(x))a'(x)+\int_{a(x)}^{b(x)}\frac{\partial f(x,t)}{\partial x}\text{d}t. \end{equation}

Therefore, for your equation

\begin{equation} \begin{split} \frac{\text{d}}{\text{d}x_{0}}\int_{x_{0}}^{\infty}\rho(x)\left(x - (x_{0} + l)\right)g \text{d}x &= - \rho(x_{0})(x_{0} - x_{0} -l)g + \int_{x_{0}}^{\infty}\frac{\partial}{\partial x_{0}} \rho(x)\left(x - (x_{0} + l)\right)g \\ &= \rho(x_{0})lg - g\int_{x_{0}}^{\infty}\rho(x) = 0 \end{split} \end{equation}

Morin does this because the problem is that if we change $x_{0}$, $\tau$ should not change, i.e. $\tau$ is not a function of $x_{0}$ and therefore should be a constant value $\tau = 0$ for all $x_{0}$ (just imagine plotting $\tau$ against $x_{0}$ - the gradient is flat).

As stated, the remainder of the problem is solved by differentiating both sides wrt $x_{0}$ once more (use the Leibniz rule again) and solving the differential equation.

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The author replaces $x_0$ with $x_0+dx_0$ both in the integral limits and in the integrand because this is a way of obtaining contributions to the derivative wrt $x_0$. You can have both contributions from the integrated result, as the latter depends on $x_0$, but also contributions from the integrand expansion.

Considering that the derivative is linked with Taylor expanding functions, this is the reason the author expands in $dx_0$ (and keeps terms of first order, with the remaining terms supposedly to vanish).

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