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This the question given in my textbook

A thin, uniform rod of length $l$ is rotated with constant angular speed $\omega$ about an axis passing through it's one end and perpendicular to one end of the rod. Find the increase in the length. $[$ Density of material is $d$ and Young's modulus is $Y$ $]$

$\bf{Solution}$

$-dT = dm\omega^2x$

$-\int dT ${upper limit T(x) lower limit (0)} = $\int \frac{M}{L}dx\omega^2x${upper limit x lower limit x=l ]

$T(x) = \frac{dA\omega^2L^2}{2}(1 - \frac{x^2}{L^2})$

[A=area of cross section of the rod which is constant for the rod] For element

$\frac{T(x)}{A} = Y\frac{dl}{dx}$

or

$dl = \frac{T(x)dx}{AY} = \frac{d\omega^2L^2}{2Y}(1 - \frac{x^2}{L^2})$

I don't understand this step $\frac{T(x)}{A} = Y\frac{dl}{dx}$ can anyone explain what is happening here thank's

akash

EDIT== I don't know how to set upper limit and lower limit in the integration that's why i have written it like this

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up vote 1 down vote accepted

That equation is basically equating Young's modulus to stress by strain ratio, i.e.

$$Y=\frac{T/A}{\Delta l/l}$$

  • As the tension is a function of $x$ on the rod, it is denoted as $T(x)$.
  • Area of cross section is $A$.
  • $\Delta l$, the change in length is denoted by $dl$.
  • $l$, the length under consideration is $dx$. Note that the textbook has considered an - infinitesimal length because tension $T(x)$ is not constant throughout the length of the rod. But we can assume it to be constant for an infinitesimally small length $dx$.

Also note that $dl$ is the change in length of only the $dx$ part, and not the whole rod.

Then you get your equation $$Y\frac{dl}{dx}=\frac {T(x)}A$$ Which gives the integral $$\int dl = \int_{x=0}^{x=l} \frac{T(x)dx}{YA}$$

You put limits as $x=0$ to $x=l$ because $x$ varies from $0\to l$. The LHS gives you $\Delta l$ which is the total increment in length of the whole rod.

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