A uniform rod of length L is rotated in a hor | vedclass

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(B) Consider a small element of the rod of length $dx$ at a distance $x$ from the axis of rotation.The mass of this element is $dm = \frac{M}{L} dx =

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$\frac{\rho \omega^2 L^3}{3 Y}$

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(B) Consider a small element of the rod of length $dx$ at a distance $x$ from the axis of rotation.The mass of this element is $dm = \frac{M}{L} dx = \rho A dx$, where $A$ is the cross-sectional area.The centrifugal force $dF$ acting on this element is $dF = (dm) x \omega^2 = \rho A \omega^2 x dx$.The tension $T(x)$ at a distance $x$ from the axis is the sum of centrifugal forces on all elements from $x$ to $L$:$T(x) = \int_x^L \rho A \omega^2 x' dx' = \rho A \omega^2 \left[ \frac{x'^2}{2} \right]_x^L = \frac{\rho A \omega^2}{2} (L^2 - x^2)$.The elongation $d\Delta L$ of the element $dx$ is given by Hooke's Law:$d\Delta L = \frac{T(x) dx}{AY} = \frac{\rho A \omega^2 (L^2 - x^2) dx}{2 AY} = \frac{\rho \omega^2}{2Y} (L^2 - x^2) dx$.The total increase in length $\Delta L$ is the integral from $0$ to $L$:$\Delta L = \int_0^L \frac{\rho \omega^2}{2Y} (L^2 - x^2) dx = \frac{\rho \omega^2}{2Y} \left[ L^2 x - \frac{x^3}{3} \right]_0^L = \frac{\rho \omega^2}{2Y} \left( L^3 - \frac{L^3}{3} \right) = \frac{\rho \omega^2}{2Y} \left( \frac{2L^3}{3} \right) = \frac{\rho \omega^2 L^3}{3Y}$.

$A$ uniform rod of length $L$ is rotated in a horizontal plane about a vertical axis through one of its ends. The angular speed of rotation is $\omega$. Find the increase in length of the rod, if $\rho$ and $Y$ are the density and Young's modulus of the rod respectively.

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