Question

[Key Concept: The tension at a section supplies the centripetal force for the whole outer part of the rod; strain = stress/Y integrated along the rod gives the elongation.]

a)

$$\begin{matrix} & T = \frac{m}{2}\omega^{2}r = \frac{m}{2}\omega^{2}\left( \frac{\mathcal{l}}{2} + \frac{\mathcal{l}}{4} \right) \\ & \ = \frac{3\text{ }m\omega^{2}\mathcal{l}}{8} = \frac{3\left\lbrack \rho(\text{ }A\mathcal{l})\omega^{2}\mathcal{l} \right\rbrack}{8} = \frac{3 \times 10^{4} \times 2 \times 10^{- 6} \times 1 \times 16 \times 10^{4}}{8} = 1200\text{ }N \end{matrix}$$

b)

$$\begin{matrix} & T = \left\lbrack \frac{m}{\mathcal{l}}(\mathcal{l} - r) \right\rbrack\omega^{2}\left\lbrack r + \frac{\mathcal{l} - r}{2} \right\rbrack \\ & \ = \frac{m(\mathcal{l} - r)}{\mathcal{l}}\omega^{2}\left( \frac{r + \mathcal{l}}{2} \right) = \frac{m\omega^{2}\left( \mathcal{l}^{2} - r^{2} \right)}{2\mathcal{l}} = \frac{\rho A\mathcal{l}\omega^{2}\left( \mathcal{l}^{2} - r^{2} \right)}{2\mathcal{l}} \\ & \text{ }T = \frac{10^{4} \times 2 \times 10^{- 6} \times 16 \times 10^{4}\left( 1 - r^{2} \right)}{2} = 1600\left( 1 - r^{2} \right) \end{matrix}$$

(c) stress $= \frac{T}{A} = \frac{1200}{2 \times 10^{- 6}} = 6 \times 10^{8}$

d)

$$\begin{matrix} & \frac{T}{A} = Y\frac{d\mathcal{l}}{dr} \\ & \ \int_{}^{}\ \frac{T}{AY}dr = \int_{}^{}\ d\mathcal{l} \\ & \ \int_{0}^{\mathcal{l}}\mspace{2mu}\mspace{2mu}\frac{1600\left( 1 - r^{2} \right)}{2 \times 10^{- 6} \times 2 \times 10^{11}}dr = \Delta\mathcal{l} \end{matrix}$$

a) & d)

$$\begin{matrix} & 4 \times 10^{- 3}\left( \mathcal{l} - \frac{\mathcal{l}^{3}}{3} \right) = \Delta\mathcal{l} \\ & \Delta\mathcal{l} = \frac{8}{3} \times 10^{- 3}\text{ }m = \frac{8}{3}\text{ }mm \end{matrix}$$