Question

A horizontally oriented copper rod of length $l=1.0m$ is rotated about a vertical axis passing through its middle. The angular speed in $rps$ at which this rod ruptures is 10x. The value of x is :

Answer 1

Let us consider an element of rod at a distance $x$ from its rotation axis (figure shown below). From Newton's second law in projection form directed towards the rotation axis
$-dT=\left(dm\right){\omega }^{2}x=\frac{m}{l}{\omega }^{2}xdx$
On integrating
$-T=\frac{m{\omega }^2}{l}\frac{x^2}{2}+C$ (constant)
But at $x=\pm \frac{l}{2}$ or free end, $T=0$
Thus $0=\frac{m{\omega }^2}{2}\frac{l^2}{4}+C$ or $C=-\frac{m{\omega }^{2}l}{6}$
Hence $T=\frac{m{\omega }^2}{2}(\frac{l}{4}-\frac{x^2}{l})$
Thus $T_{max}=\frac{m{\omega }^{2}l}{6}$ (at mid point)
Condition required for the problem is $T_{max}=S{\sigma }_m$
So, $\frac{m{\omega }^{2}l}{6}=S{\sigma }_m$ or $\omega =\frac{2}{l}\sqrt{\frac{2{\sigma }_m}{\rho }}$
Hence the sought number of rps
$n=\frac{\omega }{2\pi }=\frac{1}{\pi l}\sqrt{\frac{2{\sigma }_m}{\rho }}$

Substituting the given values, we get $n=80$