A horizontally orientied unifrom copper rod of length `l` is rotating about a vertical axis passing through its centre. Calculate the rotated frequency at which the rod ruptures. Breaking or rupture strength of copper is `sigma` and density of copper is `rho`.

Explanation

Let us consider an element of rod at a distance x form its rotation axis (figure). From Newton's second law in projection form directed towards the rotation axis
`-dT=(dm)omega^2x=m/lomega^2xdx`
On integrating
`-T=(momega^2)/(l)(x^2)/(2)+C` (constant)
But at `x=+-l/2` or free end, `T=0`
Thus `0=(momega^2)/(2)l^2/4_C` or `C=-(momega^2l)/(8)`
Hence `T=(momega^2)/(2)(l/4-x^2/l)`
Thus `T_(max)=(momega^2l)/(8)` (at mid point)
Condition required for the problem is
`T_(max)=Ssigma_m`
So, `(momega^2l)/(8)=Ssigma_m` or `omega=2/lsqrt((2sigma_m)/(rho))`
Hence the sought number of r p s
`n=(omega)/(2pi)=(1)/(pil)sqrt((2sigma_m)/(rho))` [using the table `n=0*8xx10^2rps`]