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

The stress is zero at the free surface end and maximum at the axis. Therefore the rod will ruputures at the middle. Let us consider an element of rod at a distance `x` from the axis, the mass of element `dm = rho Adx`, Applying Newton's second law
`F-(F+dF) = (dm)a_(n), -dF = (rho Adx) omega^(2) x`
`int - dF = int (rho A d x) omega^(2) x, F = -rho A omega^(2) (x^(2))/(2) + C`
at `x = (l)/(2), F = 0 therefore C = (rho A omega^(2) l^(2))/(8)`
Now `F = rho (A omega^(2))/(2) ((l^(2))/(4) - x^(2))`
at `x = 0, F = rho (A omega^(2) l^(2))/(8)` and
Stress `S = (F)/(A) = (rho omega^(2) l^(2))/(8)`
Rupture of rod will occur when `S = omega`,
`therefore (r omega^(2) l^(2))/(8) = omega rArr = sqrt((8sigma)/(rho l^(2))), n = (1)/(2pi) sqrt((8sigma)/(rho l^(2)))`