A ring of radius `r=25cm` made of lead wire is rotated about a stationary vertical axis passing through its centre and perpendicular to the plane of the ring. What is the number of rps at which the ring ruptures?

Explanation

Let us consider an element of the ring (figure). From Newton's law `F_n=mw_n` for this element, we get,
`Tdtheta=((m)/(2pi)dtheta)omega^2r`
So, `T=(m)/(2pi)omega^2r`
Condition for the problem is:
`(T)/(pir^2)lesigma_m` or, `(momega^2r)/(2pi^2r^2)lesigma_m`
or, `omega_(max)^2=(2pi^2sigma_mr)/(pir^2(2pirrho))=(sigma_m)/(rhor^2)`
Thus sought number of rps
`n=(omega_(max))/(2pi)=(1)/(2pir)sqrt((sigma_m)/(rho))`
Using the table of appendices `n=23rps`