InterviewSolution
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InterviewSolution
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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? |
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Answer» SOLUTION :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` 
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