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A rod of 1.5m length and uniform density...

A rod of `1.5m` length and uniform density `10^(4)kg//m^(3)` is rotating at an angular velocity `400 rad//sec` about its one end in a horizontal plane. Find out elongation in rod.
Given `y=2xx10^(11)N//m^(2)`

Text Solution

Verified by Experts

mass of shaded portion
`m'-(m)/(l)(l-x) [` where `m=` total mass `rhoAl]`
`T=m'omega^(2)[(l-x)/(2)+x] rArrT=(m)/(l)(l-x)omega^(2)((l+x)/(2)) T=(momega^(2))/(2l)(l^(2)-x^(2))`

this tension will be maximum at `A((m omega^(2)l)/(2))` and minum at 'B' (zero), elongation in element of width `dx'=(Tdx)/(Ay)`
Total elongation
`delta=int(Tdx)/(Ay)=underset(0)overset(l)int(m omega^(2)(l^(2)-x^(2)))/(2lAy)dx`
`delta=(momega^(2))/(2lAy)[l^(2)x-(x^(3))/(3)]_(0)^(l)=(momega^(2)xx2l^(3))/(2lAyxx3)=(momega^(2)l^(2))/(3Ay)=(rho Alomega^(2)l^(2))/(3Ay)`
`delta=(rhoomega^(2)l^(3))/(3y)=(10^(4)xx(400)xx(1.5)^(3))/(3xx2xx10^(11))=9xx10^(-3)m=9mm`
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