A stell rod of length 2l corss sectional area A and mass M is set rotating in a horizotnal plane about an axis passing through its centre and perpendicular to its length with constant angular velcoty `omega.` If Y is the Young's modulus for steel, find the extension in the lenght of the rod. (Assume the rod is uniform.)

Explanation

Consider an element of width dr at r as shown in the diagram.
Let T(r ) and T (r + dr) be the tensions at r and r + dr respectively.
Net centrifugal force on the element `= omega^(2)` rdm (where `omega` is angular velocity of the rod)
`" "=omega^(2) rmudr " "(becausemu = "mass"//"length")`
`implies " "T(r ) = T (r+dr)=mu omega^(2)rdr`
`implies " "-dT = mu omega^(2) rdr`
`implies" "[because" Tension and centrifugal forces are opposite"]`
`therefore " "-underset(T=0)overset(T)intdT = underset(r=l)overset(r=r)int muomega^(2)rdr" "[because T = 0 " at "r = l]`
`implies " "T(r ) = (mu omega^(2))/(2) (l^(2)-r^(2))`

So, Young's modulus Y `=("Stress")/("Strain")=(T(r )//A)/((Deltar)/(dr))`
`therefore" "(Deltar)/(dr)=(T(r ))/(A) = (mu omega^(2))/(2YA)(l^(2)-r^(2))`
`therefore " "Deltar = (1)/(YA) (mu omega^(2))/(2) (l^(2)-r^(2))dr`
`thereforeDelta` = change in length in right part `=(1)/(YA) (mu omega^(2))/(2)underset(0)overset(l)int(l^(2)-r^(2))dr`
`" "=((1)/(YA))(muomega^(2))/(2)[l^(3)-(l^(3))/(3)]=(1)/(3YA)muomega^(2)L^(2)`
`therefore" "` Total change in length `=2Delta = (2)/(3YA)mu omega^(2)l^(2)`