A steel uniform rod of length 2L cross sectional area A and mass Mis set rotating in a horizontal plane about an axis passing through the centre. If Y is the Young's modulus for steel, find the extension in the length of the rod.

Explanation

Let us consider an element of width dr at a distance r from the given axis of rotation as showin in the diagram.
Suppose tention in the rod at r and dr distance are `T(r ) and T (r + dr)`
Centripetal force acting on small length element
`F = dm r omega ^(2) " "[because F _(C) = mr omega ^(2) ]`
`therefore F = mu r omega ^(2) dr " "...(1)`
`[because mu = (dm )/( dr) implies dm = mu dr ]`
`therefore T (r) -T ( r + dr) =mu r omega ^(2) dr `
`[ because T(r) -T (r + dr) =` resultant force F]
`therefore -dT = mu r omega ^(2) dr`
Centripetal force and tension force are opposite to each other hence negative sign present.
Integrating on both side.
`int _(T=T) ^(T=0) dT = mu omega ^(2) int _(r ) ^(l) r dr`
` -[T] _(T) ^(0) = mu omega ^(2) [ (r ^(2))/( 2 )] _(r ) ^(l)`
`therefore -[0-T] = (mu omega ^(2))/( 2) [ l ^(2) - r ^(2) ]`
`therefore T = ( mu omega ^(2))/( 2) (l ^(2) -r ^(2)) " "...(2)`
Now if `Delta r` is the extension in the elemente of length dr
`therefore` Young.s modulus `Y = (T (r))/(A) xx (dr )/(Delta r)`
`therefore (Delta r )/( dr ) = (T (r))/(YA) = (T)/(YA)`
`therefore (Delta r )/(dr) = (mu omega ^(2))/( 2 YA) (l ^(2) - r ^(2))`
`therefore Delta r = ( mu omega ^(2))/( 2 YA) int (l ^(2) -r ^(2)) dr`
Integrating on both side,
`int Delta r = (mu omega ^(2))/( 2 YA) int ( l ^(2) -r ^(2)) dr`
`therefore Delta r = (mu omega ^(2))/( 2 YA) [ l ^(2) r - (r ^(3))/(3)]_(0) ^(l)`
`therefore Delta r = (mu omega ^(2))/( 2 YA) [[ l ^(3) - (l ^(3))/(3)]`
`therefore Deltar= (mu omega ^(2))/(2YA)xx(2l^(3))/(3)``therefore Delta r = ( mu omega ^(2) l ^(3))/( 3YA)`
`therefore ` Extension of `Delta r` in both side length, then tota extension `=2 Delta r`
`= ( 2 mu omega ^(2) l ^(3))/(3YA)`