A uniform rod of length L is rotated in a horizontal plane about a vertical axis through one of its ends. The angular speed of rotation is `omega` . Find increase in length of the rod, if `rho ` and Y are the density and Young's modulus of the rod respectively,

Options

A. `(rho ometa ^(3) Y)/(4L^(2))`
B. `(4 rho omega ^(2) L ^(3))/(3Y)`
C. `(rho omega ^(2) L ^(3))/(3Y)`
D. `(rho omega ^(3) L^(3))/( 8Y)`

Answer

C

Explanation

A uniform rod is rotated in a horizontal plane about a vertical axis through one of its ends.
Length of rod = L
Angular speed of rotation `= omega`
Density of rod `= rho`
Young's modulus of rod `= Y`
The expression for Young's mondulus is given by,
`Y= ("Stress")/("Strain")`
`Y= ((dF)/(A))/((dl)/(x))`

`Y = (dF.x)/(dl.A) `
`dF = (Y.A dl )/(x) " "...(1)`
But, centrifugal force due to mass is given by,
`dF = dm, x omega ^(2)`
`=(mdx)/(L)x omega ^(2) " "...(2)`
From equations (1) and (2),
`(YAdl)/(x) = (mdx)/(L) x omega ^(2)`
`dl = (mdx. x omega ^(2))/(L.Y.A)`
Integration on both sides,
`int dl = int_(a) ^(L) (mdx x ^(2) omega ^(2))/(L.Y.A ) `
`l = (m omega ^(2))/(L.Y.A) int _(a ) ^(1) dx x ^(2)`
` = ( m omega ^(2))/(L.Y.A) int _(a) ^(L ) dx x ^(2)`
`= (momega ^(2))/(L.Y.A) [(x ^(3))/(3) ] _(0) ^(L)`
`= ( m omega ^(2) L ^(3))/( 3L. Y.A)`
`= (m)/(AL ) (omega ^(2) L ^(3))/(3Y)`
Increase in length of rod `= (rho omega ^(2) L ^(3))/(3Y)`