A uniform rod of length `l`, mass `m`, cross-sectional area `A` and Young's modulus `Y` is rotated in horizontal plane about a fixed vertical axis passing through one end, with a constant angular velocity `omega`. Find the total extension in the rod due to the tension produced in the rod.

Options

A. `(momega^2L^2)/(AY)`
B. `(momega^2L^2)/(2AY)`
C. `(momega^2L^2)/(3AY)`
D. `(2momega^2L^2)/(2AY)`

Answer

C

Explanation

Consider a small element of length dx at a distance x from the axis of ratation.
Mass of element, `dm=(m)/(L)dx=mudx`
Where `mu=(m)/(L)`
The centripetal force acting on the element is
`dT=dmomega^2x=muomega^2xdx`
As this force is provided by tension in the rod (due to elasticity), so the tension in the rod at a distance x from the axis of rotation will be due to the cetripetal force due to all elements between `x=x` to `x=L`
`T=int_x^(L)muomega^2xdx=(muomega^2)/(2)[L^2-x^2]` .(i)
Let `dl` be increase in length of the element, then
`Y=((T)/(A))/((dl)/(dx))`
`dl=(Tdx)/(YA)=(muomega^2)/(2YA)[L^2-x^2]dx` .[using (i)]
So the total elongation of the whole rod is
`l=int_0^(L)(muomega^2)/(2YA)[L^2-x^2]dx`
`=(muomega^2)/(2YA)[L^2x-(x^3)/(3)]_0^L`
`=(1)/(3)(muomega^2L^3)/(YA)=(1)/(3)(momega^2L^2)/(YA)`