Question

A steel rod of length 2l, cross sectional area A and mass M is 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. (Assume the rod is uniform.)

Answer 1

Consider an element of width dr at t 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 = ω²rdm  ....(Where ω is the angular velocity of the rod)

= ω²rμdr   .....(∵ μ = mass/length)

⇒ T(r) – T(r + dr) = μω²rdr

⇒ – dT = μω²rdr  ......[∵ Tension and centrifugal forces are opposite]

∴ `- int_(T = 0)^T dT = int_(r = l)^(r= r) μω^2rdr`  ......[∵ T = 0 at r = l]

⇒ `T(r) = (μω^2)/2 (l^2 - r^2)`

Let the increase in length of the element dr be Δr

So, Young's modulus Y = `"Stress"/"Strain" = ((T(r))/A)/((Δr)/(dr))`

∴ `(Δr)/(dr) = (T(r))/A = (μω^2)/(2YA) (l^2 - r^2)`

∴ `Δr = 1/(YA) (μω^2)/2 (l^2 - r^2)dr`

∴ Δ = Change in length in right part = `1/(YA) (μω^2)/2 int_0^l (l^2 - r^2) dr`

= `(1/(YA)) (μω^2)/2 [t^3 - l^3/3]`

= `1/(3YA) μω^2l^2`

∴ Total change in length = 2Δ = `2/(3YA) μω^2l^2`