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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.)
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Solution
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`
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