A copper rod of length L is suspended from the ceiling by one of its ends. Find the elongation ΔL of the rod due to its own weight.
Solution:
Let m be the mass of the copper rod, g be the acceleration due to gravity and A be the cross-sectional area. It is given that L is the length of the rod then assume e to be the elongation produced. If Y be the Young's modulus of elasticity of copper, then
Y = FL/Ae
e = FL/AY
Further, F = mg and centre of mass of the rod is at the centre, so L' = L/2
Therefore, e = mgL/2AY