# 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.

Dr element of the rod is at a distance of r from the centre. We assume Tr and T(r+dr) to be the tensions at points A and B.

Dr has a centrifugal force acting on it equal to T(r+dr) - T(r)

Centrifugal force = -dT in an outward direction.

Centripetal force as a result of the rotation,

Hence,

If u is the mass per unit length, we have

On integrating both sides we get,

tension in rod,

$-T(r)=\frac{uw^{2}}{2}(l^{2}-r^{2})$ --------------------------(1)

Y = stress/strain =

Since the extension is on both sides, total extension =

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