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  • 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.

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.

Answers (1)

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, dr = dm rw^{2}

Hence, -dT = dm rw^{2}

If u is the mass per unit length, we have

-dT = u w^{2}r.dr

On integrating both sides we get,

-\int_{0}^{T}=dT=uw^{2}\int_{r}^{1}r.dr

tension in rod,

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

Y = stress/strain =

\frac{\frac{T(r)}{A}}{\frac{\delta r}{dr}}

{\frac{\delta r}{dr}}=\frac{T(r)}{AY}=\frac{uw^{2}}{2AY}\left ( l^{2} -r^{2}\right )dr

\int_{0}^{\delta }\delta r=\int_{0}^{l}\frac{(uw^{2})}{2AY}(l^{2}-r^{2})dr

\delta =\left ( \frac{uw^{2}}{2AY} \right )\left ( \frac{2}{3} \right )l^{3}

\delta =\frac{uw^{2}}{3AY}(l^{3})Since the extension is on both sides, total extension =

2\delta =2\frac{uw^{2}}{3AY}(l^{3})

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