A rod of mass m length l and modulus Y is rotating about an axis passing through one of its end perpendicular to its length with angular velocity[Eqn]\omega[/Eqn]. Find the strain at the center of the rod.

Dear Pratik,

    We know that centrifugl force= mass * radius * angular velocity²

so we get F = m * l/2 * ω²

radius = l/2 beacuse the centre of mass of the rod is at l/2 distance from one the end.

we know that Y= longitudinal stress/longitudinal strain

long strain = long stress/ Y

    Δl/l           = F/ (A*Y)

        Where A is cross sectional area of the rod....

So there is some data still missing..

I hope this may help u to some extent...

I suggest u to post ur questions properly...

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Adapa Bharath

I forgot to mention it........the cross sectional area is A. But the 4 options are.......

1) (mω²L)/(4AY)

2) (3mω²L)/(4AY)

3) (mω²L)/(8AY)

4) (3mω²L)/(8AY)