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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...
Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation. All the best !!!Regards,Askiitians ExpertsAdapa 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)
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