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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.

23 Points
11 years ago

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

All the best  !!!

Regards,

Pratik Somani
16 Points
11 years ago

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)