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The bar with length l, density r, diametr d, Young's modulus E, Poisson's ratio mu, is spinning around the cross-section, what is the change in the width of this bar?

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This sounds like a homework question... –  Nic Nov 24 '11 at 10:38

1 Answer 1

The answer i arrived at, assuming angular velocity=w, Poisson's ratio=v, area of crossection=A; is:-

Change in diameter = -dvr$w^{2}$$l^{2}$/2E
Additionally, the change in the length of the rod = r$w^{2}$$l^{3}$/2E.

Getting to the answer:-
The diagram

The increase in length of the dx element in the figure is given by( we can get it from the formula E=FL/A(dL):-
dy = dm $w^{2}$l x/A
Putting,
dm = rA dx ,
we get, dy = r$w^{2}$l x dx/E
Integrating it we get,
change in length of rod= y = r$w^{2}$$l^{3}$/2E.

If we put dy in the equation of change in diameter due to axial stretching(as given on Wikipedia ),
change in diameter in dx element = du = -dv dy/l
hence,
du = -dvr$w^{2}$ x dx/E
Integrating we get,
change in diameter = -dvr$w^{2}$ $l^{2}$/2E

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