Can π be contracted in relativistic situation?

 `\pi`, a universal constant I guess, is defined as ratio of circumference and diameter of a circle in Euclidean geometry. But...

The Problem:

A disk(2D) of radius `R` purely rotates about its axis with the speed of `v(=R \omega)` which is comparable to `\c`. Now find the ratio of circumference and diameter of the disk in relativistic situation. 

Please neglect the centrifugal force. So it is an gedanken experiment.

My Solution:

I am smart enough that I am able to recognise that the length along the radius will not change at all as the length is perpendicular to velocity by STR.

So,  `\frac{Circumference}{diameter}=\frac{{2\pi R}/\gamma}{2R}`

 As length will be contracted for the moving observer.

`\implies \frac{Circumference}{diameter}= \frac{\pi} {\gamma}`

Oh my god! Lorentz factor can also change the constant.

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