Relativistic Rotating Disk

The relative velocity at any point is given by $\omega R$ Hence the circumference will have a lorentz expansion given by

$\sqrt { 1-\frac { { (\omega R) }^{ 2 } }{ { c }^{ 2 } } }$

But the radius won't undergo any expansion because it is perpendicular to the direction of motion Thus the ratio becomes

$\frac { 2\pi R\sqrt { 1-\frac { { (\omega R) }^{ 2 } }{ { c }^{ 2 } } } }{ R } =2\pi \sqrt { 1-\frac { { (\omega R) }^{ 2 } }{ { c }^{ 2 } } } =\quad 3.47$

Okay, right, if one measures the radius of the spinning disk, it's still 1 m, but if one measures the perimeter, it's contracted. But again, that really depends on how one attempts to measure the perimeter. A lot of relativistic paradoxes or problems stem from inconsistent protocol for measurement. For instance, if a shadow were to be cast directly below the spinning disk of "radius 1 m but perimeter 3.47 m", and then use a measuring tape to find the perimeter of the shadow, it's still 6.28 m, not 3.47 m. It would be wonderful if such an experiment could be done.

The formula for relativistic length contraction is L'=L/(Gamma), where: L' is the length contraction due to relativistic effects L is the original length Gamma is the Lorentz Transformation factor, which is equal to 1/sqrt(1-(v/c)^2) C is the speed of light The ratio of the disk's circumference to its radius will be L'/r Evaluating L', substitute r times the angular velocity for v in the Gamma expression and evaluate. L'/r can then be expressed as (2 pi) sqrt(1-((r*omega)/c)^2) Plugging in the numbers, the ratio is found to be 3.473

Bonus Thought: I'm not sure if this is correct but I believe Einstein's mass-energy equivalence is E=mc^2(gamma-1). As the angular velocity approaches a value comparable to the speed of light, mass is uniformly lost to provide for the energy necessary to keep the velocity at its established value. Because mass is lost, the object shrinks uniformly.

If we apply the so called Lorentz transformations then we find the ratio= 3.47.But here I have a question. The idea of rigid body is meaningless in relativity since no signal can propagate with infinite speed. So this answer can't be right.

the lenghts parallel to the motion of the disc will undergo length contraction. but the radius,being perpendicular to the rotational motion will remain as such for the observer. hence using the relativistic formula, we get the ratio as ratio=\frac { 2\pi r\sqrt { 1-{ (\frac { \omega r }{ c } ) }^{ 2 } } }{ r } =2\pi \sqrt { 1-{ (\frac { \omega r }{ c } ) }^{ 2 } } =3.473161359 after plugging in the variables Note that when \omega r<<c,this ratio tends to its classical value of 2\pi