The Newtonian centripetal force is well known as: F_{N} = m r ω^{2} = m v^{2}/r , which is the force needed to move a mass m in a circle with radius r at an angular velocity of ω.
Let mass m be at the end of a long string with negligible own mass and length r. Anchor the string at the origin of an inertial coordinate system in free space. Without altering the string length, insert two small force gauges (of negligible mass) inline, one at the 'origin end' and one at the 'mass end'. Sling the mass around the origin. According to Newton, the gauges will both measure the centripetal force as given above. Now sling the mass at a relativistic transverse velocity (v=rω). What happen to the readings on the two gauges (given that the string doesn't break)?
One can halfexpect that they will record higher values than given by the above Newtonian formula. What is very surprising is that Special Relativity (SR) predicts that the gauge at the 'mass end' will record a higher force (F_{m}) than the gauge at the 'origin end' (F_{o}). This is despite the fact that they measure forces in the same ('massless') string!
I will give the SR formulas first and then discuss how it works.
F_{o} = m v^{2}/[r √(1  v^{2}/c^{2})] and F_{m} = m v^{2}/[r (1  v^{2}/c^{2})]
Note the square root below the line for F_{o}. That's the normal relativistic velocity time dilation factor. For F_{m} there are hence two time dilation factors below the line. For readability, replace 1/√(1  v^{2}/c^{2}) with γ, the Lorentz factor, so that:
F_{o} = γ m v^{2}/r and F_{m} = γ^{2}m v^{2}/r
As a numerical example, let's use 1 kg at the end of a 1 km long string and swing it 1% of c, or 3,000,000 m/s, giving γ = 1.00005. If my sums are correct, the forces will be (in kiloNewton):
F_{N} ≈ 9,000,000 kN
F_{o} ≈ 9,000,450 kN
F_{m} ≈ 9,000,900 kN
OK, there are surely no strings that can take these forces, but it illustrates the relative values. The 'handwaving' argument for the difference between F_{o} and F_{m} is simply that the gauge at the mass moves at relativistic speed relative to the gauge at the origin, so it measures a higher force (by the time dilation factor). But, that's not good enough  some clarifications are needed.
To start with, measuring force essentially means measuring acceleration. Unitwise, acceleration is a distance interval divided by the square of a time interval. As in the famous 'twinparadox', a clock moving noninertially (accelerated) is time dilated or slowed down by a factor γ relative to an inertial clock. This means the circling clock measures the said time interval as too short by this factor. Square that and we have the reason why the massgauge will experience an acceleration that is a factor γ^{2} larger than what Newton would have predicted.
Secondly, why would the origingauge register a force of only a factor γ larger than the Newton value? We now have to think 'moving mass' for the circling object. As far as the origin observer is concerned, the 1 kg mass now weighs γ kg and would hence need a force γ times what Newton predicts. This is smaller than the γ^{2} times Newton that the massgauge registers. This is still not very rigorous, but it gives the general idea.
Confusing, but true according to SR; and who would wish to argue with Einstein on that?
J
Notes.
1. The whole thing is so counterintuitive: if the string had a small, but significant mass, Newton would predict that the gauge at the massend will record a smaller force than the one at the origin. A relativistic treatment becomes very complex in such a case, but normally the mass at the end will dictate and the endgauge will still measure a larger force than the one at the origin.
2. Relativity 4 Engineers appendix C treats relativistic accelerations in a general way, but also leads up to fairly technical discussions.
3. If you have the inclination, here is a paper by Hrvoje Nikolic, with a seriously technical treatment of relativistically rotating frames.

Re: Relativistic Sling Forces