Relativistic centripetal force

Because we use magnetic fields to bend the path of particles in accelerators and E&M is Lorentz invariant by construction, we just apply the bending radius in a magnetic field equation in the lab frame and never bother to compute the force. The radius of curvature is

$$ R = \frac{p}{qB} .$$

Note that for a ring like the LHC, the bending is not actually uniform, but only in the bending magnets, but not in the quadrupoles or cavities (if any), so it is locally tighter than you would get by naively apply the above equation with a 27 mile radius.

If you insist on finding the force you'll note that over the course of one cycle the momentum changes by $2\pi |p|$, and it takes $(27\text{ km})/c$ seconds to get there so the mean force is

$$ F = \frac{p c}{r} = \frac{2 \pi p c}{27 \,\mathrm{Km}}. $$

Again, in any given magnet it will be larger by a factor of less than 10 because the magnets don't cover the whole beam line.

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Aside: If you spend much time doing particle physics you'll come to love ultra-relativistic mechanics: it's even easier than non-relativistic mechanics.

I'm not someone who inderstands the physics well, so I'm not 100% sure of this. Comments on this will be greatly appreciated. (Update: This approach seems to be correct)

F=dp/dt works as well as $v^2/r$. As long as you want to spin it at a constant velocity, your lorentz factor will be constant, and since $p=\gamma m_0v$, your force will become $m_0\gamma d\vec{v}/dt$. Then you can solve it normally using vectors (exact same proof a the classical one for CPF). Your end result will be $\gamma m_0 v^2/r$.

When dealing with accelerations due to forces, again use $F=d(\gamma m_0 v)/dt$. If $m_0$ is constant, then we get $F=\frac{m_0a}{(1-\frac{v^2}{c^2})^\frac{3}{2}}=\gamma^3m_0a$. Since we still have $\gamma$, the accelerations will be very much different.

So if a proton has twice the energy of another, the proton has twice the $\gamma$ (from $E=\gamma m_0c^2$). So, the acceleration will be eight times less.