I see a lot of questions regarding situations what would happen if the world would stop spinning. This got me to wondering how much energy it would actually take to stop the world from spinning.
[Physics] How much energy would it take to stop Earth’s rotation on its axis
earthestimationhomework-and-exercisesrotational-dynamics
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The acceleration of an object spinning with angular velocity $\omega$ at a distance $r$ is given by:
$$ a = r\omega^2 $$
The angular velocity of the earth is 2$\pi$ radians per day or $7.3 \times 10^{-5}$ per second, and the Earth's radius is $6.378 \times 10^6$ metres, so the acceleration is 0.034 ms$^{-2}$ or 0.0034g. So as a person standing on the surface you wouldn't notice.
However the acceleration does affect the shape of the Earth. Rock is viscous and will flow in response to a force but it does so very slowly. As a result of the Earth's rotation it's radius is about 31km greater at the equator than at the poles. When the rotation stopped it would gradually settle back to a sphere, though it would take a million years or so.
The Milankovic cycles are at leastly partly due to the fact the Earth is not a perfect sphere, and these would stop or at least be changed when the Earth settled back to a sphere. Assuming you believe the Milankovic cycles cause ice ages, one effct of rotationg stopping would be no more ice ages. Having said that, stopping rotation would play havoc with the weather as you'd get no Coriolis force so no jet stream. Mind you, there'd be no hurricanes either.
The kinetic energy of the bullet is $\frac12 mv^2 \approx 1$ kilojoule.
If the deceleration is continuous over $x=1$ meter, energy conservation gives us an acceleration $a = v_\text{initial}^2/2x \approx 65,000\,\mathrm{m/s^2} \approx 6600\,g$, and the stopping time is $t = v_\text{initial}/a \approx 5.6$ milliseconds.
Spreading the bullet's energy over the stopping time gives an average power of 175 kilowatts. If you make some hand-waving assumptions that the mechanism for stopping the bullet is inefficient you might multiply this power by a factor of 10–100.
This is a lot of power! But the time interval is very brief. And it's certainly not prima facie unphysical—after all, the gunpowder explosion that launched the bullet involved the same energy transfer and an acceleration length of much less than a meter.
After some thinking, and a silly mistake, I can make an order-of-magnitude estimate of the magnetic field that would have to be involved.
I would expect that the main effect involved in rapidly stopping a bullet would not be diamagnetism, a small effect where the magnetic field strength inside a "non-magnetic" material is changed in its fourth or fifth decimal place (and thus the energy density of the field $E\propto B^2$ is changed in its eighth or tenth decimal place).
The predominant factor on introduction of a strong magnetic field to a bullet would be eddy currents in the material. Wikipedia gives me a formula for energy loss due to eddy currents in a material, $$ P = \frac{\pi^2 B^2 d^2 f^2}{6k\rho D} $$ where $P$ is the power in watts per kilogram, $B$ is the peak field, $d$ is the thickness of the conductor, $f$ is the frequency, $k$ is a dimensionless constant which depends on the geometry, $\rho$ is the resistivity, and $D$ is the mass density. Wiki gives $k=1$ for a thin plane and $k=2$ for a thin wire, so I wild-guess $k=3$ for a zero-dimensional bullet. Using values for the lead core of the bullet, we find the rate of field change \begin{align*} (Bf)^2 &= \mathrm{ \frac{18}{\pi^2} \frac{2\times10^{-7}\,\Omega\,m \cdot 10^4\,kg/m^3}{(10^{-2}\,m)^2} \frac{2\times10^5\,W}{8\times10^{-3}\,kg} } = \mathrm{10^{9} \frac{N^2}{C^2\,m^2} }\\ {}\\ Bf &= \mathrm{ \pi\times10^4\,T/s } \end{align*}
The simplest assumption about the frequency is that the field is being ramped up to its maximum while the bullet stops, so we've seen a quarter-oscillation and $1/f = 20\,\mathrm{ms}$. This gives us a peak field of 600 tesla, which is large, but not absurdly large.
On the other hand, if Magneto is actually an FM radio broadcaster at 100 MHz, he'd need only a field of $$\mathrm{ \frac{ \pi\times10^4\,T/s }{ 10^8\,Hz } = \pi\times10^{-4}\,T. }$$ I don't think that radio engineers ordinarily think about local peak magnetic field strengths, but this isn't outrageous either. My college NPR station has a 100 kW transmitter. However, their antenna isn't shaped correctly to put that entire power into a one-cc volume.
Best Answer
The rotational kinetic energy of a (uniform) solid sphere rotating about an axis passing through the center of mass is given by $\frac{1}{2}I\omega^{2}$, where $I=\frac{2}{5}MR^{2}$. So $K=\frac{1}{5}MR^{2}\omega^{2}$. Using $M=6\times10^{24}\,\mbox{kg}$, $R=6400\,\mbox{km}$, and $\omega=\frac{2\pi}{T}$, with $T=24\,\mbox{hrs}$, we get $$K\approx2.6\times10^{29}\,\mbox{J}.$$