# Will two trains running along the equator in opposite direction experience same wear out?

Two identical trains, at the equator start travelling round the world in opposite directions. They start together, run at the same speed and are on different tracks. Which train will wear out its wheel treads first? Will their weight change?

Answer is given as : the train travelling against the spin of the earth. This train will wear out its wheels more quickly because the centrifugal force is less on this train.

How do the forces change when the frame of reference is same for both trains?

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Wouldn't one train experience more daylight hours than the other, also? Shouldn't that have some effect via heat exposure? Or, are we only addressing centrifugal force here? –  Iszi Jan 3 '13 at 14:52
In the long run, the daylight hours will be the same for both trains. However, the eastbound train will experience more day/night cycles and therefore undergo more thermal wear. This will probably be a much stronger effect than those associated with different centripetal force requirements. –  Johannes Jan 3 '13 at 15:09
They'll get wet, though! –  Emilio Pisanty Jan 3 '13 at 23:57

As the question is probably more about reference frames, than actual wearing out of train wheels, i'll try to answer it.

The reference frame is not intertial, but rotating reference frame. There is an centrifugal force associated with movement of the reference frame: $F=\frac{mv^2}{r}$.

Assuming non rotating reference frame, there is an centrifugal force acting on both trains, but when one train moves in one direction and the other in other direction, their speeds sum with earths rotation speed with different signs: $\tilde{v}_1=v_e+v_t$ and $\tilde{v}_2=v_e-v_t$. ($v_e$ - trains speed in non-moving reference frame because of earths rotation, $v_t$ - trains speed against rotating reference frame.)

This means the force acting on trains in the direction of earths center will be:

$F_1 = mg - \frac{m(v_e+v_t)^2}{r}$ and $F_2 = mg - \frac{m(v_e-v_t)^2}{r}$. And as $F_1 \neq F_2$, wheels will wear out differently.

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The Earth's equatorial radius is about 3963 miles (6378 km) (see this) and the velocity at the equator due to the rotation is about 1040 miles per hour (1674 km/h) (see this). The centripetal acceleration formula is $v^2/r$ which gives a centripetal acceleration of $0.111 feet/sec^2 (0.0339 m/s^2)$ at the equator (see this link to see how Wolfram|Alpha makes it easy to calculate quantities like this). Therefore this effect decreases the acceleration of gravity at the surface of the earth which is nominally $32.2 feet/sec^2 (9.815 m/s^2)$. Thus the force of gravity is 0.3% weaker at the equator due to the rotation of the earth.

If we can ignore the prevailing winds for the two trains, then the change in the force of gravity for the two trains and thus the change in the wear of the wheels will be due to the different centripetal accelerations due to their different speeds. Assume the trains travel at 100 miles per hour (160.9 km/h), then the differences in their centripetal accelerations will be $0.134$ versus $0.0909 feet/sec^2$ ($0.0407$ versus $0.0277 m/s^2$) which means that the difference in the net gravitational forces experienced by the two trains will only be about 0.1% and it will be smaller for the train traveling in the direction of rotation of the Earth - this is a very small difference.

Thus it would be very difficult to do this experiment in the real world since the trains themselves are likely to differ by more than 0.1% and In addition, the differences in the prevailing winds is likely to be much more significant than this small effect.

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I would have liked to see SI units here. Remember that there are only 3 countries which don't have adopted SI: Burma, Liberia, and the US. (Should we count the US as a developing country henceforth?) –  stevenvh Mar 5 '13 at 10:36
@stevenvh - Thanks for the great suggestion, I added the SI units. –  FrankH Mar 6 '13 at 6:53