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Lets say you have a wheel of radius 100m. The wheel has a corridor at its center, such that the top of the corridor is at the center of the wheel, while the bottom of the corridor is at the edge of the wheel. Lets also say the wheel is spinning at constant angular velocity, such that at the edge of the wheel, you experience 1g of centripetal acceleration. If you dropped an apple at the center of the wheel, what velocity would the apple be going at when it hit the edge of the wheel ?

I thought of this question when watching 8.01x Lecture 5 by Walter Lewin. He mentioned at about 31:25 that if you started at the center of the wheel, and tried to go down the corridor, that "it would be suicide". Would you move fast enough for this to occur?

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This is a good question, and can be answered, but a simple numerical answer is misleading. As it turns out, the "ballistics" of objects in a rotating reference frame are not at all like those of falling objects in an ordinary gravitational field.

The centrifugal force on an object in a rotating reference frame is $m\omega^2 r$.

In addition, there is a Coriolis force on moving objects. (There is no gravitational equivalent to this; ballistics in a rotating reference frame are really quite unlike those in normal gravity.) This force will cause the apple to hit the side of the corridor rather than fall straight down. The Coriolis force does no work, so if the apple's interactions with the sides of the corridor are frictionless and perfectly elastic, they can be ignored in determining the final kinetic energy (and thus velocity) of the apple. However, they cannot be ignored in determining the direction of its motion, which I'll discuss below.

Integrating the centrifugal force along the whole path gives the work done. If $R$ is the radius of the wheel: $$\Delta E = \int_0^R m \omega^2 r\,dr = \frac{1}{2} m \omega^2 R^2.$$

The final velocity is then $\omega R$, or 30 m/s using the 100-meter radius and 0.3 rad/s angular velocity from Lewin's video. This is equivalent to falling from approximately 45 m, if the final velocity is perpendicular to the rim of the wheel. This is definitely lethal for those of us who are not as perfectly elastic as this oddly bouncy apple.

However, the final velocity is not guaranteed to be perpendicular to the rim, and in fact it won't be. First of all, the Coriolis force has caused the direction of the apple's motion to change. Second, the apple has bounced, possibly multiple times, off the sides of the corridors. If the corridor walls are smooth and radial, the impact with the rim should actually be close to tangential, making the eventual collision more like falling off a motorcycle than falling down an elevator shaft: dangerous, and certainly damaging, but not necessarily lethal.

Of course, all of this is premised on the idea that the apple collides elastically and frictionlessly with the corridor walls. In practice, apples tend to collide inelastically with most things. The apple will likely collide inelastically with one wall, then begin to slide down that wall, pinned to it by the Coriolis force. Friction with that wall will slow its "fall." It is an amusing differential equations exercise to determine the final velocity for large $R$ in terms of $\omega$, $R$, and the coefficient of dynamic friction $\mu$.

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  • $\begingroup$ The Coriolis force ($\vec F =2m\vec v \times \vec{\omega}$) looks a lot like $\vec F=m\vec v \times \vec B_g$ from gravity (en.wikipedia.org/wiki/Gravitoelectromagnetism) where $\vec B_g$ circulates around mass current via $\vec{\nabla}\times\vec B_g=-4\pi G\vec J/c^2$. $\endgroup$
    – JEB
    Commented Jun 30 at 1:43
  • $\begingroup$ @JEB Gravitomagnetism is basically the relativistic Coriolis force, but when people think of "gravity" I'd say they don't think about the Lense-Thirring effect and orbital precession. $\endgroup$ Commented Jun 30 at 3:21
  • $\begingroup$ @controlgroup I'm just disagreeing with the statement that there "There is no gravitational equivalent to this", when all fictitious forces, including $\vec F = m\vec g$, are GR manifestations of choosing a coordinate system that is not a geodesic. You examples involve paths, and I am not even going there. $\endgroup$
    – JEB
    Commented Jul 1 at 18:06
  • $\begingroup$ @JEB Yes, I'm agreeing with you. I also disagree with that statement. $\endgroup$ Commented Jul 1 at 19:29

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