Obviously, if it is fastened, its radius is not changing. But, suppose that there is no friction in the corridor. What will happen to the radius? The corridor acts as an arm that will push the object perpendicularly to itself.

As Lewin explains earlier in this lecture

Not only absolute speed changes, but direction also because, as arm turns, it will push our object right ahead the radius with the speed of corridor in that point, v = w*R, so that object never stays behind the corridor's wall. Particularly, our object will acquire the radial component very soon, in the next moment of time.

Pic. 2 |

That is, I think that corridor arm will keep the tangent component of object's velocity, v

_{y}= W*R(t), constant. Radial component, v

_{x}= v * sin(w*R1, v) stays intact after interaction with corridor:

Pic. 3 |

Here, angle

*ɑ*denotes relationship between vertical and horizontal component

*before interaction*with the corridor wall. Angle

*Ɣ*denotes the same relationship between vertical and horizontal component

*after interaction*with the arm. We then give infinitesimal time Δt for the object to move freely. Then, interaction happens again. On the next time step, W*R2 will have a horizontal component to the left (see Pic. 2) and object will go after the walls, never staying behind them.

For the process to start up, initial R=R0 must not be zero. V0 can be either zero or a vector in any direction.

In discrete time simulation, I first correct the direction,

*ɑ*

*→ Ɣ.*Next, time is given an increment and radius is updated. With arm rotated β°, position of the object moving at speed V1 can be found from the following picture (we are interested only in next radius, R2).

The system of equations

(1) R

_{2}

^{2}= (R

_{1}+Δx)

^{2}+ Δy

^{2}

(2) Δx/Δy=tg(Ɣ)

(3) (R

_{1}+Δx)/Δy = ctg(β)

Can be resolved wrt R

_{2}= R

_{1}/(ctg(β)-tg(Ɣ))*sqrt(1+ctg

^{2}(β)). This gives the new tangent component, V2

_{y}= W*R

_{2}. At the same time we can compute the new angle, ɑ

_{2}between new the new radius and V1, by summing β and Ɣ computed in the previous iteration.

Given current velocity V1 and tangent velocity in R2, we can "correct" the speed again, the same way as described in Pic. 3 (radial component of v1 plus W*R2):

Making Δt smaller and smaller, we should be able to approximate the true, continuous, acceleration. In Octave, I have achieved the converging results at 250 steps per round of corridor rotation. In all simulations, we have the same initial radius and constant W.

v1=0 |

v1=5, a1=60° |

v1=5, a1=99.9° |

Lets have initial speed of the order R*W. On the 2nd diagram, initial angle is 60 degrees from vertical. The numbers have doubled. Making angle 99.9 gives the 3rd picture. We see that effect of starting in the direction right away from the center does not add up very much.

Increasing initial speed ten times while keeping initial direction away from the center,

v1=50, a1=99.9 |

v1=500, a1=99.9 |

we first mention the angle overshoot now: the horizontal component is much larger than vertical initially. But, as radius increases, it returns to the stable 45 deg. The speed and radius respond with 5-fold increase per lap. Additional 10x increase in initial speed, adds 10x speed increase after a loop of acceleration.

Using orders of magnitude more discretization dots per round, does not change the numbers. I still could not express the system in the form of differential equations to compute the exact trajectory. Does the problem have a shortcut solution?

Computations are animated below (JavaScript + HTML Canvas). The arrow origin is the location of our astronaut. The arrow size and direction his velocity (with respect to space station frame).

You can see that angle stabilizes at 45° and speed and radius grow exponentially.

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