Question

2. It is well-known that living on the inside surface of a cylinder of radius R when rotating at a rate of o about its axis in the outer-space, can simulate the same gravitational g, as living on earth if Ro2-g Show that, inside such a cylinder, when projectiles are shot in any arbitrary direction at low speeds (v<< Ro) and to low altitudes at nearby points (Ar<R), the equation of motion of such trajectories (as observed by a an observer in the rotating frame) are identical to those for a similarly limited projectiles as observed on earth. Hint: You may start with equation 4.89 in the text book which describes the space acceleration of an object versus acceleration observed in the rotating body. Solve for ar and show that the sum of the three terms leads to the same acceleration as on earth for a projectile shot in an arbitrary direction. Note that the origin of the rotating coordinates is best placed at the projectile launch point, with the z,-axis pointing to the center and xr-axis tangent, and yraxis along the axis of cylinder ZI yr body

Answer 1

When projectiles are thrown with $\Delta r<<R$ , it basically can be imagined like this:

Let this condition be achieved by making R very large so that when we are standing on the inner surface of the cylinder, the surface on which we are standing essentially is a flat surface.

Now, since we are standing on the inner surface inside , we will feel a downwards force. (This is actually the centrifugal force, that is directed radially outwards.)

Since we made Rw^2 =g, this downwards force will feel just like the Gravitational force of attraction of earth that produces an acceleration(constant) on all objects near the surface of the cylinder.

There is no math required anymore here. We have just elaborated , without equations how the situation is the same as on earth.