Rotating Disk and vertical projectile

Question

I have a question regarding classical mechanics and rigid bodies. Suppose we have a rotating disk with the plane of the disk perpendicular to the z-axis. The disk rotates with an angular velocity about the z-axis. A particle that lies along the disk is then projected vertically upward in the z-axis direction away from the plane of the disk. What would the particle experience and what effects would play a role in its trajectory? Consider situations with and without gravity.

I first thought rotating frames so the Coriolis effect, but then also considered centrifugal force. I am thinking that the initial velocity of the particle would include not only a vertical velocity, but also horizontal components due to the disks rotation. We therefore experience coriolis acceleration, but I am not sure where the particle would travel and how it behaves. How would the effect of gravity change either one of these component, and for that matter how would gravity change the situation at all?

Thanks for any response!

[EDIT] If possible, can we work with Coriolis effect even if it is more complicated?

Answer 1

We could consider the problem in a rotating reference frame as you have alluded to. Presumably, we would pick a frame where the disk itself is not moving, and then introduce the fictitious forces which you mentioned. But your scenario is simple enough that we don't need to do all that; once the particle has been launched from the disk, it is no longer under the influence of the disk and can be treated like any other projectile. Therefore the problem is answered in two steps:

1. Find the velocity of the particle at the moment of launch.
2. Solve for the trajectory.

The particle is launched straight up (relative to the disk) at a speed $v_z$, which is known beforehand. Now we need to find $v_x$ and $v_y$. How we find them will depend on what other information we are given initially. If the disk rotates at an angular frequency $\omega$ and the launch position is at a distance $r$ from the disk's center, then the initial speed of the particle's image in the x-y plane will be $r \omega$, and its direction will be tangential to the curve it was following at the moment of launch. The image of the particle in the x-y plane will travel at this constant velocity (neglecting air resistance) until it lands. The height of the particle at any point in time can be found from the expression: $$z(t) = v_z t - 4.9 \text{ m } \text{s}^{-2} t^2$$ Where $t$ is the time elapsed since the launch. This last expression also assumes that the particle isn't going too far or too fast; if it goes large distances relative to the size of the earth, things get more complicated.

Answer 2

Once the particle leaves the disk, the only forces applied would be gravity (if any). So what is the motion of a particle under the influence of gravity?

The problem boils down to what is the initial conditions, which you correctly stated would be vertical (+z direction) as well as tangential to disk due to the rotation. That is all you need to solve this problem.

Initial Conditions + Gravity = Solution

Answer 3

There are two possible interpretations of this question. In each case, once the particle leaves the disk it stops being affected by the disk. If there's gravity it eventually falls back. If not, it keeps going.

One interpretation is that, on the disk, there is a launch device that is pointing straight up. The particle will be released with the vertical speed given, and also retaining the speed it was moving with along with the disk. This seems to be what Maxwell Aifer has done. And his answer is fine. The particle will trace a path along the disk that generally moves toward the outside. It's a fun path to graph. The OP should graph it, being careful to be careful about the usual things about slope and limits and so on.

Another interpretation is that the particle is launched such that in the x-y-z coords it is moving straight up at launch. The launcher would have to be pointing backwards against the rotation of the disk. In that case, it goes straight up in the x-y-z system, and falls back. From the point of view of the disk it will then trace a rising path. If there's gravity it eventually falls back. The path along the surface of the disk will be a nice circle, possibly covering several turns of the disk.