As seen on the merry-go-round:
As seen in the inertial reference frame:
The merry-go-round has an angular velocity Ω. A red stripe is painted on the flat surface of the merry-go-round. Imagine yourself positioned at the rotating end of the stripe. It is possible to "play catch with yourself". With a frictionless hockey puck, the trajectory in the inertial frame is not affected by the rotation. It slides in a straight line on the surface. A trajectory beginning at x=C and y=0 can be written as:
x(t)=A Ωt + C
(Although Ω does not affect the dynamics directly, it is convenient to use the dimensionless quantity Ωt as the time variable).
The coordinate transformation from the inertial frame to merry-go-round frame is:
X(t)= cos(Ωt)x(t) + sin(Ωt)y(t)
Y(t)=-sin(Ωt)x(t) + cos(Ωt)y(t)
In the X-Y coordinate system of the merry-go-round,
the trajectory of the hockey puck is:
X(t)=A Ωt cos(Ωt) + B Ωt sin(Ωt) + C cos(Ωt)
Y(t)=-A Ωt sin(Ωt) + B Ωt cos(Ωt) - C sin(Ωt)
Here we take C=10 (for no special reason). For certain combinations of A and B, we can successfully "play catch with ourselves" and have the hockey puck return to us at a specified time t. The physlet shows one such trajectory as the default example. In an assignment, students are asked to find other such examples, and to check their anwsers with the physlet.
Here is something to consider. To mathematically solve for the above trajectory in the rotating coordinate system, by invoking the fictitious forces, is extremely tedious. Both the Coriolis and centrifugal forces must be invoked. In this case, it is much easier to "do dynamics" in the inertial frame, and transform the straight-line, forceless, trajectory into that seen in the rotating frame. In dynamical meteorology that is not the case, and doing dynamics in the rotating spherical frame, with a Coriolis force, is very often more fruitful than analyis in a non-rotating spherical frame.