
Legend for arrows:

As seen on the merrygoround: 
As seen in the inertial reference frame: 
z(r)=Ω^{2}r^{2}/(2g)
here z is the height of the surface, r is the distance from the rotation axis, Ω is the angular velocity and g is the acceleration due to gravity. Even if the surface is frictionless, the normal force of reaction would have a component directed radially inward. With curvature as specified above, the horizontal component of the normal force would balance the centrifugal force for an object at rest on the surface. "At rest" means being motionless as seen within the rotating reference frame.
Another way of analyzing this balance is to analyze forces tangent to the surface. Analyzed that way, the downhill component of gravity balances the uphill component of the centrifugal force.
Objects at rest on such a surface could remain at rest: they would not accelerate. However, in the rotating reference frame, objects that move on such a surface would experience a Coriolis force orthogonal to the velocity vector. For the anticlockwise motion seen here, the force is rightwardturning.
Now, as seen in the inertial reference frame, the rotation of the merrygoround surface has no direct effect on the dynamics. All that is presented to the sliding hockey puck is a frictionless surface. The effect of rotation is indirect; the slope of the side has been "tuned" to the rotation rate, as specified above. Assuming the curvature is shallow, meaning dz/dr<<1, the hockey puck will experience simple harmonic motion in both directions. The combined motion will be an ellipse. For simplicity of the mathematical analysis, we consider the axes of the ellipse aligned with the Cartesian coordinates:
x(t)=A cos(Ωt)
y(t)=B sin(Ωt)
Here A=10 and B is adjustable. The magnitude of A is indicated by a red stripe painted on the surface of the merrygoround, B is indicated by a blue stripe .
As seen in the rotating reference frame, the trajectory is an anticylconic circle. The Coriolis force constantly deflects the motion to the right. The speed is uniform. A mathematical analysis of the motion within the rotating frame is rather easy. Let the speed of the puck be V. With the trajectory being anticyclonic, the rightward Coriolis force will be directed toward the center of a trajectory circle, which we can define to have a radius R. In this case R=(AB)/2, but the fact is not important in the analysis. The equation of motion in the rotating frame is that the centripetal acceleration is provided by Coriolis force:
V^{2}/R = 2ΩV
The radius of the trajectory there depends on the initial speed,
R=V/(2Ω)
The angular frequency about the cicle is V/R, or 2Ω, which is twice the angular frequency of the merrygoround.
Recall the coordinate transformation from the inertial frame to merrygoround frame is:
X(t)= cos(Ωt)x(t) + sin(Ωt)y(t)
Y(t)=sin(Ωt)x(t) + cos(Ωt)y(t)
We can convert the elliptical trajectory of the inertial frame to the merrygoround frame by using this transformation:
X(t)= Acos^{2}(Ωt) + Bsin^{2}(Ωt)
Y(t)=Asin(Ωt)cos(Ωt) + Bcos(Ωt)sin(Ωt)
With use of trigonometric identities, this becomes:
X(t)= (A+B)/2 + (AB)/2 cos(2Ωt)
Y(t)= (AB)/2 sin(2Ωt)
We should recognize the above to be an anticyclonic circle about a center at X=(A+B)/2.
Though either reference frame can be used to obtain the trajectory as seen on the merrygoround, the trajectory obtained with within the rotating frame, with the use of the Coriolis force, requires less labor. Also, analysis within the rotating frame allows for general principles about the trajectory to be easily deduced.
By the way, this anticylonic orbit, with the Coriolis force providing the centripetal acceleration, is called an inertial oscillation.