Forces of a conical pendulum in the reference frame of the ball (of the pendulum) I'm a bit confused about the forces (specifically their magnitude), involved of a conical pendulum in the reference frame of the ball of the pendulum. I'd like to consider this ball to be in contact with the floor.
So the ball will experience the weight, normal force (touching the ground), tension force (along the string), and a centrifugal force (radially outwards). Would the horizontal component of the tension force be called the centripetal force or could you equate calling the tension force the centripetal force? 
Would this centrifugal force have a magnitude of mw^2 r? Or would this be the magnitude of the centripetal force? Saying that the centrifugal force is equal to mw^2 r seems to get the same answer as the regular conical pendulum problem.
Any clarifications would be appreciated, and thank you for your time!
 A: First, I would like to quote the definition of centripetal force given on Wikipedia:

A centripetal force is a force that makes a body follow a curved path. Its direction is always orthogonal to the motion of the body and towards the fixed point of the instantaneous center of curvature of the path

Now, in the given case, looking from ground frame the ball is moving in circle, and so there must be a center seeking force on it. Here only a component of tension is towards center (normal from ground, it's weight and tension's other component are perpendicular to the plane of circle, and these are the only forces on it.)
Talking about centrifugal force, it's Not a real force but a Pseudo force. The ball has variable velocity ( the direction is changing) , implying it has an acceleration ( which is the centripetal acceleration). So when we look in ball's frame of reference, we are in a non-inertial frame of reference and need to use a Pseudo force, which we call the Centrifugal Force.
Now coming on magnitude, $ \mathit m\omega^2r $ is the magnitude of centripetal force, which in this case is the component of  tension that's towards circle's center. When you change frame from ground to the ball's, the Pseudo force, centrifugal force, has magnitude equal to Mass times it's acceleration, which is precisely equal to $ \mathit m\omega^2r $, and exactly in opposite direction to the centripetal force.
Hope this helps.
A: Assuming that in the frame of the laboratory the mass is undergoing circular motion at constant speed in the horizontal plane then the horizontal component of the tension pointing inwards towards the centre of the horizontal circle is producing the centripetal acceleration of the mass.
The horizontal component of the tension force is $mr\omega^2$ where $m$ is the mass of the body, $r$ the radius of the horizontal circle and $\omega$ the angular speed of the mass.  
In the frame of the mass the mass is not moving and certainly not accelerating yet the forces acting on the mass have a net component which is equal to the horizontal component of the tension towards the centre of the horizontal circle as seen in the laboratory frame.  
In order to be able to use Newton’s laws in the frame of the mass and extra force is added in effect to convert the situation into a statics problem with no net force on the mass.
This fictitious/pseudo force, which you have called the centrifugal force, has a magnitude equal to the horizontal component of the tension, $mr\omega^2$, and is directed away from the centre of the horizontal circle.
Adding this force to the mass means that now the net force on the mass is zero.
