I think the answer is that the second diagram you drew won't happen. I just picked up a string and tried this. What happened is that the first diagram is easy. For the second, I have to twirl the string faster, and I can't quite get it to stay above my hand. The best I can do is to get the mass to swing in a plane almost even with my hand.

Note: it's a little difficult to do this fairly with your hand. When I tried it, I had a tendency to slightly adjust the plane of motion of the mass, so that it oscillated slightly. This was particularly bad when trying the second situation. If I didn't do that, the sting would hit one of my knuckles every time it passed. That's another indication that the plane of revolution is actually below my hand.

Your force diagrams are qualitatively correct. Gravity points down towards the floor, and the tension points along the string at some angle. It's easiest to break the tension into a vertical component (which will either add to or subtract from gravity), and a radial component, which lies in the plane of the ball's orbit and provides the centripetal force.

To be concrete, take $\theta$ to be the angle between the string and the vertical. The string hanging under just gravity means $\theta = 0$, and the ball orbiting in the horizontal plane is $\theta = 90^\circ$. You want the vertical component to cancel out gravity, so $w = T\cos{\theta}$. As you increase $\theta$, $\cos{\theta}$ decreases, so you have to increase $T$ to keep the vertical component balanced. The centripetal force is $F_c = T\sin{\theta}$, which increases both because you increased $\theta$ and because you increased $T$. The ball will then need to move faster to account for the increased centripetal force. I'll let you work out the actual details, but you should get that the ball will need to move infinitely fast to get to $\theta = 90^\circ$.

An illustration of the wobble effect I noticed is when cowboys attempt to lasso animals. You can watch this video starting at 1:20 to see this.

You have to consider the two axes separately: lets call $x$ the horizontal plane, and $y$ the up and down (direction in which gravity acts). In the "free-body diagram" of the ball, there are also **three** forces to consider: 1) gravity, 2) centripetal, and **3) the tension in the string**. The tension in the string needs to balance the combination of gravity and centripetal acceleration.

The centripetal acceleration itself, has nothing to do with gravity, it only has to do with the ball's circular motion. Instead of the ball spinning nearly-horizontally, you can imagine spinning it ever so slowly---and it remaining nearly vertical. Gravity hasn't changes, only the centripetal acceleration, and thus tension in the string---via force balance in the $x$ direction.

Because there is gravity in the $y$ direction, **the string can never be fully horizontal, because some amount of the string tension must act in the $+y$ direction to counteract gravity**. Thus, only the $y$ component of the string-tension is determined by gravity, and this occurs regardless of the centripetal force.

## Best Answer

As you suspect the wire cannot be perfectly horizontal nor, if its weight is considered, perfectly straight. At any point in time there are two forces acting on the ball,( ignoring the weight of the rope) : gravity (mg) and the tension due rope .

The vector sum of these must be equal to $mv^2/r$ in the radial direction. To counter the vertical gravitational force, the tension force must have an opposite vertical component hence the rope cannot be exactly horizontal.

The faster the mass is spun around, the smaller will be the angle . Also, in the absence of resistive losses, the motion of the mass is perfectly horizontal. The same is true if the losses are exactly balanced by the boy swinging the rope.