Why doesn't a spinning object in the air fall? Let's say I have a ball attached to a string and I'm spinning it above my head. If it's going fast enough, it doesn't fall. I know there's centripetal acceleration that's causing the ball to stay in a circle but this doesn't have to do with the force of gravity from what I understand. Shouldn't the object still be falling due to the force of gravity?
 A: The string is at a slight angle to horizontal $\theta$. It is not exactly horizontal. The slight angle is such that the tension in the string exactly counteracts gravity, $T\sin(\theta)=m g$. So, there is actually a force acting upwards that counteracts gravity, and it is supplied by the string.
You're right that if $\theta=0$ exactly, there would be a problem and the object would necessarily fall a bit.
A: 
We have the ball orbiting at a distance $R$ from the centre of rotation and the string inclined at angle $\theta$ with respect to the horizontal.
Two main forces act on the ball: gravity $mg$ ($m$ is the mass of the ball, $g$ the Earth's gravitational acceleration) and $F_c$, the centripetal force needed to keep the ball spinning at constant rate. $F_c$ is given by:
$$F_c=\frac{mv^2}{R},$$
where $v$ is the orbital velocity, i.e. the speed of the ball on its circular trajectory.
Trigonometry also tells us that if $T$ is the tension in the string, then:
$$T\cos\theta=F_c.$$
Similarly, as the ball is not moving in the vertical direction, thus $F_{up}$:
$$T\sin\theta=F_{up}=mg.$$
From this relation we can infer:
$$T=\frac{mg}{\sin\theta}.$$
And so:
$$\frac{mg}{\tan\theta}=F_c=\frac{mv^2}{R}.$$
Or:
$$\tan\theta=\frac{gR}{v^2}.$$
From this follows that for small $\tan\theta$ and thus small $\theta$ we need large $v$. But at lower $v$, $\theta$ increases. Also note that $\theta$ is invariant to mass $m$.
A: I appreciate that this has already been answered correctly, but I thought it may be worth adding a simplistic summary:
When the ball is spinning, there is a force acting on it which pushes it away from the centre of rotation.  The only way it can get further away from that point is by moving upwards (because the string stops it from moving outwards without moving upwards).  So if the force pushing the ball out is greater than the force pulling it down (gravity), it will rise.
A: I differ with all of the explanations above. If you are spinning a ball horizontally and leave the string, it would fall at once if it is in vacuum/airless place. In other scenario which is the realistic one, it does not fall because the ball stirs the air around it away, thus creating lower air pressure zone in the plane where it is being rotated. So the air below it creates an upward pressure to hold the rotating object. 
This is the theory behind how a helicopter works. By the way, have you heard of the Ninja weapon Shuriken?
