Ball swung in a horizontal circle, can the string ever by exactly horizontal to the ground? You swing a ball on the end of a lightweight string in a horizontal circle at constant speed.  Can the string ever by truly horizontal?  If not, would it slope above the horizontal or below the horizontal?  Why?
 A: In a situation wherein the ball is completely horizontal, the downward gravitational force acting on the ball would be unbalanced and it would cause the ball to dip downwards. 
This would cause the string to slant upwards towards the person spinning the ball. Due to this, there would now be a component of the centripetal force in the upwards direction. In the vertical equilibrium position of the ball, this upwards component of the centripetal force would balance the downward gravitational force. 
A: 
Let's assume the ball is rotating in a circle (in the horizontal plane) with radius $R$ at angular velocity $\omega$.
In order for the ball to stay on its circular path, a centripetal force $F_c$ has to act on it, given by:
$$F_c=mR\omega^2$$
This force is the resultant of the tension in the string  $T$ and the weight $mg$. With some simple trigonometry:
$$F_c=T\cos \theta$$
and:
$$T\sin \theta=mg$$
Reworked we get:
$$F_c=\frac{mg}{\tan\theta}=mR\omega^2$$
So that:
$$\tan\theta=\frac{g}{R\omega^2}$$
So $\theta \to 0$, for:
$$R\to \infty$$
and/or:
$$\omega \to \infty$$
So the angle $\theta$ will always be positive.
Note, interestingly, how the angle is independent from the mass.
A: Hint: Draw a free body diagram for the ball on the end of the string to understand if such a true horizontal circular set up is allowed by Newton's laws. 
