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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, with $F_{up}$:

$$T\sin\theta=F_{up}=mg.$$

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 $m$.

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 $m$.

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) 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, with $F_{up}$:

$$T\sin\theta=F_{up}=mg.$$

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 $m$.

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, with $F_{up}$:

$$T\sin\theta=F_{up}=mg.$$

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 $m$.

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) 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, with $F_{up}$:

$$T\sin\theta=F_{up}=mg.$$

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. But note that $\theta$ is invariant to mass $m$.

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) 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, with $F_{up}$:

$$T\sin\theta=F_{up}=mg.$$

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 $m$.