# Regarding gravity force during circular motion

If I spin a ball attached to a string so that the string is pointing diagonally upwards, the centripetal force would be:

$$T \cos\theta,$$

where $\theta$ is the angle between the horizontal and the string, and $T$ is the tension. Then the vertical force would be: $$T \sin\theta + mg.$$

Given that the ball spins in a horizontal circle, how does it deal with an unbalanced vertical force and not fall downwards?

• Are you certain that is the centripetal force? – Johnathan Gross Oct 4 '17 at 22:25
• If your ball is spinning in a horizontal plane, the string will make an angle to the horizontal. – Floris Oct 4 '17 at 22:25

## 4 Answers

You should really start with the constraint $T \sin \theta = mg$. You know that the vertical forces have to be balanced! Then you can express $T$ as $T = \frac{mg}{\sin \theta}$.

And since the centripetal force is $F_c = T \sin \theta$, you can then say $F_c = mg \frac{\cos \theta}{\sin \theta} = mg \cot \theta$.

You can check that this makes sense! If $\theta = 90$, then the ball is at rest and there should be no centripetal force. Indeed $\cot 90 = 0$ so $F_c = 0$.

If $\theta = 0$, then the ball is completely horizontal. That's actually impossible, and you can see $\cot 90 \to \infty$, so the force would have to be infinite.

It does fall downward, a cycle with the angle $\theta$ maintained all the way around is not achievable.

If, however, it falls to $-\theta$ at an azimuth of $180^o$, (neglecting losses due to air resistance/friction and so on) it accelerates enough during that fall to return to $+\theta$ at an azimuth of $0^o$

This diagram should help: The force of gravity is countered by the vertical component of the tension in the string, $T\sin\theta$; the centripetal force is provided by the horizontal components, $T\cos\theta$.

The string cannot be horizontal on a stable orbit with gravity.

You have a mistake. If you define the positive forces to up, rhen the gravity is negative.

$$T\sin\theta - mg=0$$ Or if you define the positive forces to down, $$-T\sin\theta + mg=0$$

Anyway, $$T\sin\theta = mg$$

Which means that the forces are in opposite directions. So, the force is not unbalanced force in the vertical direction.

• I believe you meant to write “$T sin \theta$”? – Kieran Moynihan Oct 5 '17 at 6:21
• Why you say that? – Gabriel Sandoval Oct 5 '17 at 15:13
• Because $T cos \theta$ is the component of the tension perpendicular to the gravitational force, whereas $T sin \theta$ is the component parallel (and opposite in direction) to the gravitational force. – Kieran Moynihan Oct 5 '17 at 15:17
• Ohh, yes. You're right. I was thinking in a different angle. – Gabriel Sandoval Oct 5 '17 at 15:35

## protected by Qmechanic♦Oct 5 '17 at 5:48

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?