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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?

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

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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 $0^o$

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This diagram should help:

enter image description here

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.

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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.

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    $\begingroup$ I believe you meant to write “$T sin \theta$”? $\endgroup$ – Kieran Moynihan Oct 5 '17 at 6:21
  • $\begingroup$ Why you say that? $\endgroup$ – Gabriel Sandoval Oct 5 '17 at 15:13
  • $\begingroup$ 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. $\endgroup$ – Kieran Moynihan Oct 5 '17 at 15:17
  • $\begingroup$ Ohh, yes. You're right. I was thinking in a different angle. $\endgroup$ – Gabriel Sandoval Oct 5 '17 at 15:35

protected by Qmechanic Oct 5 '17 at 5:48

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