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Is the explanation shown in the diagram right?

Diagram

This is: the net force F1 = tangential + tension is way much bigger than the weight of the ball and, therefore, the resulting force F1 + weight is F1 so that the ball does not fall down.

UPDATE Thank you very much for your answers.

Actual situation:

enter image description here

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  • $\begingroup$ There is no tangential force, just tangential speed which is outcome of centripetal tension force. $\endgroup$ May 14 at 22:09
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    $\begingroup$ No, it is incorrect. Assuming uniform rotational motion, then there's no $F_{tangential}$, because there's no tangential acceleration. Also, are you aware that the object and the centre of the circle can never lie in the same horizontal plane? $\endgroup$
    – Gert
    May 14 at 22:10
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    $\begingroup$ $F_{tangential}$ would not prevent the ball from being pulled to the center (assume it was$> 0$), because it isn't radial. $\endgroup$
    – Gert
    May 15 at 17:40
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    $\begingroup$ is against a radial outward force (that comes from ...?) It's the centrifugal force. What makes the ball spin? Initially forces acted to make it spin. But in uniform rotation there are no NET forces acting on it (see Newton's 2nd Law) any more. $F_{tangential} \neq 0$ is only true if $v_{tangential} \neq 0$ (non-uniform rotation) $\endgroup$
    – Gert
    May 15 at 19:01
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    $\begingroup$ It is Real in the correct frame of reference (one that rotates with the ball) It is this FoR that's used to calculate string tension and angle of string with the horizontal. $\endgroup$
    – Gert
    May 15 at 20:08
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If you were to spin a ball about a vertical axis using a string, it would never rise up to a level above the horizontal level. In fact the angle upto which it will rise could also be calculated. Suppose the length of the string is $l$ and at equilibrium the angle it makes with the vertical axis is $\theta$, $\omega$ be the angular velocity of the axis, and the tension in the string is $T$. Then, we have: $$T\cos \theta =mg$$

$$T\sin \theta =m{\omega}^2 l\sin\theta $$ From these two equations, we have: $$\cos\theta=\frac {g}{{\omega}^2 l}$$ Clearly, if $\omega\to \infty$, $\theta\to 90°$, which means $\theta$ can never exceed $90°$. Also note that $w>\sqrt {\frac gl}$ is required if ball is to rise up even a little.

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  • $\begingroup$ Actually, it cannot reach $90^{\circ}$ either. That would indeed require $\omega \to \infty$. $\endgroup$
    – Gert
    May 15 at 17:43
  • $\begingroup$ True, that's why I wrote that $\theta {\to 90°}$ instead of $\theta=90°$ $\endgroup$ May 15 at 17:44
  • $\begingroup$ I came to theses equations too. But I need to understand the diagram of forces. $ T sin( \theta)$ is the x projection/component of T (tension). The right hand of that equation $m \omega^2 r sin(\theta)$ is also a force. Then, must I assume that is in the opposite direction than $T_x$? Why? $\endgroup$ May 15 at 18:34
  • $\begingroup$ Can someone comment about what is wrong in the following when PE and KE are equated? $h$ is height of the rotating conical pendulum. $ mg h (\sec \theta-1) = \frac12m(\omega h \tan \theta)^2\to \omega^2= \dfrac{g}{h} \dfrac{2 \cos \theta}{1+ \cos \theta}$ which is valid only for small $\theta?$. $\endgroup$
    – Narasimham
    May 15 at 19:09
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    $\begingroup$ @cibercitizen1 not at all. $\omega^2 r$ is the centripetal acceleration. $\endgroup$ May 15 at 19:29

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