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See image below:

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Say I have a mass 'm' on a string or a very thin rod being twirled around on some pivot point by a motorised axle. Originally its twirling around in equilibrium in a horizontal plane at height A.

I then increase the rotational speed of the motor which will cause the 'string/thin rod' to rise up from height A to B .

I can understand how the increasing vertical component of tension T1 to T2 might cause the ball to rise height 'y' but I cannot understand the forces (if any ) that would move it 'x' and further away from the vertical spinning rod.

Am I missing something obvious?

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    $\begingroup$ The length of the string is not changing, so moving up in y implies moving out in x. $\endgroup$
    – mike stone
    Commented Oct 22, 2021 at 16:46

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The horizontal component of the tension in the rod/string provides the centripetal force which keeps the mass moving in a circle. When the rod/string makes an angle $\theta$ with the vertical the mass is moving in a circle with radius $r = L \sin(\theta)$ where $L$ is the length of the rod/string. So if the tension in the rod is $T$ we have

$T \sin (\theta) = mr \omega^2 = mL \sin (\theta) \omega^2 \\ \Rightarrow T = mL \omega^2$

Note that since $m$ and $L$ are fixed, $T$ depends only on $\omega$. But if the mass is rotating at a constant height, the vertical component of tension must equal the weight of the mass. So we have

$T \cos (\theta) = mg \\ \Rightarrow mL \omega^2 \cos (\theta) = mg \\ \displaystyle \Rightarrow \cos (\theta) = \frac g {L \omega^2}$

So $\theta$ depends on $\omega$ as well. If $\omega$ increases, $T$ must increase and $\theta$ must also increase i.e. the rod/string moves away from the vertical axis. The rod/string has a fixed length, so as it moves away from the vertical the mass (fixed on the end) will move higher and move outwards from the axis at the same time.

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  • $\begingroup$ Many thanks for providing the maths to show that θ must increase but I was wondering in terms of real forces how the ball will move outwards if all we have is the tension force with no components directed outwards, while the mg force is just vertically down. $\endgroup$
    – Dubious
    Commented Oct 23, 2021 at 23:48
  • $\begingroup$ @Dubious The inertia of the ball makes it move away from the axis until the horizontal component of the tension is sufficient to provide the centripetal force to keep it moving in a circle (if there were no rod or string the ball would move away from the axis in a tangential straight line). Alternatively, from the point of view of the ball's reference frame, it is centrifugal force that moves it outwards until it is balanced by the horizontal component of tension. $\endgroup$
    – gandalf61
    Commented Oct 24, 2021 at 7:58
  • $\begingroup$ Many thanks gandalf61 - I did have an inkling about using the CF inertial force but was unsure. $\endgroup$
    – Dubious
    Commented Oct 25, 2021 at 10:55
  • $\begingroup$ I don't have the technical finely engrained answer but forces cause movement and the minimum energy positions(valleys) are the horizonal and vertical like for frisbees and fidget spinners moving forward. In space they flip flop between both minimum energy positions when spinning still. In this case the tension had a vertical component while it's horizontal had to equal mv^2/r $\endgroup$
    – ChemEng
    Commented Dec 12, 2022 at 14:19

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