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See image below:

enter image description here

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?

Qmechanic
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Dubious
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1 Answers1

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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.

gandalf61
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