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This is a rock tied to a string spinning vertically. Here, $T+mgsin\theta = mv_1^2/r => T = mv_1^2/r-mgsin\theta$ Suppose I give it a velocity $v$ at the bottom.

1) At what angle $\theta$ will the tension become zero?

2) If the velocity ends up $=0$ at $\theta = 0$, then the tension $T = m0^2/2-mg$ which would end up giving tension a negative value. How is this possible?

3) If the velocity at any point ends up zero, does the tension necessarily have to end up equalling zero as well?

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  • $\begingroup$ If you provide to the bottom an initial speed $\:v_{bottom}\:$ then the kinetic energy would be $\:E_{bottom}^{kin}=(1/2)\;mv_{bottom}^2\:$. If $\:E_{bottom}^{kin} >E_{top}^{dyn}=mg(2r) \:$ that is $\:v_{bottom} > 2\sqrt{gr} \:$, then the rock will execute this vertical nonuniform circular motion for ever. To the contrary, if $\:v_{bottom} <\sqrt{2gr}<2\sqrt{gr} \:$ then the rock will be a pendulum and execute a simple harmonic oscillation with "angular amplitude" $\:\cos\theta_{max}=1-(v^2/2gr)\:$. $\endgroup$
    – Frobenius
    Commented May 16, 2016 at 13:09
  • $\begingroup$ In case that $\: \sqrt{2gr}< v_{bottom} < 2\sqrt{gr} \:$ I think that the rock will execute a periodic motion with an orbit that would require a figure to be explained. $\endgroup$
    – Frobenius
    Commented May 16, 2016 at 13:09

3 Answers 3

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[...] which would end up giving tension a negative value. How is this possible?

It isn't. If you set zero speed $v=0$, then you will no longer have circular motion, and the object will accelerate downwards. A non-zero speed $v$ is a requirement for circular motion to happen, because a radial acceleration towards the center can be present only as such. Otherwise it would be like assuming that the object would continue moving around the center even if you stop pulling in the string which obviously isn't the case.

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  • $\begingroup$ That answers 2) and 3). How would I solve 1)? $\endgroup$
    – xasthor
    Commented May 16, 2016 at 8:00
  • $\begingroup$ Well, put T=0 into the equation $\endgroup$
    – Steeven
    Commented May 16, 2016 at 8:05
  • $\begingroup$ But the eq uses $v_1$. I want to solve for $v$ i.e the initial velocity I need to provide at the bottom $\endgroup$
    – xasthor
    Commented May 16, 2016 at 8:10
  • $\begingroup$ Do you want to solve for v when the question asks for the angle? $\endgroup$
    – Steeven
    Commented May 16, 2016 at 8:12
  • $\begingroup$ Also, note that in order to have a uniform circular motion the speed is constant. The only acceleration is towards the centre which is always perpendicular to the direction, so the speed is never changed. So initial speed (which would be the speed when the circular motion is initially obtained) is no different from v1. $\endgroup$
    – Steeven
    Commented May 16, 2016 at 8:23
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Tension in the string will never become zero, as long as rock is moving along the circular path.

Also, speed of the rock also, will never become zero in this case.

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As the stone moves in a vertical orbit, the velocity will change with height (as kinetic energy converts to potential energy).

To maintain the circular orbit, the radial component of force (tension plus component of gravity) must equal the centripetal force required.

If the stone has insufficient velocity at the bottom, there will come a point where you would need "negative tension" in order to maintain the orbit. If you have a rigid rod (rather than a string) this just means that the rod is in compression rather than tension at that point; but if you have a string, the string would go slack and the stone would no longer follow the circular path.

enter image description here

As shown in the diagram, there is a component of force of the gravity that can point either towards the center or away from it; when that component is larger than the force needed to maintain the circular orbit, $\frac{mv^2}{r}$, the stone will continue along the parabolic path as shown (unless the string is rigid so it can provide "push", which is what "negative tension" would be).

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