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Normally, if a body is attached to a string & is rotated in a vertical circle, then to loop the whole circle, the rope must not slack at the topmost point & there should be a velocity for having centripetal force. For this, the minimum initial velocity required is $\sqrt{5gr}$, $r$ being the radius of the vertical circle.

However there is another theory , which has baffled my sense & is quoted :

If a particle of mass $m$ is connected to a light rod & is whirled in a vertical circle of radius $r$, then to complete the circle, the minimum velocity of the particle at the bottommost point is not $\sqrt{5gr}$. This is because, the velocity of the particle can be zero also. Using conservation of mechanical energy, we get the minimum value as $2\sqrt{rg}$.

Ok, it is understandable that the condition that the rope must not slack is not required here. However, if the velocity at the topmost point is zero, there must be no centripetal force, which would make it rotate. So, isn't this theory wrong? Or, am I mistaking? If so, Where am I mistaken?

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The topmost point would be in unstable equilibrium, so the speed at the top can be arbitrarily close to zero, and the mass will start to fall. At the top the potential energy is 2mgr (relative to the bottom of the circle). QED

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  • $\begingroup$ Oh! you are very fast. Just can you explain why at the top , it will be at unstable equilibrium? $\endgroup$
    – user36790
    Commented May 30, 2015 at 10:09
  • $\begingroup$ Assuming the light rod is inextensible, it doesn't matter what N is - unless you need to know it. It's in unstable equilibrium because en infinitesimal move away from the equilibrium point will result in the mass moving further away from that point, not back to it. $\endgroup$
    – Dr Chuck
    Commented May 30, 2015 at 10:15
  • $\begingroup$ Chuck: Really sorry for the above comment;I thought you answered my other question. However, I would ask still why at the top, it is at unstable equilibrium. BTW, the derivation was done putting the velocity at the top 0. So, why are you telling that it is approximately close to zero? $\endgroup$
    – user36790
    Commented May 30, 2015 at 10:21
  • $\begingroup$ Yes, you are right in your definition but see the ball at the top is balanced by the rod against gravity. So, why should it be unstable? And why should it move away from the top if it has zero velocity? $\endgroup$
    – user36790
    Commented May 30, 2015 at 10:26
  • $\begingroup$ I said arbitrarily close to zero, not approximately, so strictly speaking the velocity would need to be greater than the value cited (and subject to other caveats). The equilibrium point is when the mass is at its highest point. $\endgroup$
    – Dr Chuck
    Commented May 30, 2015 at 10:34
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I think you are a little confused. In case of a string, velocity at topmost point is not zero since weight needs to be balanced by centrifugal force but tension may tend to zero.

In case of rod, velocity may be zero since weight can be balanced by normal force from the rod.

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  • $\begingroup$ If the velocity is zero, how can there be centripetal force to rotate it? $\endgroup$
    – user36790
    Commented May 31, 2015 at 0:57
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    $\begingroup$ If velocity becomes zero a topmost point, then the rod is in a position of unstable equilibrium(it won't rotate). But if the velocity just tends to zero the rod will rotate. $\endgroup$
    – user2369284
    Commented May 31, 2015 at 4:16
  • $\begingroup$ So, I was correct, not confused. If it were zero, then there would be no centripetal force at all. The theory is wrong in this sense. $\endgroup$
    – user36790
    Commented May 31, 2015 at 6:33

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