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Consider a particle attached to the centre of a circle via an inextensible string undergoing vertical circular motion. If the initial velocity $u$ is such that $\sqrt{2gl}<u<\sqrt{5gl}$, the tension becomes 0 at a point where $\theta$ is obtuse. Now, the only force acting on the particle is the force of gravity, and the particle should undergo projectile motion.

But since the particle is still tethered to the centre, the string must become taut when the particle attempts to move this way, and the particle should continue moving in a circular path. How then does the particle undergo projectile motion, i.e., how is such a situation possible:

enter image description here
Wouldn't this also mean that the minimum initial velocity for vertical circular motion should be $\sqrt{2gl}$ rather than $\sqrt{5gl}$, because the particle would be able to complete a circular path? And if the path is not parabolic, what is the actual path followed by the particle?

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  • $\begingroup$ It could do that only if the string stretches. (many materials do). Realisticly, the deformation would be at the point where gravity and centrifugal force coincide, with a similar deformation inwards at the top of the circle. Stretching (rubber vs steel) may not be desirable in this system. $\endgroup$ Nov 27, 2020 at 16:59

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The particle does follow a parabolic trajectory, but the parabola will lie inside the circle, not outside. For example, if $u = \sqrt{5gl/2}$, then $v = \sqrt{gl/2}$ when the tension goes to zero at $\theta = 120^\circ$. The subsequent trajectory can be plotted, and looks like this (using units where $m = g = l = 1$):

enter image description here

(The fact that the parabola intersects the bottom of the circle is, I think, just a coincidence.)

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  • $\begingroup$ I also use Desmos :-) $\endgroup$
    – Sebastiano
    Dec 10, 2020 at 21:54
  • $\begingroup$ @Sebastiano: It's often easier than firing up Mathematica, which is my other go-to. :-) $\endgroup$ Dec 11, 2020 at 13:45
  • $\begingroup$ Sir one thing i would like to know its not obvious that the kinematics equations we know would be valid here unless we write the energy conservation equation at anytime in vertical circular motion right Sir ? As tension could have do something , which only from energy consideration we got to see the equations of motion remain valid. $\endgroup$
    – Orion_Pax
    Apr 12, 2022 at 14:28
  • $\begingroup$ @Orion_Pax: It's not clear to me what you're asking. But if you're asking "is energy conserved in vertical circular motion even though there is a tension force", then the answer is "yes" because a string can only exert a radial force, which will be at right angles to the motion and so it will do no work. $\endgroup$ Apr 12, 2022 at 15:08
  • $\begingroup$ Yeah sir that i understood , i mean the equations v^2 = u^2 +2as, v= u +at ,s=ut+1/2at^2 validity here , we can only be sure of that after conserving energy . ( If imagine we just use this equation without proving it from energy conservation , its not trivial isnt Sir ) unless we derive it . As such why i am asking because in other kinematics problems the object is not constraint by some other things like throwing a stone etc . So there we can easily say whether a remains constant etc . Here we cannot conclude anything right Sir unlesd we derive from Energy ? $\endgroup$
    – Orion_Pax
    Apr 12, 2022 at 17:10

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