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A circular hoop of radius is stood vertically on a flat surface and held in place. A smooth wire is tightly stretched between the upmost point of the hoop and a point on the hoop at height ℎ. A bead of mass is held at rest at the top of this wire, upon which it is threaded.(Ignore air resistance)

Is it possible to use conservation of energy to solve find time taken? Start with the following energy relationship, $$ mg(2r)=mgh+\frac{1}{2}mv^2$$ Then solving for v, $$v=\sqrt{4gr-2gh}$$ As this is the velocity at the final point, this should be usable in a suvat, where a=g, u=0. $$v=u+at$$ $$t=\frac{v}{a}$$ $$t=\sqrt{\frac{4r-2h}{g}}$$ The suggested method is to use kinematics only, but I don't see why mine is not correct.

jamie
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    Please clarify exactly what you are trying to find. – tomi Jan 30 '20 at 16:27
  • Don't assume that acceleration is constant. – David Quinn Jan 30 '20 at 16:40
  • As the wire is not vertical, your assumption that $a = g$ is false. First $a$ is directed along the wire, while $g$ is directed vertically, and second, the wire exerts force on the bead, so gravity is not the only force in play. – Paul Sinclair Jan 31 '20 at 04:10

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