English

An interesting track for the Brachistochrone

Classical Physics 2020-10-30 v1

Abstract

If a particle has to fall first vertically 1 m from A and then move horizontally 1 m to B, it takes a time t(=τ1+τ2=τ3=3/2g)=0.67t(=\tau_1+\tau_2=\tau_3=3/\sqrt{2g})=0.67 s. Under gravity and without friction, if it sides down on a linear track inclined at 45045^0 between two points A and B of 1 m height, it takes time t(=τ4=2/g)=0.63t(=\tau_4=2/\sqrt{g})=0.63 s. Between these two extremes, historically, Bernoulli (1718) proved that the fastest track between these points A and B is cycloid with the least time of descent t=τB=0.58t=\tau_B=0.58 s. Apart from other interesting cases, here we study the frictionless motion of a particle/bead on an interesting track/wire between A and B given by y(x)=(1xν)1/ν.y(x)=(1-x^{\nu})^{1/\nu}. For ν>1\nu > 1 the track becomes convex and t>>τ4t>>\tau_4, and when ν>1.22\nu >1.22, the motion with zero initial speed is not possible. We find that when ν(0.09653,0.31749),τ4<t<τ3\nu \in (0.09653, 0.31749), \tau_4<t <\tau_3 and when ν(0.31749,1),τB<t<τ4\nu \in ( 0.31749, 1),\tau_B < t < \tau_4. But most remarkably, the concave curve becomes very steep/deep if ν(0,νc=0.09653)\nu \in (0, \nu_c=0.09653), then t=0.2258t=0.2258 s <τB< \tau_B, this is as though a particle would travel 1 meter horizontally with a speed equal 2g\sqrt{2g} m/sec to take the time (=1/2g=τ2)<τB=1/\sqrt{2g}=\tau_2) < \tau_B. The function t(νt(\nu) suffers a jump discontinuity at ν=νc\nu=\nu_c, we offer some resolution.

Cite

@article{arxiv.2010.15514,
  title  = {An interesting track for the Brachistochrone},
  author = {Zafar Ahmed and Amal Nathan Joseph},
  journal= {arXiv preprint arXiv:2010.15514},
  year   = {2020}
}

Comments

5 pages and 3 figures

R2 v1 2026-06-23T19:44:31.096Z