Question

A ball released from a height h along an incline, slides along a circular track of radius R (at the end of the incline) without falling vertically downwards. Show that h_min = \(\frac52\) R.

Answer 1

To just loop-the-loop, the ball must have a speed v₂ = \(\sqrt{2 Rg}\) at the bottom of the circular track.

If h_min is the minimum height above the bottom of the circular track from which the ball must be released, by the principle of conservation of energy, we have, mgh_min

= \(\frac12\)\(mv^2_2\) = \(\frac12\)m(5Rg)

_{\(\therefore\)} h_min = \(\frac52\)R = 2.5 R

Note : 1f the ball rolls all along the track without slipping, its total energy at the top of the circular track should take into account the rotational kinetic energy of the ball.