Vertical loop R = 0.4 m: speed at the top

A small ball moves on the smooth inner wall of a vertical circular track of radius R = 0.4 m (g = 10 m/s²). It just manages to pass the highest point. Find its speed at the highest point and at the lowest point.

Step-by-step

1. mg = mv_top²/r → v_top = √(gr) = √(10×0.4) = 2
   Just passing the top: gravity alone provides all the centripetal force (rope/track force is 0)
2. ½mv_bottom² = ½mv_top² + mg·2r → v_bottom = √(5gr) = 4.4721
   The height difference from bottom to top is 2r, and mechanical energy is conserved

Why it works

At the highest point of a just-passing loop gravity alone supplies the centripetal force and the track pushes with nothing, giving v = √(gR); the lowest point is 2R below, so energy conservation raises the speed there.

Common mistakes

• Thinking the ball can pass the top with a speed near zero, when it needs at least √(gR).
• Using a height difference of R between the lowest and highest points instead of 2R.

Related

vertical-throw projectile-motion vertical-loop-r0-2-just-pass