Vertical loop R = 1.6 m: top and bottom speed

A small ball moves on the smooth inner wall of a vertical circular track of radius R = 1.6 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) = √(10×1.6) = 4
   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) = 8.9443
   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 is the centripetal force, so v_top = √(gR); from there the ball drops 2R to the lowest point and speeds up while its mechanical energy is conserved.

Common mistakes

• Swapping the two speeds, since the lowest point is always the faster one.
• Ignoring energy conservation and giving the ball the same speed at the top and at the bottom.

Related

vertical-throw projectile-motion vertical-loop-r0-9-just-pass