Vertical loop R = 0.9 m: top speed

A small ball moves on the smooth inner wall of a vertical circular track of radius R = 0.9 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) = √(10×0.9) = 3
   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) = 6.7082
   The height difference from bottom to top is 2r, and mechanical energy is conserved

Why it works

The just-passing condition means the track exerts no force at the top, so mg = mv²/R there; a larger radius needs a larger speed, since the required centripetal force grows with R.

Common mistakes

• Taking gR instead of √(gR) for the speed at the highest point.
• Thinking the highest-point speed does not depend on the radius.

Related

vertical-throw projectile-motion vertical-loop-r0-8-just-pass