Vertical loop R = 0.8 m: top and bottom speed

A small ball moves on the smooth inner wall of a vertical circular track of radius R = 0.8 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) = √(10×0.8) = 2.8284
   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) = 6.3246
   The height difference from bottom to top is 2r, and mechanical energy is conserved

Why it works

The just-passing condition means the normal force vanishes at the top, so the ball's weight alone keeps it on the circle: mg = mv_top²/R. Sliding down to the bottom from a height 2R adds 4gR to the square of the speed.

Common mistakes

• Squaring the radius instead of multiplying g by R, which gives 0.64 instead of 8 under the root.
• Forgetting that the speed at the bottom must be larger than at the top.

Related

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