Vertical loop R = 0.2 m: top and bottom speed

A small ball moves on the smooth inner wall of a vertical circular track of radius R = 0.2 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) = √(10×0.2) = 1.4142
   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) = 3.1623
   The height difference from bottom to top is 2r, and mechanical energy is conserved

Why it works

Just passing the top means the normal force from the track is zero, so the weight alone curves the path: mg = mv_top²/R, giving v_top = √(gR). Between the lowest and highest points the ball rises by one diameter, 2R, and energy conservation turns that height into extra speed.

Common mistakes

• Multiplying g by the diameter 2R instead of the radius when finding the top speed.
• Believing the ball keeps the same speed all round the loop.

Related

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