Vertical loop R = 1 m: top and bottom speeds

A small ball moves on the smooth inner wall of a vertical circular track of radius R = 1 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) = 3.1623
   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) = 7.0711
   The height difference from bottom to top is 2r, and mechanical energy is conserved

Why it works

Just passing the highest point means the track exerts no force there, so mg = mv_top²/R and v_top = √(gR). Between the lowest and highest points the ball rises through a height 2R, so v_bottom² = v_top² + 4gR.

Common mistakes

• Using the radius instead of the diameter for the height climbed between the two points.
• Treating the speed at the lowest point as the same as the speed at the highest point.

Related

vertical-throw vertical-loop-r1-6-just-pass vertical-loop-r2-5-just-pass