Vertical loop R = 2.5 m: top and bottom speed

A small ball moves on the smooth inner wall of a vertical circular track of radius R = 2.5 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) = √(10×2.5) = 5
   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) = 11.1803
   The height difference from bottom to top is 2r, and mechanical energy is conserved

Why it works

At the just-passing point the track exerts no force at the top, so mg = mv_top²/R and v_top = √(gR). Energy conservation between the bottom and the top, a height 2R apart, gives v_bottom² = v_top² + 4gR.

Common mistakes

• Forgetting the square root and reporting v_top = 25 m/s instead of 5 m/s.
• Thinking the speed at the highest point is independent of the radius.

Related

vertical-throw vertical-loop-r1m-just-pass vertical-loop-r0-2-just-pass