A.2 · IB Physics
Vertical circle: loop-the-loop
Release a cart from height h onto a frictionless track with a vertical loop of radius r. Watch the speed, normal force and centripetal acceleration change around the loop, and find the release height that keeps the cart on the track at the top.
Ready
- Height y:
- Speed v:
- Centripetal a = v²/r:
- Normal force N:
- KE:
- GPE:
Energy: ½mv² + mgy = mgh
On the loop: N = mv²/r + mg cos θ (θ measured from the bottom)
Top: N ≥ 0 needs v ≥ √(gr), so h ≥ 2.5r
g = 9.8 m s⁻², no friction
On the loop: N = mv²/r + mg cos θ (θ measured from the bottom)
Top: N ≥ 0 needs v ≥ √(gr), so h ≥ 2.5r
g = 9.8 m s⁻², no friction
Try this
- Set r = 2.0 m and lower h step by step. Find the smallest h that still completes the loop and compare it with 2.5r.
- Set r = 2.0 m and h = 4.0 m. Note the height where the cart leaves the track and check it against (r + 2h)/3.
- For any loop that is completed, compare the predicted N at the bottom and at the top: show that the difference is always 6mg, whatever h is.
Exam tip: at the top of the loop both the weight and the normal force point down towards the centre, so write mg + N = mv²/r and set N = 0 for the minimum speed; never draw “centripetal force” as an extra arrow on the free-body diagram.