# Circular Motion Simulator - Centripetal Force, Banked Curve, Loop, Conical Pendulum

> Animated circular motion: velocity and centripetal vectors, cut the string, centrifugal force, banked curves, coaster loops, conical pendulum.

- URL: https://qilingtools.com/en/math/circular-motion
- Category: Physics Tools (https://qilingtools.com/en/physics)
- Tags: [Physics](https://qilingtools.com/en/tags/physics), [Education](https://qilingtools.com/en/tags/education)
- Last updated: 2026-10-26
- Price: Free, no sign-up
- Privacy: Runs locally in the browser; files are never uploaded

The circular-motion situations students most often get wrong, as animations. Uniform circle shows the velocity vector and the acceleration pointing at the centre; press “cut the string” and the ball flies off along the tangent in a straight line, not outward along the radius. Switch to the frame rotating with the string and the ball sits still, held between tension and the centrifugal force, while the straight escape path turns into a spiral — a good way to understand centrifugal and Coriolis forces. Banked curve gives the minimum, design and maximum safe speeds with friction, draws normal force, friction and the net force towards the centre, and plots the friction needed against speed. Vertical loop integrates a roller-coaster car along the track with RK4; choose a track, a string or a rigid rod, and when the normal force reaches zero the car leaves the track as a projectile. Conical pendulum gives the cone angle, period and tension, compared with a simple pendulum.

## How to use

1. Pick a scene: uniform circle, banked curve, vertical loop or conical pendulum
2. Set the radius, speed, bank angle, friction, release height or string length, or click a preset
3. Press play; cut the string, switch to the rotating frame or scrub the time slider
4. Read the period, centripetal acceleration, safe speeds, normal force (g-force) at each point and minimum height

## Features

- Animated velocity and centripetal acceleration vectors; tangential escape when the string is cut
- Inertial vs rotating frame: centrifugal balance and the spiral path after the cut
- Banked curve with friction: minimum, design and maximum speeds with a force diagram
- Friction-needed vs speed chart, with highway ramp, Daytona and velodrome presets
- Roller-coaster loop simulated with RK4: N/mg around the loop, detachment point and projectile path
- Conical pendulum angle, period and tension; set the angle from rpm

## FAQ

### When the string breaks, does the ball fly straight outward?

No. The moment the string breaks the centripetal force disappears and the ball keeps the velocity it had, which points along the tangent. So it moves in a straight line along the tangent (ignoring gravity and drag); its distance from the centre grows, but it doesn’t move radially. The feeling of being “thrown out” is the centrifugal force of the rotating frame — in the ground frame nothing pushes the ball outward, there’s just nothing pulling it in any more.

### Why must a roller coaster start at least 2.5R high?

At the top of the loop the required centripetal force mv²/R is supplied by gravity plus the track’s normal force, which can’t be negative, so v² ≥ gR. Energy conservation from rest at height h gives v² = 2g(h − 2R) at the top (height 2R), hence h ≥ 2.5R without friction. For R < h < 2.5R the normal force reaches zero somewhere in the upper half and the car leaves the track as a projectile; for h ≤ R it never rises above the centre and simply rolls back.

### At what speed does a banked curve need no friction?

When the horizontal component of the normal force alone supplies the centripetal force: tanθ = v²/(gR), so the design speed is v = √(gR tanθ). Slower than that the car tends to slide down and friction acts up the slope; faster, it tends to skid outward and friction acts down the slope. Maximum static friction μN widens this to a safe range — the minimum is zero on a gentle enough bank, and there is no upper limit once μ > cotθ. The car’s mass cancels in all these speeds.

### What sets the period of a conical pendulum?

Vertically T cosθ = mg and horizontally T sinθ = mω²L sinθ, so ω² = g/(L cosθ) and the period is 2π√(L cosθ/g). L cosθ is the height h of the pivot above the circle, so the period depends only on h, not on the mass. A wider cone means a smaller h and faster rotation; as the angle tends to zero the period approaches the small-angle period 2π√(L/g) of the same pendulum.

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