Vertical Circular Motion: The Minimum Speed at the Top of a Loop

Ride the inside of a vertical loop and find out why there's a minimum speed at the top.

This interactive AP Physics 1 simulation shows the forces in vertical circular motion — how the normal force (N) and weight (mg) combine to supply the centripetal force that bends the path into a circle. At the top, the slowest safe pass is when the track force drops to zero and gravity alone turns the rider, giving the critical speed vₘᵢₙ = √(gr) = 8.0 m/s (for r = 6.4 m, g = 10 m/s²). Drag the speed slider to watch the free-body diagrams respond in real time, see how the normal force at the bottom grows to several times the rider's weight (N = m(g + v²/r)), and prove that the minimum speed is independent of mass. A visual, exam-ready way to master loop-the-loop problems.

Why is there a minimum speed at the top of a vertical loop? At the top, gravity already points toward the centre. The slowest safe pass is when the normal force is zero and gravity alone provides the centripetal force: mg = mv²/r, so vₘᵢₙ = √(gr). Any slower and the rider leaves the track.

What is the minimum speed at the top of a loop of radius 6.4 m? vₘᵢₙ = √(gr) = √(10 × 6.4) = 8.0 m/s.

Why is the normal force largest at the bottom of a loop? At the bottom the normal force must support the weight and supply the centripetal force: N = m(g + v²/r), making it much larger than mg — this is the rider's apparent weight.

Does the minimum loop speed depend on mass? No. Because vₘᵢₙ = √(gr) has no mass term, heavier and lighter riders need the same minimum speed — mass scales gravity and inertia equally, so it cancels.