➰FRQ: Inside the Loop

Assessments aligned to 2026 AP Physics 1 standards

📋Scenario

A stunt rider and motorcycle (combined mass 200 kg) travel around the inside of a smooth vertical circular loop of radius 6.4 m at a stunt show. The speed of the motorcycle is not constant around the loop. Take g = 10 m/s² and neglect air resistance and friction losses.

Motion occurs in a vertical plane. Use an inertial reference frame fixed to the ground. On the radial axis at any point of the loop, take the direction toward the center of the loop as positive.

📝Free response questions

(a) Draw (or describe completely) the free-body diagram of the rider–motorcycle system (i) at the top of the loop and (ii) at the bottom of the loop, naming each force and its direction.

(b) Starting from Newton’s second law applied at the top of the loop, derive a symbolic expression for the minimum speed v_min at the top required to maintain contact with the track. Then calculate v_min for this loop.

(c) At the bottom of the loop the motorcycle moves at 20 m/s. Calculate the magnitude of the normal force exerted by the track on the rider–motorcycle system at that instant, and compare it to the system’s weight (express as a ratio).

(d) Rider A (total mass 200 kg) and Rider B (total mass 100 kg) argue before the stunt. Rider A claims: “I am heavier, so I need a higher speed at the top to avoid falling.” Rider B claims both need the same minimum speed. Use your derivation from part (b) to decide who is correct, and explain physically why the mass does not (or does) matter.

🗝️Answer Key & Scoring Guide

Part (a) — Model Answer

(i) At the top: two forces, both directed vertically downward (toward the center): the gravitational force mg exerted by the Earth, and the normal force N exerted by the track. The track is above the system, and a normal force can only push perpendicular to and away from the surface.

(ii) At the bottom: the gravitational force mg downward and the normal force N upward (toward the center). At both points the net force points toward the center of the loop.

Scoring (2 points):

✔️1 point: Top-of-loop diagram with BOTH gravity and normal force directed downward (no upward normal force, no separate “centripetal force” arrow).

✔️1 point: Bottom-of-loop diagram with gravity downward and normal force upward, with N necessarily larger than mg implied or stated (net force toward center).

⚠️Common error: Drawing the normal force upward at the top of the loop — loses the first point; a surface cannot pull.

Part (b) — Model Answer

At the top, taking toward the center (downward) as positive, Newton’s second law gives:

N + mg = mv²/r

The system maintains contact while N ≥ 0. The minimum speed corresponds to N = 0, where gravity alone provides the centripetal force:

mg = m(v_min² / r) ⇒ v_min = √(gr)

v_min = √(10 m/s² × 6.4 m) = √64 = 8.0 m/s

The minimum speed at the top of the loop is 8.0 m/s.

Scoring (3 points):

✔️1 point: Correct Newton’s second law equation at the top with both N and mg toward the center.

✔️1 point: Applies the contact condition N = 0 (or N ≥ 0) to obtain v_min = √(gr).

✔️1 point: Correct numerical value 8.0 m/s with units.

⚠️Common error: Setting N = mg at the top (transferring the flat-ground result) leads to v = √(2gr) ≈ 11.3 m/s — earns the first point only if the second-law equation was written correctly before the wrong condition was imposed.

Part (c) — Model Answer

At the bottom, taking toward the center (upward) as positive:

N - mg = mv²/r ⇒ N = m(g + v²/r)

N = 200 kg × (10 m/s² + (20 m/s)² / 6.4 m) = 14500 N

The normal force is 14500 N. The weight is mg = 2000 N, so N/mg = 7.25 — the track pushes on the system with 7.25 times its weight, which is why riders feel heaviest at the bottom of the loop.

Scoring (3 points):

✔️1 point: Correct second-law equation at the bottom with N upward and mg downward, net force toward the center: N - mg = mv²/r.

✔️1 point: Correct numerical value N = 14500 N with units.

✔️1 point: Correct comparison N/mg = 7.25 (or “7.25 times the weight”).

⚠️Common error: N = mv²/r alone (forgetting gravity) gives 12500 N — earns the numerical point only if consistent with an explicitly stated (incorrect) force equation, per standard consistency scoring; the equation point is lost.

Part (d) — Model Answer

Rider B is correct. From part (b), v_min = √(gr): the mass does not appear. Physically, both the gravitational force (mg) and the required net centripetal force (mv²/r) are proportional to mass, so mass divides out of the contact condition. Doubling the mass doubles the gravitational force available to curve the path, but it also doubles the force required for the same circular motion — the two effects cancel exactly. This is the same reason free-fall acceleration is independent of mass.

Scoring (2 points):

✔️1 point: Correctly selects Rider B, citing v_min = √(gr) with no mass dependence.

✔️1 point: Physical explanation that both the available force (mg) and the required force (mv²/r) scale with m, so mass cancels.

⚠️0-credit note: “B is correct because the formula has no m” with no physical reasoning earns 1 point (claim + evidence) but not the second (reasoning) point.