FRQ: Vertical Circular Motion — Inside the Loop

Assessments aligned to 2026 AP Physics 1 standards — 10 points

Question Type: Qualitative/Quantitative Translation (QQT)  |  MID-LEVEL  

▤ 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 centre 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, the object exerting it, 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 the comparison 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 riders need the same minimum speed. In a paragraph-length response, state which rider is correct and construct a coherent argument that connects your symbolic result from part (b) and the numerical value you calculated there to the physical reason the two riders do or do not differ. Your argument must reference both the force available to curve the path and the force required for circular motion.

❖ Answer Key & Scoring Guide

▸ earns credit   ⚠︎ common error, partial credit   ✗ common error, no credit

Part (a) — Model Answer

(i) At the top of the loop there are exactly two forces on the system, both directed vertically downward, which at that location is toward the centre of the loop: the gravitational force F_g = mg exerted by the Earth, and the normal force F_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 — it can never pull.

(ii) At the bottom of the loop the gravitational force F_g = mg exerted by the Earth points downward and the normal force F_N exerted by the track points upward, which at that location is toward the centre. Since the net force must point toward the centre, F_N is necessarily larger than mg. At both locations the net force points toward the centre of the loop; no separate “centripetal force” is exerted on the system.

Free-body diagrams of the rider–motorcycle system at the top and the bottom of the loop. The dashed amber arrow marks the positive radial direction (toward the centre) at each location.

Scoring (2 points):

▸ 1 point: Top-of-loop diagram showing both the gravitational force and the normal force directed downward, with no upward normal force.

▸ 1 point: Bottom-of-loop diagram showing the gravitational force downward and the normal force upward, with F_N stated or drawn larger than mg.

⚠︎ Common error (partial credit): drawing the normal force upward at the top of the loop — loses the first point but the second point is still available; a surface can push but cannot pull.

✗ Common error (no credit): adding a third arrow labelled “centripetal force” on either diagram earns 0 of 2 points for that diagram. The centripetal force is the name for the net of the real forces already drawn, not an additional force exerted by anything.

Part (b) — Model Answer

At the top of the loop, taking toward the centre (downward at that location) as positive, Newton’s second law along the radial axis gives:

F_N + mg = m v² / r

The system maintains contact with the track while F_N ≥ 0. The minimum speed corresponds to the limiting case F_N = 0, in which the gravitational force alone supplies the net centripetal force:

mg = m v_min² / r  ⇒  v_min = √(g r)

v_min = √(10 m/s² × 6.4 m) = √(64 m²/s²) = 8.0 m/s

The minimum speed at the top of the loop is 8.0 m/s. Below this speed the required centripetal force is smaller than mg, the track cannot pull inward to make up the difference, and the system leaves the track.

Scoring (3 points):

▸ 1 point: Begins the derivation from Newton’s second law rather than from a rearranged result — writes F_N + mg = m v² / r at the top, with both forces taken toward the centre.

▸ 1 point: Applies the contact condition F_N = 0 (or F_N ≥ 0) to obtain the symbolic result v_min = √(g r).

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

⚠︎ Common error (partial credit): setting F_N = mg at the top, transferring the flat-ground result, gives v = √(2 g r) ≈ 11.3 m/s — earns the first point only, and only if the second-law equation was written correctly before the wrong condition was imposed.

Part (c) — Model Answer

At the bottom of the loop, taking toward the centre (upward at that location) as positive, Newton’s second law along the radial axis gives:

F_N − mg = m v² / r  ⇒  F_N = m (g + v² / r)

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

The normal force is 14500 N, directed upward. The weight of the system is mg = 2000 N, so F_N / mg = 7.25. The track pushes on the system with 7.25 times the system’s weight, which is why a rider feels heaviest at the bottom of a loop.

Scoring (3 points):

▸ 1 point: Correct second-law equation at the bottom with F_N upward and mg downward and the net force toward the centre: F_N − mg = m v² / r.

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

▸ 1 point: Correct comparison F_N / mg = 7.25, or the equivalent statement “7.25 times the system’s weight”.

⚠︎ Common error (partial credit): writing F_N = m v² / r alone, omitting gravity, gives 12500 N — the equation point is lost, and the numerical point is earned only if the value follows consistently from the stated (incorrect) equation.

Part (d) — Model Answer

Rider B is correct: both riders need the same minimum speed of 8.0 m/s at the top of the loop. The symbolic result from part (b), v_min = √(g r), contains only g and r — the mass cancels before the numerical stage, which is why substituting 200 kg or 100 kg makes no difference to the 8.0 m/s value. The physical reason is that mass enters the contact condition twice, on both sides of the same equation. The force available to curve the path at the top is the gravitational force mg, which is proportional to mass. The net force required to hold the system on a circle of radius r at speed v is m v² / r, which is proportional to mass in exactly the same way. Halving the mass from Rider A to Rider B halves the gravitational force available, but it also halves the force required for the same circular path, so the two changes cancel and the limiting speed is unchanged. Rider A’s intuition holds the required force fixed while only the available force grows; both scale together. This is the same cancellation that makes free-fall acceleration independent of mass.

Scoring (2 points):

▸ 1 point: Selects Rider B and supports the claim with the symbolic result v_min = √(g r), noting explicitly that mass does not appear and that 8.0 m/s therefore applies to both riders.

▸ 1 point: Paragraph-length physical reasoning naming both scalings — the available gravitational force mg and the required net force m v² / r are each proportional to mass, so mass cancels in the contact condition.

⚠︎ Common error (partial credit): “Rider B is correct because the formula has no m in it” with no physical reasoning earns the first point only — claim and evidence are present, but the required reasoning that links the two mass-proportional forces is absent.

✗ Common error (no credit): selecting Rider A because a heavier system “needs more force” earns 0 of 2 points — the gravitational force supplying that requirement grows by exactly the same factor.